[EM] CLDMMPO
Mike, Here's why I think that the CLD part is not necessary when we limit MMPO to three slots: The most likely situation where the CL wins is the case in which there is a clone cycle of three candidates that generate a lot of opposition among themselves, more opposition than any of them generate against the CL. When we limit to two slots of approval (and two or fewer slots of disapproval) then there can be no clone cycle, assuming that clones are mostly approved together or disapproved together. So that basically takes care of the CL problem. AS for Kevin's bad example, I have suggested including the disapprovals as oppositions as well as symmetric completion at the bottom. Either of these by itself will solve the problem, but I think that the disapproval idea is easier to sell than explaining why we want symmetric completion at the bottom . but not at the top. 49 C 03 A 24 AB 24 B (A?) With the disapprovals included (along the diagonal) with the other pairwise oppositions we get Oppositions to A are [ 73, 24, 49] Oppositions to B are [ 27, 52, 49] Oppositions to C are [ 27, 48, 51], so C wins. But if the B supporters give as much support to A as the A supporters have given to B, then the 73 disapproval opposition reduces to 49 and A wins with room to spare (a one percent margin). It also solves the other Kevin bad example 49 A 01 A=C 01 B=C 49 B The disapproval opposition to C is 98, which makes C the MMPO loser when we include disapproval as an opposition, i.s. as the opposition of the approval cutoff ideal candidate/level of acceptance. What do you think? Forest From: MIKE OSSIPOFF To: Subject: [EM] CLDMMPO Forest-- You wrote: I wonder if it is possible for a CL to win three slot MMPO when the number of ballots on which X appears in the bottom slot is counted as an oppsitions to X. In other words, I wonder if the CL disqualification is redudant in that context. Also, how does the CLD rule affect the FBC in general? [endquote] I too have been concerned that FBC compliance could be affected by CLD, or the other disqualification and completion proposals that I've speculatively suggested. I suggest that when one method is completed by another, or when there are disqualifications, the , relation should be used instead of the // relation. So, when applying the 2nd method--the completion method, or the method used after the disqualifications--the entire initial set of candidates would be used in calculating the scores for the completion or post- disqualification method, even though that method is applied only to the post-disqualification candidates. Doesn't that do a lot to protect FBC compliance. I found that CLDMMPO wouldn't avoid Kevin's MMPO bad-example (I mentioned that in my other post today). But, as Ted suggested, maybe 3-slot methods can avoid many of the problems that can happen with unlimited-ranking methods. So that's another thing to investigate. Might 3-slot MMPO be easier to protect from Kevin's bad-example? Is there some easy way to achieve that? Mike Ossipoff Election-Methods mailing list - see http://electorama.com/em for list info
[EM] CLDMM{O
Mike, I wonder if it is possible for a CL to win three slot MMPO when the number of ballots on which X appears in the bottom slot is counted as an oppsitions to X. In other words, I wonder if the CL disqualification is redudant in that context. Also, how does the CLD rule affect the FBC in general? Forest Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Kristofer: MMPO objections
Kristopher, I agree that Plurality failure is bad in a public proposal and hard to defend in any case. In the case of MMPO the question is moot because Plurality failure is so easily fixed by either of the following natural tweaks: 1. Put 50 percent in each of the diagonal positions. (A candidate would beat a clone of itself half of the time.) 2. Put the respective truncation totals down the diagonal positions. (These totals are the pairwise oppositions of the Minimum Acceptable Candidate.) With this second fix, you can also create a list of oppositions against MAC, and if MAC's max opposition is smaller than any other candidate's max opposition, then various possible courses of action exist: (a) throw out these candidates and start over. (b) elect the approval winner (i.e. the one with min opposition from MAC, which is the same as the one with most opposition against MAC). (c) use the fall back lottery to elect the winner. Date: Wed, 04 Jan 2012 17:16:26 +0100 From: Kristofer Munsterhjelm To: MIKE OSSIPOFF Cc: election-meth...@electorama.com Subject: Re: [EM] Kristofer: MMPO objections On 01/04/2012 04:56 PM, Kristofer Munsterhjelm wrote: The Plurality criterion isn't just failed by methods that return an un-Plurality-method-like result. It is also failed by methods that return an un-Approval-like result. Recall that the Plurality method says if A is ranked first on more ballots than B is ranked at all, B shouldn't win. That should of course say the Plurality *criterion*. Sorry about that. Election-Methods mailing list - see http://electorama.com/em for list info
[EM] Three Slot Voting Equipment
That's very interesting, Mike. I didn't know that three slot voting equipment was already in place; I never knew how exactly they handled ballot iniatives. All the more reason to narrow down to the best three slot methods! Election-Methods mailing list - see http://electorama.com/em for list info
[EM] MAMD (Max Assent Min Dissent)
Date: Mon, 2 Jan 2012 19:44:48 +
From: MIKE OSSIPOFF
To:
Subject: [EM] Forest: MAMD
Forest--
MMPO has several big advantages:
1. The unmatched brevity of its definition
2. Its full-rankings flexibility, which allows the full sincere-
expressivity benefit of AERLO
3. Its ability to elect unfavorite middle CWS better than do
most or all of the conditional methods
(Of course, though, in fairness to the conditional methods, the
election of an unfavorite CW would be seen by most people as a
bad-example)
A modification of MMPO that would avoid Kevin's MMPO bad-
example, without gaining some other
un-plurality-like outcome that would bother people, without
losing FBC or ABE-success, and without
significant added complication or wording--that would be a good
thing to find.
Tell me if this is MaxAssentMinDissent (MAMD):
MMPO, except that each ballot that doesn't rank x is counted as
voting x over himself, which is treated
as any other pairwise opposition in the MMPO count
[end of definition of MAMD?]
Pretty close. The other difference from regular MMPO is due to symmetric
completion of the truncations. This counts half of the ballots in which x and y
are both truncated as opposition of x against y and as opposition of y against
x.
Because MAMD has just been posted, I haven't yet had the
opportunity to evaluate it. Tell anything
you find out about MAMD and the matters discussed above.
In your post, you mentioned symmetric completion, at bottom, and
everywhere but top. That isn't used in
MAMD, is it?
Actually MAMD does make use of one of these symmetric completion options. Which
one you use makes no difference in examples with only three candidates. If you
like MMPO better than MinMax(margins), then you should use symmetric completion
only at bottom. Otherwise, you should use symmetric completion for all levels
except top. A compromise version might use symmetric completion only below the
approval cutoff, which would reduce to the first option in the case of implicit
approval.
MMPO with symmetric completion at bottom, while avoiding Kevin's
bad-example, also sometimes loses
MMPO's ABE-success:
60: AB
55: B
100: C
Here is the pairwise opposition matrix for MAMD:
[[155, 110, 87.5],
[105, 100, 115],
[127.5, 100, 115]] .
The max dissent against B is from the 110 A supporters. This is the minimum of
the max dissents, since A has is disapproved by 155, and B has a complaint of
115 against C , not to mention the 115 disapproval against C.
So yes, B wins.
B wins.
Maybe MAMD is what can make MMPO one of the best, criticism-
invulnerable, FBC/ABE methods.
I hope so.
INow here's what I meant about the MinMax(margins) clone problem:
17 ABCD
17 BCAD
17 CABD
16 DABC
16 DBCA
16 DCAB
The clone cycle {A, B, C} beats D by a margin of three.
But the pairwise defeats within the clone cycle are all by margins of 33, so the
MinMax(margins) winner is the Condorcet Loser D.
It's not nice to have a CL be the method winner is the failure of the Clone
Winner property: If the clone cycle were replaced by a single candidate, that
candidate would be the method winner.
That's why there should be no more than two slots on either side of the approval
cutoff. True clones would then be restricted to two adjacent slots per ballot.
In that case a clone cycle becomes impossible. The problem disappears.
If the method holds up to critical scrutiny, then the big problem is in
packaging for public consumption.
The method must be stated as simply, positively, and clearly as possible with
reasonable heuristic motivations for its peculiar characteristics, such as
including the disapprovals among the dissents, and the symmetric completion
feature at the bottom ... something beyond saying that these are kludges to make
the method work in Kevin's bad examples, etc.
Forest
Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] MAMD (Max Assent Min Dissent)
Mike wrote .. MMPO with symmetric completion at bottom, while avoiding Kevin's bad-example, also sometimes loses MMPO's ABE-success: 60: AB 55: B 100: C Forest replied Here is the pairwise opposition matrix for MAMD: [[155, 110, 87.5], [105, 100, 115], [127.5, 100, 115]] . The max dissent against B is from the 110 A supporters. This is the minimum of the max dissents, since A has is disapproved by 155, and B has a complaint of 115 against C , not to mention the 115 disapproval against C. So yes, B wins. Of course, if the A faction knows that the B faction sincere order is BAC, the 60 AB voters can split up to give 15 A 45 AB Then if the B voters stubbornly bullet, C will win. But if as many (45) of them vote sincerely as the A voters, then A will win. I think this is a pretty good resolution of the defection problem. Election-Methods mailing list - see http://electorama.com/em for list info
[EM] MMPO objections
The ordinary MMPO pairwise opposition matrix has blanks down the main diagonal. If you put the respective disapprovals in those positions, then the Plurality problem goes away. Filling in the diagonal elements with disapprovals is tantamount to incorporating a virtual Minimum Acceptable Candidate (MAC) as the approval cutoff. The disapproval is simply the opposition by MAC. The simplest way to incorporate this feature is to use implicit approval. Then the diagonal element in position (i, i) is simply the number of ballots on which candidate i is unranked (or unrated). With this convention, when all ballots rate candidates only at the extremes, the method elects the approval winner. In Kevin's example 49 A 01 A=C 01 B=C 49 B , the MAC opposition to C is 98, which is much larger than any other opposition, so the approval loser C is also the MMPO loser for this version of MMPO. Now consider these facts: 1. When there are complete rankings ordinary MMPO elects the same candidate as MinMax(margins). 2. In case of incomplete rankings MinMax(margins) elects the same candidate whether or not the ballots are treated with symmetric completion. 3. When incomplete ballots are treated with symmetric completion, regular MMPO elects the same candidate as MinMax(margins). 4. Under MMPO with symmetric completion, exempting the equal top position from symmetric completion trades Condorcet Criterion compliance for FBC compliance. 5. This partial symmetric completion version of MMPO resolves Kevin's approval bad example (ABE). 6. Introducing the virtual candidate MAC to this version of MMPO does not change this satisfactory resolution. 7. If we limit the method to three slots (or four slots with MAC between the two middle slots), then the clone winner failure that MMPO shares with MinMax(margins) goes away. In sum, I propose this version of MMPO. The problem is how to package it for public approval. I suggest calling it MaxAssentMinDissent. People are familiar with the concept of decisions rendered by deliberative bodies like the supreme court being accompanied by a count of concurring votes, dissenting votes, and abstentions. If we lump the concurring and abstentions together into the category of assent, then our method maximizes the minimum assent and minimizes the maximun dissent from all of the pairwise decisions relative to the method winner. If candidate X is elected there will be dissentions relative to each of the other candidates including MAC. Suppose that the largest of these dissents is 43 percent, i.e. 43 percent of the ballots show a preference of some candidate Y over X. The largest dissent against Y will be larger than this 43 percent dissent against X, so Y has no better claim than X to be elected. Likewise the minimum assent for X would be 100-43=67, and this is greater than the minimum assent for Y. Etc. Thoughts? Forest In sum MMPO with MAC and Bottom Symmetric Completion is the ri Election-Methods mailing list - see http://electorama.com/em for list info
[EM] ACF grade voting
Suppose the ballot limits grade options to A, C, and F, but a sizeable faction would like to award a grade of B to a particular candidate. If half of them voted a grade of A and the other half a grde of C, the resulting grade points would be the same. So in elections with large electorates there is no need to have grade ballots with all five grade options. Those who want to award a B grade can flip a coin to decide between A and C. Those who would like to award a grade of D can decide between C and F with a coin toss. The grade averages will come out the same as if the higher resolution grade ballots were used. If two or more candidates are statistically tied, the tied candidate with the greatest number of A's and C's should be elected. Election-Methods mailing list - see http://electorama.com/em for list info
[EM] Proportional Range Voting via Honest Approval PAV
Now that we have a good definition of honest approval strategy, we can automatically adapt methods (like PAV) that are based on approval style ballots to cardinal ratings style ballots. Definition: Honest Approval Strategy: Approve your k top ranked candidates, where k is the sum of your cardinal ratings (on a scale of zero to one) rounded to the nearest whole number. So given a set of Range/Score/Grade ballots, first convert them to cardinal ratings on a scale of zero to one. Then use honest approval strategy to convert these ratings ballots to approval style ballots. Finally apply PAV (or whatever approval method you like) to determine who gets elected. Election-Methods mailing list - see http://electorama.com/em for list info
[EM] SODA, negotiation, and weak CWs (Jameson Quinn)
Jameson, could you please submit this again in a plain text format that doesn't put in extra form feeds? Election-Methods mailing list - see http://electorama.com/em for list info
[EM] Approval Strategy
Mike wrote As for myself, in Score-Voting, I'd probably use non-extreme points assignments only in two instances: 1. The excellent diplomatic ABE solution that you suggested for Score-Voting Forest replied Excellent except that satisfaction of the FBC is in doubt. I assumed you meant my Smith//Range proposal (for which FBC is in doubt). But you probably meant my more recent MMMPO aka LRV which does satisfy the FBC. In the first method zero info strategy does take care of the ABE defection problem, but potential failure of the FBC is bad for a public proposal. Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Fwd: SODA, negotiation, and weak CWs
Jameson asked for thoughts. My first thought is that this kind of analysis is exactly what we need. My second thought is that so far SODA has held up well under all the probes for weakness that anybody has come up with. SODA seems to be a very robust method. My third thought is that I have never seen this kind of soul searching or probing directed at IRV by IRV enthusiasts. Date: Sun, 25 Dec 2011 11:28:24 -0600 From: Jameson Quinn To: EM Subject: [EM] Fwd: SODA, negotiation, and weak CWs Message-ID: Content-Type: text/plain; charset=iso-8859-1 I'm resubmitting this in a text-friendly format, at Forest's request. I'll also take the opportunity to add one paragraph about how rated methods can fail to find the highest-utility candidates in scenarios like this. Added text is marked ADDED. -- Forwarded message -- From: Jameson Quinn Date: 2011/12/25 Subject: SODA, negotiation, and weak CWs To: EM In order to have optimum Bayesian Regret, a voting system should be able to not elect a Weak Condorcet Winner (WCW), that is, a CW whose utility is lower than the other candidates. Consider the following payout matrices:Group Size Candidate Utilities Scenario 1 (zero sum) A B C a 4 4 1 0 b 2 0 3 2 c 3 0 2 4 Total utility 16 16 16 Scenario 2 (pos. sum) A B C a 4 3 1 0 b 2 0 3 1.5 c 3 0 2 3 Total utility 12 16 12 Scenario 3 (neg. sum) A B C a 4 4 0.5 0 b 2 0 3 2 c 3 0 1 4 Total utility 16 11 16 All three scenarios consist of 3 groups of voters: groups a, b, and c, with 4, 2, and 3 voters respectively, for a total of 9 voters. All scenarioshave 3 candidates: A, B, and C, who favor their respective groups. And in all three scenarios, candidate B is the CW, because the preference matrix is always 4: AB 2: BC 3: CB But in scenario 1, the utilities of the three candidates are balanced; in scenario 2, B has the highest utility; and in scenario 3, A and C have the highest utilities. Obviously, any purely preferential system will tend to give the same result in all three scenarios. This might not be 100% true if strategy propensitydepended on the utility payoff of a strategy; but the strategicpossibilities would have to be just right for a method to get it right for this reason. It's easy to see how Range could get it right in scenarios 2 and 3. With just a bit of strategy, it's also easy to see how it could successfullyfind the CW in scenario 1. You can also construct plausible stories of how Approval or MJ could get it right in all 3 scenarios, although it probably involves adding some random noise to voting patterns rather than assuming pure honest votes. ADDED: Of course, Range, Approval, and MJ can all get these scenarios wrong too. Because the scenarios present a classic chicken dilemma between B and C, these rated systems could all end up electing A, regardless of utility. But what about SODA? As a primarily preferential system, it seems that it should give the same result in all three scenarios. If candidates all rationally pursue the interests of their primary constituency, then A will approve B to prevent B from having to approve C, leaving a win for B. But if candidate A decides to make an ultimatum, things could go differently. A says to B: Make some promise that transfers 0.5 point of utility to each member of group a, or I will not approve you. Assume that B can make a promise to transfer utility from one group to another at 80% efficiency; and that such promises are not strictly enforceable. Thus, if A gets too greedy, B can simply promise the moon and not keep the promise;but if A asks for something reasonable, B will see honesty as worth it. B could promise to transfer 0.5 point of utility from groups b and c to group a. Since utility transfers are assumed to be only 80% efficient, that transfer of 2.5 utility points would result in a net loss of 0.5. So the payoffs would be: Group Size Candidate Utilities Scenario 1a(zero sum) A B C a 4 4 1.5 0 b 2 0 2.5 2 c 3 0 1.5 4 Total utility 16 15.5 16 Group Size Candidate Utilities Scenario 1b(zero sum) A B C a 4 4 1.5 0 b 2 0 3 2 c 3 0 1.1 4 Total utility 16 15.3 16 Scenario 2a(pos. sum) A B C a 4 3 1.5 0 b 2 0 2.5 1.5 c 3 0 1.5 3 Total utility 12 15.5 12 Scenario 3a(neg. sum) A B C a 4 4 1 0 b 2 0 2.5 2 c 3 0 0.5 4 Total utility 16 10.5 16 Scenario 3b(neg. sum) A B C a 4 4 1 0 b 2 0 3 2 c 3 0 0.1 4 Total utility 16 10.3 16 Note that in scenarios 1a and 2a, this utility transfer has left B giving the same utility to groups a and c, while in scenario 3a, B has switchedfrom favoring group c over group a, to favoring group a over group c. Also, note that in scenario 2a, group b still gets a full point of advantage with candidate B versus what they would get with candidate C, whereas in the other two Xa scenarios, group
[EM] Proportional Range Voting via Honest Approval PAV
While writing the below it occurred to me that we could construct another Proportional Representation method based on ordinal ballots (ranked preferences) by the following technique: (1) Convert the ordinal rankings into cardinal ratings via the monotonic, clone free techinque that I outlined under the title Borda Done Right a few months back. (2) Convert these ratings into approval style ballots via the honest approval strategy. (3) Apply PAV to the resulting ballot set. Here's a brief review of the ballot conversion at the heart of Borda Done Right: For each candidate C on the ballot in question let p(C) be the probability that a candidate ranked equal to or behind C would be elected in a random favorite election. Then treating these probabilities as scores, normalize them to a cardinal ratings scale of zero to one. [Any other monotone, clone free lottery could be used in place of the random favorite lottery in the definition of p(C).] I would be interested to see some simulations comparing this ordinal--cardinal--approval---PAV method of PR with other PR methods based on rankings. Since all of the steps are monotone and clone proof, the composite method should share these properties to the extent that PAV does. My Best to You All on this holiday weekend! Forest - Date: Sun, 25 Dec 2011 16:40:50 + (GMT) From: fsimm...@pcc.edu To: election-methods@lists.electorama.com Subject: [EM] Proportional Range Voting via Honest Approval PAV Now that we have a good definition of honest approval strategy, we can automatically adapt methods (like PAV) that are based on approval style ballots to cardinal ratings style ballots. Definition: Honest Approval Strategy: Approve your k top ranked candidates, where k is the sum of your cardinal ratings (on a scale of zero to one) rounded to the nearest whole number. So given a set of Range/Score/Grade ballots, first convert them to cardinal ratings on a scale of zero to one. Then use honest approval strategy to convert these ratings ballots to approval style ballots. Finally apply PAV (or whatever approval method you like) to determine who gets elected. Election-Methods mailing list - see http://electorama.com/em for list info
[EM] Approval Strategy
From: MIKE OSSIPOFF To: Subject: [EM] Approval strategy Message-ID: Content-Type: text/plain; charset=iso-8859-1 Forest-- You wrote: Also, going back to what you metioned before about the value of showing support for losers that you like better than the winner (given they have the chance of the proverbial snowflake), I think that this is perhaps the main rationale for extending approval all of the way down to the candidate most likely to win (and include that candidate only if the runner up is below it). [That's another way to state strategy A.] [endquote] Ok, sure, I understand the justification. To show support for candidates better than the frontrunner you're voting for, and who are unlikely to be a serious rival to hir, one could approve all of the inbetween candidates better than the frontrunner that one approves. It comes down to a question of whether you want to show that support, or whether you want to do _strictly_ instrumental voting, where you assume that even the non- frontrunner inbetween candidates have a finite chance of being a rival to a frontrunner. So I don't disagree with strategy A--It's merely a question of whether one wants to deal with the inbetween candidates expressively or strictly instrumentally. I think that the honest Approval strategy is a fascinating finding--Effective score-voting in Approval, without randomization. That means that score voting is especially easily available in Approval. So, having proposed Score-Voting, one could then point out that it can be easily achieved with the more modestly- demanding Approval balloting. As for myself, in Score-Voting, I'd probably use non-extreme points assignments only in two instances: 1. The excellent diplomatic ABE solution that you suggested for Score-Voting Excellent except that satisfaction of the FBC is in doubt. 2. When a candidate is about as bad as a candidate can be and still be barely acceptable, ...and so I feel that it's questionable whether s/he deserves a full approval. In an Approval election, I'd randomize in those instancesunless honest Approval would work there too. Honest Approval stratgegy is probably only for someone who wants to do score voting for all the candidates. Or maybe not. Maybe some intended maximum and minimum ratings wouldn't interfere with voting an honest Approval ballot. Then, honest Approval voting would avoid for me, too, the task of randomizing. The main difficulty with honest approval voting is that when there are three candidates generally one will be rated 100%, one zero, and the other somewhere in between, so the expected number of approvals (as the sum of the ratings) will be somewhere between one and two. On this basis you know that you should approve at least one candidate (which is already obvious) but nor more than two (which is already obvious as well). So in the very practical case of three candidates the method still requires something like randomization or gut feeling to decide whether or not to approve the middle candidate. By the way, of course your Score-Voting ABE solution only works if it's known how many votes A and B, combined, are going to get. This is true, so the method requires knowledge when the defection strategy is likely. But the more likely the knowledge is sufficient to make defection strategy possible, the more likely there is enough knowledge for the other faction to make it risky. But it is better to have a method such that zero info strategy by the naive betrayed faction will automatically defend against defection on the part of the cunning faction.. In a first election by Approval or Score, that might not be known. Then, maybe an A voter might have to just hold hir nose and fully approve B, to keep C from wining. Later, when the numbers are more predictable, the A voters could use your Score/Approval ABE solution. But that shows a big advantage of MMT, GMAT, MGMAT, MTAOC and MMABucklin Mike Ossipoff Election-Methods mailing list - see http://electorama.com/em for list info
[EM] Better Than Expectation Approval Voting
Mike's exposition of basic Approval and Range strategy as variations on the theme of Better Than Expectation strategy was very interesting and valuable, including the recommendation of introducintg Approval after score or grade voting, which are much more familiar to most people. That was probably the most important part of his message, but I want to make a few more remarks about Approval strategy. 1. When there are candidates between the two front runners and you are not sure where to draw the approval line, put it adjacent to the candidate with the greatest likelihood of winning. In other words put your approval cutoff adjacent to the candidate most likely to win on the side of the candidate next most likely to win. This is what Rob LeGrand calls strategy A. 2. Suppose that order is easier than ratings for you. Joe Weinstein's idea is to approve candidate X if and only if it is more likely that the winner will be someone that you rank behind candidate X than someone that you rank ahead of candidate X. Note that when there are two obvious frontrunners Joe's strategy reduces to Rob's strategy A. 3. Suppose that on principle someone would never use approval strategy on a score/grade/range ballot, but is forced to use an approval ballot anyway. How could they vote as close as possible to their scruples? For example suppose that you would give candidate X a score of 37 percent on a high resolution score ballot, but are forced to vote approval style. In this case you can have a random number generator pick a number between zero and 100. If the random number is less than 37, then approve the candidate, otherwise do not. If all like minded voters used this same strategy, 37 percent of them would approve candidate X, and the result would be the same as if all of them had voted 37 on a scale from zero to one hundred. Now for the interesting part: if you use this strategy on your approval ballot, the expected number of candidates that you would approve is simply the sum of the probabilities of your approving the individual candiates, i.e. the total score of all the candidates on your score ballot divided by the maximum possible score (100 in the example). Suppose that there are n candidates, and that the expected number that you will approve is k. Then instead of going through the random number rigamarole, just approve your top k candidates. Forest Election-Methods mailing list - see http://electorama.com/em for list info
[EM] Better Than Expectation Approval Voting (2nd try readable format)
Mike's exposition of basic Approval and Range strategy as variations on the theme of Better Than Expectation strategy was very interesting and valuable, including the recommendation of introducintg Approval after score or grade voting, which are much more familiar to most people. That was probably the most important part of his message, but I want to make a few more remarks about Approval strategy. 1. When there are candidates between the two front runners and you are not sure where to draw the approval line, put it adjacent to the candidate with the greatest likelihood of winning. In other words put your approval cutoff adjacent to the candidate most likely to win on the side of the candidate next most likely to win. This is what Rob LeGrand calls strategy A. 2. Suppose that order is easier than ratings for you. Joe Weinstein's idea is to approve candidate X if and only if it is more likely that the winner will be someone that you rank behind candidate X than someone that you rank ahead of candidate X. Note that when there are two obvious frontrunners Joe's strategy reduces to Rob's strategy A. 3. Suppose that on principle someone would never use approval strategy on a score/grade/range ballot, but is forced to use an approval ballot anyway. How could they vote as close as possible to their scruples? For example suppose that you would give candidate X a score of 37 percent on a high resolution score ballot, but are forced to vote approval style. In this case you can have a random number generator pick a number between zero and 100. If the random number is less than 37, then approve the candidate, otherwise do not. If all like minded voters used this same strategy, 37 percent of them would approve candidate X, and the result would be the same as if all of them had voted 37 on a scale from zero to one hundred. Now for the interesting part: if you use this strategy on your approval ballot, the expected number of candidates that you would approve is simply the sum of the probabilities of your approving the individual candiates, i.e. the total score of all the candidates on your score ballot divided by the maximum possible score (100 in the example). Suppose that there are n candidates, and that the expected number that you will approve is k. Then instead of going through the random number rigamarole, just approve your top k candidates. Forest Election-Methods mailing list - see http://electorama.com/em for list info
[EM] FairVote in _Science_ magazine (MIKE OSSIPOFF)
Mike, Right ON! But I tripped up for a second on an unintentional typo concerning Richie's second claim... 2. The article said that the best strategy in Approval is to rank the candidates sincerely. Replace Approval with IRV in the above statement: Forest From: MIKE OSSIPOFF To: Subject: [EM] FairVote in _Science_ magazine Message-ID: Content-Type: text/plain; charset=iso-8859-1 Looking at some back-pages of electology discussion, I was reminded of Richie's article in _Science_ magazine, published some time ago. First, it's astonishing that someone like Richie was able to publish in _Science_. But equally astonishing was that he could make the statements that he made there, and they were published without being checked for accuracy. The postings pointed out two really silly statements made in the article: 1. The article said that, according to (unnamed?) experts, voters in Approval elections will tend to approve only one candidate. That statement was answered in the electology posting. I'd answered it for Richie decades ago. Regarding the very many people who now think that they need to vote (in Plurality) for the Democrat, and who say that that's necessary as a pragmatic vote, to avoid wasting their vote, and who say that it's necessary to hold your nose and vote for Democrat, though you don't really like her--Richie thinks that those people are suddenly going to start voting only for their favorite? :-)? No, those hold-their-nose lesser-of-2-evils Democrat voters, if we switched to Approval, would continue voting for the Democrat in Approval. The difference is that, with Approval, they can also vote for everyone whom they like better than the Democrat. Of course, if it turns out, based on the Approval election vote- totals, or from (newly) honest and relevant polling, that those voters' favorite can beat the Republican, then of course, at that time, they might very well stop voting for the Democrat, and might start voting only for one or more candidates whom they like better than the Democrats. One thing that Richie doesn't understand is that, if a voter, in Approval, votes only for hir favorite, that's because s/he feels that hir favorite has a win, or that s/he doesn't consider anyone else to be acceptable. That's not a disadvantage of Approval. That's good strategy. Maybe Approval vote totals will soon show that progressive, better-than-Democrat candidates have a win and that their supporters needn't vote for a Democrat in Approval. That could result in well-informed, good- strategy plumping, bullet-voting. But, more likely, people will vote, in Approval, for a set of progressives, who are similar, and similarly-good candidates. ...unless there's only one that they consider acceptable, or unless their favorite appears to have a clear win over all the others. Approval strategy, when the election has completely unacceptable candidates who could win, is to vote for all of the acceptable candidates and for none of the unacceptable candidates. But regarding the person who now holds their nose and votes for a Democrat whom s/he doesn't like, though s/he likes others more--That person will, in Approval, vote for that same Democrat, and for everyone whom s/he likes more.? ...until Approval's vote totals, or genuinely worthwhile polling, show that there's no need to vote for the unliked Democrat. 2. The article said that the best strategy in Approval is to rank the candidates sincerely. ...and that was published in _Science_ magazine :-) It's common knowledge that strategy incentive is present in all nonprobabilistic voting systems. Richie's statement is hardly surprising, coming, as it does, from Richie. But it's indeed surprising that no one at _Science_ questioned the accuracy of that statement before publishing it. But, then, that could be said of statement #1, above, too. Just as with the other statement, I and others had explained the incorrectness of that statement to Richie decades ago. As is common knowledge among everyone who discusses voting systems (except for Richie, evidently), your needed compromise can be eliminated because s/he didn't have your vote yet, when s/he needed it, because your vote was on your favorite instead. Voting for your favorite instead of insincerely voting your compromise in 1st place, has given the election to someone who is worse than your compromise. How to avoid that? Rank your compromise in 1st place, burying your favorite. It's been reported that, when IRV is used in in national elections, many voters say that they vote for a lesser-of-2-evils compromise in 1st place, burying their favorite, so as not to waste [their] vote.? Maybe not coincidentally, the use of IRV there coincides with two- party dominance. It has been pointed out that, if a
Re: [EM] Least Expected Umbrage, a new lottery method
Jobst, Yes, your Condorcet Lottery was the first of this kind, as I pointed out on the EM list when the Rivest paper first came to our attention. Suppose that we replace each entry in the margins matrix with its sign (-1, 0, or 1 depending on whether it is negative zero or positive). we could call this matrix the Copeland matrix. Then the Condorcet Lottery is determined from the Copeland Matrix in the same way that the Rivest Shen Lottery is determined from the margins matrix. Since these matrices are anti-symmetric and the game is zero sum, the row and column players have identical optimum (mixed) strategies, i.e. the games are symmetrical. In the case of the Least Expected Umbrage lottery it turns out that the row and column player lotteries are not usually the same. Take the very ballot set used as an example in the Rivest Paper: (40) A B C D (30) B C A D (20) C A B D (10) C B A D The respective rows of the pairwise matrix are given by A: 0 60 40 100 B: 40 0 70 100 C: 60 30 0 100 D: 0 0 0 0 while the respective rows of the margins matrix are A 0 20 -20 100 B -20 040 100 C 20 -40 0100 D -100 -100 -100 0 The common strategy of both row and column players for this game is the lottery A/4+B/2+C/4, meaning that the respective probabilities for A, B, C and D are 25%, 50%, 25%, and zero. For LEU we go back to the basic pairwise matrix and put 50 in the lower right hand corner to represent half of the number of ballots on which C was ranked bottom with itself: 0 60 40 100 40 0 70 100 60 30 0 100 0 0 0 50 The bottom row of this matrix is dominated by each of the other rows, so it can be removed from the row player's strategy, and the (opposite of) the last column is dominated by the (opposite of) each of the other columns, so it can be removed from the column player's strategy. We are left with the matrix 0 60 40 40 0 70 60 30 0 For the zero sum game represented by this matrix the optimal strategy of the row player turns out to be (33*A+24*B+34*C)/91, and the optimal strategy for the column player is (33*A+34*B+24*C)/91 . The row player's expected payoff is 3000/91. If the row player adopts the Rivest strategy of (A+2*B+C)/4 instead of the optimal row player strategy (33*A+24*B+34*C)/91, and the column player sticks to the optimal strategy (33*A+34*B+24*C)/91, then the expected payoff for the row player will be less. [Exercise: calculate this expectd payoff; I'm out of time.] My Best, Forest From: Jobst Heitzig Date: Wednesday, December 21, 2011 4:25 am Subject: Re: [EM] Least Expected Umbrage, a new lottery method To: fsimm...@pcc.edu I just realized that this is quite similar to Condorcet Lottery (http://lists.electorama.com/htdig.cgi/election-methods- electorama.com/2005-January/014449.html) where I even mention in the end that the payoff matrix could be chosen to reflect defeat strengths rather than just defeats (i.e., having entry 1 where i beats j). So maybe we should compare Rivest-Shen and LEU to Condorcet Lottery? Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] FairVote in _Science_ magazine (MIKE OSSIPOFF)
So basically Richie was stubbornly repeating his lies, but we cannot fault Science for propagating them. - Original Message - From: Jameson Quinn I believe the only time FairVote has been published in Science was as a response to an editorial by Steven Brams. Brams also got a chance to respond, and while he didn't refute every one of their distortions (and couldn't refute the empirical predictions which Burlington later provedfalse), I believe that you can't really fault Science for providing a forum for both sides of the debate. (On both sides, it was editorial content, not subject to peer review.) Jameson 2011/12/21 Mike, Right ON! But I tripped up for a second on an unintentional typo concerning Richie's second claim... 2. The article said that the best strategy in Approval is to rank the candidates sincerely. Replace Approval with IRV in the above statement: Forest From: MIKE OSSIPOFF To: Subject: [EM] FairVote in _Science_ magazine Message-ID: Content-Type: text/plain; charset=iso-8859-1 Looking at some back-pages of electology discussion, I was reminded of Richie's article in _Science_ magazine, published some time ago. First, it's astonishing that someone like Richie was able to publish in _Science_. But equally astonishing was that he could make the statements that he made there, and they were published without being checked for accuracy. The postings pointed out two really silly statements made in the article: 1. The article said that, according to (unnamed?) experts, voters in Approval elections will tend to approve only one candidate. That statement was answered in the electology posting. I'd answered it for Richie decades ago. Regarding the very many people who now think that they need to vote (in Plurality) for the Democrat, and who say that that's necessary as a pragmatic vote, to avoid wasting their vote, and who say that it's necessary to hold your nose and vote for Democrat, though you don't really like her--Richie thinks that those people are suddenly going to start voting only for their favorite? :-)? No, those hold-their-nose lesser-of-2-evils Democrat voters, if we switched to Approval, would continue voting for the Democrat in Approval. The difference is that, with Approval, they can also vote for everyone whom they like better than the Democrat. Of course, if it turns out, based on the Approval election vote- totals, or from (newly) honest and relevant polling, that those voters' favorite can beat the Republican, then of course, at that time, they might very well stop voting for the Democrat, and might start voting only for one or more candidates whom they like better than the Democrats. One thing that Richie doesn't understand is that, if a voter, in Approval, votes only for hir favorite, that's because s/he feels that hir favorite has a win, or that s/he doesn't consider anyone else to be acceptable. That's not a disadvantage of Approval. That's good strategy. Maybe Approval vote totals will soon show that progressive, better-than-Democrat candidates have a win and that their supporters needn't vote for a Democrat in Approval. That could result in well-informed, good- strategy plumping, bullet-voting. But, more likely, people will vote, in Approval, for a set of progressives, who are similar, and similarly-good candidates. ...unless there's only one that they consider acceptable, or unless their favorite appears to have a clear win over all the others. Approval strategy, when the election has completely unacceptable candidates who could win, is to vote for all of the acceptable candidates and for none of the unacceptable candidates. But regarding the person who now holds their nose and votes for a Democrat whom s/he doesn't like, though s/he likes others more--That person will, in Approval, vote for that same Democrat, and for everyone whom s/he likes more.? ...until Approval's vote totals, or genuinely worthwhile polling, show that there's no need to vote for the unliked Democrat. 2. The article said that the best strategy in Approval is to rank the candidates sincerely. ...and that was published in _Science_ magazine :-) It's common knowledge that strategy incentive is present in all nonprobabilistic voting systems. Richie's statement is hardly surprising, coming, as it does, from Richie. But it's indeed surprising that no one at _Science_ questioned the accuracy of that statement before publishing it. But, then, that could be said of statement #1, above, too. Just as with the other statement, I and others had explained the incorrectness of that statement to Richie decades ago.
[EM] Least Expected Umbrage, a new lottery method
Let M be the matrix whose row i column j element M(i,j) is the number of ballots on which i is ranked strictly above j plus half the number of ballots on which neither i nor j is ranked. In particular, for each k the diagonal element M(k , k) is half the number of ballots on which candidate k is unranked. Now think of M as the payoff matrix for the row player in a zero sum game. Elect the candidate that would be chosen by the optimal strategy of the row player. [End of Method Definition] Remarks: If there is a saddle point (i, j) such that the element M(i,j) in that position is the lowest in its row and the highest in its column, then the game is deterministic, and the winner is candidate j. In this case candidate j is the same as the MMPO winner under the Symmetric Completion Bottom rule, i.e. the Least Resentment Voting (LRV) winner. However in general the optimal strategies are mixed, which means that the players' moves are determined by probability distributions or lotteries. In this case, the column player's optimal lottery is used to pick the winner. By definition this method chooses the winner in a way that minimizes its expected opposition, so on average it accomplishes more completely the heuristic justification of LRV than LRV itself does. In other words use of this method will (on average) distress the opposition less than any other method. So let's call it the Least Expected Umbrage method or LEU. We need to check that it passes all of the tests and satisfies all of the properties that we think are most important. I know that most people are prejudiced against chance, so determinism is high on their list of importance. But I hope that future generations will be more enlightened on this score, and embrace the judicious use of chance. When they look back and see that we anticipated some of their ideas, they might forgive us for some of our other oversights. Forest Election-Methods mailing list - see http://electorama.com/em for list info
[EM] SODA strategy
If voters think that SODA is complex, then it's because they have been exposed unnecessarily or prematurely to the niceties of strategy considerations. Let's take a lesson from IRV supporters. They don't get anybody worried about IRV's monotonicity failure or FBC failure by bringing them up to unsophisticated voters. We need to emphasize the simplicity of SODA voting to the public, and answer the strategy questions to the experts. Election-Methods mailing list - see http://electorama.com/em for list info
[EM] FBC failure for acquiescing coalition methods
Mike,
I think your example applies to all acquiescing coalition methods that we have
considered. The failure is
caused by someone leap frogging over others to get to the top position.
But I think that most of these methods satisfy this FBC like property:
If the winner changes when (on some ballot) candidate X is moved to the top
slot along with all of the
candidates that were ranked above X, then the new winner will be X or one of
the other candidates that
were raised on that ballot.
This seems like a reasonable substitue for the FBC, since it builds into it a
consistency requirement,
namely that if you raise X. then sincerity requires raising to the same level
or higher all candidates that
you prefer over X.
Forest
From: MIKE OSSIPOFF
To:
Subject: [EM] Forest: I found an FBC failure for Minimal Aquiescing
Majorities-Top
Message-ID:
Content-Type: text/plain; charset=iso-8859-1
Forest--
Say it's like the ABE, except that there's one more candidate, D.
In the ABE, you were an A voter, but now, with D in the
election, you like D best,
with A your 2nd choice.
(Say all the A voters vote as you do)
The B voters, while willing to middle-rate A for a majority
coalition, wouldn't
be willing to miiddle-rate D.
If you vote A D together in 1st place, then your top-rating
for D means that
{A,B} is no longer a winning set, because you vote D over B.
If you vote in that way, C wins.
But you can at least make A win, because the B voters are
willing to middle-rate A.
You can do that by top-rating only A. You can middle-rate D if
you want to.
Then, {A,B} wins, and, in that set, A wins with the most top votes.
You can get your best possible outcome (the election of A) only
by voting someone over
your favorite.
Mike Ossipoff
Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] SODA might be the method we've been looking for.
Like Andy I prefer SODA as well, especially for a deterministic method. In some settings I prefer certain stochastic methods to deterministic methods. But my curiosity impels me to see what can be done while ignoring or putting aside the advantages of both chance and delegation. Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Forest: MAMT
Chris and Mike,
I think I finally have the right version which I will call MSAC for Majority
Support Acquiescing Coalitions:
Definitions:
A coalition is a subset of the candidates.
A ballot acquiesces to a coalition of candidates iff it rates no candidate
outside the coalition higher than
any member of the coalition.
A Majority Support Acquiescing Coalition is a coalition that is acquiesced to
by more than half of the
ballots.
A Majority Support Acquiescing Coalition is minimal if it ceases to be a
Majority Support Acquiescing
Coalition when any one of its members is removed.
Method Definition for MSAC:
(1) First find all of the Minimal Majority Support Acquiescing Coalition.
(2) Among these call the one with the greatest support G.
(3) Elect the member of G with the greatest average rating.
- Original Message -
From: C.Benham
Date: Tuesday, December 13, 2011 11:19 am
Subject: Forest: MAMT
To: em
Mike,
I can see how MAMT tries to meet Mono-add-Plump, but it fails.
49: C
27: AB
24: BA
MAMT (like all reasonable methods) elects A,
MSAC also elects A.
But say we add 20
ballots
that plump for A.
49: C
27: AB
24: BA
20: A (new ballots)
(120 ballots, majority threshold 61).
Now there are two minimal subset acquiescing majorities: {AB}
71 and
{CA}69. All three candidates are qualified so the most top-rated
candidate (C) wins.
But under MSAC candidate A still wins, because the {A, B} coalition is the set
G, and the range winner
of this set is A.
Note also that the MSAC winner is C for the following set of ballots:
49: C
27: AB
24: B ,
because in this case the coalition {B, C} is the value of G, and C is the range
winner of this coalition.
Forest
Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] This might be the method we've been looking for:
Thanks for checking the details. In traditional game theory the rational stratetgies are based on the assumption of perfect knowledge, so the A faction would know if the B faction was lying about its real preferences. Even knowing that the other faction knew that they were lying they could still threaten to defect, and even carry out their threat. There is no absolute way out of that. - Original Message - From: Andy Jennings Date: Monday, December 12, 2011 12:40 pm Subject: Re: [EM] This might be the method we've been looking for: To: Jameson Quinn Cc: fsimm...@pcc.edu, election-methods@lists.electorama.com You're right. I've drawn out the game theory matrix and the honest outcome: 49 C 27 AB 24 BA is indeed the stable one, with A winning. So the only way for B to win is for his supporters to say they are indifferent between A and C and threaten to bullet vote B. Then the A supporters fall for it and vote A=B to prevent C from winning. B wins. I wonder if this is sequence of events is likely at all. ~ Andy On Fri, Dec 9, 2011 at 2:31 PM, Jameson Quinn wrote: No, the B group has nothing to gain by defecting; all they can do is bring about a C win. Honestly, A group doesn't have a lot to gain from defecting, either; either they win anyway, or they misread the election and they're actually the B's. Jameson 2011/12/9 Andy Jennings Here’s a method that seems to have the important properties that we have been worrying about lately: (1) For each ballot beta, construct two matrices M1 and M2: In row X and column Y of matrix M1, enter a one if ballot beta rates X above Y or if beta gives a top rating to X. Otherwise enter a zero. IN row X and column y of matrix M2, enter a 1 if y is rated strictly above x on beta. Otherwise enter a zero. (2) Sum the matrices M1 and M2 over all ballots beta. (3) Let M be the difference of these respective sums . (4) Elect the candidate who has the (algebraically) greatest minimum row value in matrix M. Consider the scenario 49 C 27 AB 24 BA Since there are no equal top ratings, the method elects the same candidate A as minmax margins would. In the case 49 C 27 AB 24 B There are no equal top ratings, so the method gives the same result as minmax margins, namely C wins (by the tie breaking rule based on second lowest row value between B and C). Now for 49 C 27 A=B 24 B In this case B wins, so the A supporters have a way of stopping C from being elected when they know that the B voters really are indifferent between A and C. The equal top rule for matrix M1 essentially transforms minmax into a method satisfying the FBC. Thoughts? To me, it doesn't seem like this fully solves our Approval Bad Example. There still seems to be a chicken dilemma. Couldn't you also say that the B voters should equal-top-rank A to stop C from being elected: 49 C 27 A 24 B=A Then A wins, right? But now the A and B groups have a chicken dilemma. They should equal-top-rank each other to prevent C from winning, but if one group defects and doesn't equal-top-rank the other, then they get the outright win. Am I wrong? ~ Andy Election-Methods mailing list - see http://electorama.com/em for list info Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Mike: variations on Kevin's ABE and LRV
Mike asked ... You can reach the person managing the list at Forest? Could LRV, due to the bottom symmetrical completion, sometimes have an ABE-like problem, if the numbers were somewhat differentfrom those of the usual ABE? Could it have a co- operation/defection problem for that reason? It works exactly like MinMax(margins) when there is no equal top ranking as is the case for x: C y: AB z: B (or BA), MinMax(margins) handles the ABE just fine, but it doesn't satisfy the FBC, which is the main contribution of LRV. What doe you think about the example 33 A 17A=C 17B=C 33 B approval scenario? LRV elects C, but gives a toss up to A and B in the scenario 34 A 16A=C 16B=C 34 B . Forest Election-Methods mailing list - see http://electorama.com/em for list info
[EM] Mike Ossipoff: LRV properties
Mike, LRV is just another equivalent way of describing MMMPO. The are just MMPO with symmetric completion at the bottom level but not at the top. In addition to FBC it satisfies MAP, KMBE, and U, but not LNHe. MAP means Mono-Add-Plump KMBE means Kevin's MMPO bad-example LNHe means Later-No-Help U means that a majority can be established unilaterally LRV/MMPO goes beyond MAP to satsify mono-add-equal-top; if a new ballot ranks the old winner equal top, then the winner stays the same. The FBC is a corollary of the following property.: If (on a ballot) a candidate is moved up from a position above the winner or to a position that is not above the winner, then the winner either stays the same or is changed to the candidate that was moved up. Now here's my attempt at Minimal Mutual Acquiescing Majority Top: A ballot acquiesces to a set S of candidates if no candidate outside of S is ranked above any candidate inside the set S on the ballot in question. A set S is a Mutual Acquiescing Majority set if a majority of ballots acquiesce to S. A set with a certain property is minimal with respect to that property if it ceases to have that property when any of its members is removed. A Minimal Mutual Acquiescing Majority set is a Mutual Acquiescing Majority set which would cease to be a Mutual Acquiescing Majority set if any of its candidates were removed. Minimal Mutual Acquiescing Majority, Top method: Of all the candidates that belong to some Minimal Mutual Acquiescing Majority set, elect the one rated top (or equal top) on the greatest number of ballots. End of method definition. Note that there is always a Minimal Mutual Acquiescing Majority set because if no proper subset of the candidates is such a set, then the entirre set of candidates is a Minimal Mutual Acquiescing Majority set. Forest Election-Methods mailing list - see http://electorama.com/em for list info
[EM] MAM evaluation. Summary of FBC/ABE methods.
Date: Sat, 10 Dec 2011 18:44:15 + From: MIKE OSSIPOFF To: Subject: [EM] MAM evaluation. Summary of FBC/ABE methods. Message-ID: Content-Type: text/plain; charset=iso-8859-1 Evaluation of MAM: Forest-- I've been looking at various ways of doing Mutual Acquiescing Majorities (MAM). (I think that was the name that you used). Initially I thought it worked, and that it had a better set of properties than any comparably simple method. But then I realized that I couldn't make that method work. Any success? My definition of the MAM set: A set of candidates whom every member of the same majority vote equal to or over every candidate outside the set. [end of MAM set definition] How about changing set to minimal subset:. A minimal subset of candidates such that for some majority of voters, no member of this majority votes any candidate outside the subset over any member of the subset. Minimal means that if any candidate is removed from the subset it will no longer have the desired property. This solves the problem and simplifies the method description since if no other subset is minimal, then the entire set of candidates is minimal. Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] MMPO and Symmetric Completion
Jameson, good idea and valuable comments. However, I'm not sure that regret is the right word. I regret something after I make a bad choice. I resent something when I make a good choice that is over-ridden by somebody with the power to do so. I suggest Least Resentment Voting, LRV. Forest - Original Message - From: Jameson Quinn Date: Thursday, December 8, 2011 4:46 pm Subject: Re: [EM] MMPO and Symmetric Completion To: fsimm...@pcc.edu Cc: election-methods@lists.electorama.com Now let's come up with a good name for this MMPO with partial symmetric completion. Actually we need a good technical name as well as a catchy name for public proposal. This is indeed a good method. In simple parlance, you want the candidatewho is least disliked against any other candidate, counting equal bottom as half-disliked. So I suggest Least Regret Voting as a name. Unfortunately, this philosophy -- choosing the least-worst -- is less intuitively-appealing to most people than majoritarian philosophies. If I didn't know how the two systems worked, I'd probably be more inclined to like a system called Majority Judgment than a system called Least Regret Voting. (Actually, even knowing their content, I'd still probably pick MJ; though I'd pick LRV if you modified it slightly by using a ballot with meaningfulrating categories, which is perfectly compatible with the system. I'll still be pushing MJ and SODA over rated-LRV, though, until there's at least a few dozen people in the world not on this list who've heard of LRV. Anyway, regardless of which methods I support, I still think it's a pity that the naming has an inherent bias towards MJ over LRV; the systems should win or lose on their merits, not on their names.) As for a technical name... I guess I'd choose BSC-MMPO, Bottom-symmetrically-completed etc. But I'm one who's perfectly happy to call MJ, MJ, rather than Cardinal Median//Median Drop Median or some such technically-descriptive name. So I don't really care about the technical names. Jameson ps. I know that BR fans might object to the different definition of regret implicit in calling it LRV; but I was never a fan of the BR name myself so I can't say I worry about that. BR is a great idea but, I feel, a poor name. Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Acquiescing majority MMT (MIKE OSSIPOFF)
Mike,
yes it is the same as the one you repeated at the end of your reply below.
But notice that in the ballot set
49 C
27 AB
24 B
there are two Acquiescing Majorities, namely both {A, B} and {B, C}, and that C
has more top votes than B.
Forest
From: MIKE OSSIPOFF
To:
Subject: [EM] Acquiescing majority MMT
Message-ID:
Content-Type: text/plain; charset=iso-8859-1
Forest--
Let's find out what its properties are.
Just preliminarily, it sounds like one of the MMT ideas that I
considered.
But it seemed to me, at the time (if it's the same method I was
considering) that, if a ballot can be counted in that majority
merely by rating each candidate
in the set equal to or over every candidate outside the set,
then, in the ABE, the B votes could
rate A at bottom, with C, and still be part of the relevant
majority. So there there is
the set required by the acquiescing rule, and there is one
candidate rated above bottom
by everyone in that set: Candidate B.
So, the method that I'd considered wouldn't pass in the ABE. I
don't know if the method
you describe is the same one, but, preliminarily, it sounds similar.
But maybe not. Any possibility could yield improvement.
My definition of that set was something like this:
A set of candidates rated equal to or over everyone outside the
set by each member of the
same majority of the voters.
Mike Ossipoff
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[EM] Least Resentment Voting (LRV) in the context of Approval
If I remember correctly Kevin Venzke's first post to this list was a geometric argument that the MMPO winner was apt to be closer to the voter median position in Approval than the Approval winner. The scenario he had in mind was something like this Scenario One: 26 A 24 A=C 24 B=C 26 B The geometry was something like this: AB _ It weemed pretty obvious that the MMPO winner C was more likely the true majority choice than than either of the tied approval winners A.or B. Not long after that Kevin came up with his MMPO bad example: Senario Two: 48 A 2 A=C 2 B=C 48 B. The MMPO winner C lacked too much approval to be the real winner. Kevin took back his proposal, and went on the bigger and better things like trying to convince Juho that MinMax (margins) was not an acceptable public proposal. either. But as I mentioned in a recent posting MMPO with symmetric completion is the same as MinMax(margins), and by exempting the top rank or slot from the symmetric completion, we get a really nice compromise between MMPO and MinMax(margins), and what's more this version gives the right answer in both scenarios above. Kevin was ever so close to proposing what I now call LRV Least Resentment Voting or more technically Bottom- Symmetric-Completion MMPO. In the first scenario above both A and B have max opposition of fifty, while the max opposition of C is 26 + 13 = 39, so C wins. In the second scenario, A and B still have max opposition of 50 each, but now C's max opposition is 48 + 24 = 72, so C loses, to a tie betwen A and B. Where is the cutoff where all three are tied? Scenario Three: 34 A 17 A=C 17 B=C 34 B In this scenario the max opposition (with the bottom symmetric completion rule in effect) is 34+17 for A, 34+17 for B, and 34 +17 for C. The second largest opposition to either A or B is 17+17=34, while the second largest opposition to C is still 34+17, so C loses the tied situation , and the winner is a toss up between A and B. Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] MMT
Mike, what about a version of MMT that we could call MAMT: Define a Mutual Acquiescing Majority set S as a set of candidates that are acquiesced to on a majority of ballots, i.e. for each ballot of the given majority set of ballots, no candidate outside the set S is ranked or rated above any member of S. Of all the candidates that are members of one or more Mutual Acquiescing Majority sets, elect the one with the most top ratings or rankings. If no candidate is in a MAM set, elect the one with the most top ratings or rankings. Forest Election-Methods mailing list - see http://electorama.com/em for list info
[EM] MMPO and Symmetric Completion
MinMax Pairwise Opposition satisfies the FBC but not the Condorcet Criterion. MinMax(margins) satisfies the Condorcet Criterion, but not the FBC. MMPO combined with symmetric completion of all equal rankings and truncations is exactly equivalent to MinMax(margins), so symmetric completion of MMPO trades in the FBC for the CC. But if we exempt the equal top rank from symmetric completion MMPO retains the FBC. As before the CC is sacrificed, but the only thing standing between this version of MMPO and the CC is symmetric completion of the top ranked candidates. Since we value the FBC more than the CC, we refrain from symmetric completion of the top rank. How far does this take us from the CC? I believe that the resulting method (MMPO based on symmetric completion of all equal ranks except top) still has great Condorcet efficiency whether the voting is zero info or perfect info. Perhaps it will even satisfy Juho, since it is as close as you can get to MinMax(margins) while satisfying the FBC. Personally, I prefer to go a little further and refrain from symmetric completion of any ranks except equal bottom. This move further towards classical MMPO drastically reduces any incentive for insincere order reversals at any level (as in MMPO). Why not go all the way to MMPO? Because the symmetric completion at the bottom is necessary (and sufficient) for resolving the Approval Bad Example and Kevin's MMPO bad example. Unfortunately, symmetric completion at the equal bottom level destroys Later No Harm compliance, but the method still satisfies the following (nameless?) criterion: If candidate X is advanced on a ballot by moving from a position above the winner to a higher position or by moving to a position below (or equal) to the winner from a lower position, then either the winner is unchanged or the new winner is the candidate X. Put this property together with mono-raise and mono-add-equal-top and you have a nice set of complementary monotonicity properties. Now let's come up with a good name for this MMPO with partial symmetric completion. Actually we need a good technical name as well as a catchy name for public proposal. Forest Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Forest's FBC/ABC method
Here's an equivalent but simpler description of the FBC/ABE compliant method that I have been calling (since Mike's pointer about MaxMin vs. MinMax) MaxMin(EqualRankPairwiseRule): Let M be a matrix whose entry in row i and column j is the number of ballots on which candidate i is rated or ranked strictly above candidate j plus half the number of ballots on which candidates i and j are ranked or rated equal bottom (i.e. neither one is ranked or rated ahead of any other candidate on the ballot). Find the maximum value in each column of M. Find the column j whose maximum value is minimal. Elect candidate j. In other words, elect the MinMax(ModifiedPairwiseOpposition) candidate, where the modified pairwwise opposition of candidate x against candidate y includes half the number of ballots on which neither x nor y is ranked or rated above any other candidate. In other words, it is the same as MMPO after symmetrical completion of the order among the equal bottom candidates on each ballot (but not among other equally ranked or rated candidates). Note that this method is a modification of MMPO that retains the FBC compliance as well as the mono- add-equal-top compliance. In addition it finesses the Approval Bad Example (defection problem) and does not elect the weak middle candidate of Kevin's Bad MMPO example. Note that mono-add-equal-top is stronger than mono-add-top, which in turn is stronger than mono-add- plump. Shall we abbreviate MinMax(ModifiedPairwiseOpposition) with MMMPO? It still needs a fancier name for public consumption. From: fsimm...@pcc.edu To: election-methods@lists.electorama.com Subject: [EM] Chris: Forest's FBC/ABC method (MIKE OSSIPOFF) Message-ID: Content-Type: text/plain; charset=us-ascii Mike is right; it should be called MaxMin instead of MinMax. From: MIKE OSSIPOFF To: Subject: [EM] Chris: Forest's FBC/ABC method Message-ID: Content-Type: text/plain; charset=iso-8859-1 Chris-- I'll describe Forest's proposal briefly: It's minmax margins (but it's defined as maxmin, with respect to xy - yx), looking at all pairwise comparisons, rather than just at defeats. But, instead of just xy - yx, it's x top or y - yx. As I said in my other posting, it seems to have the same properties as MMT. In other words, FBC, LNHa, 3P, and the (unnecessary) Mono-Add-Plump and the (unnecessary) avoidance of electing C in Kevin's MMPO bad-example. Though Mono-Add-Plump and complying in Kevin's example are unnecessary,they avoid misguided or dishonest criticism by opponents of a reform proposal. As I've said, maybe it's better to ask for a little less than MMPO and MDDTR, in order to avoid the distraction that such criticisms could cause, during an enactment campaign. --especially given that the opponents are likely to have a lot more media money than the proponents. Mike Ossipoff Election-Methods mailing list - see http://electorama.com/em for list info
[EM] Chris: Forest's FBC/ABC method (MIKE OSSIPOFF)
Mike is right; it should be called MaxMin instead of MinMax. From: MIKE OSSIPOFF To: Subject: [EM] Chris: Forest's FBC/ABC method Message-ID: Content-Type: text/plain; charset=iso-8859-1 Chris-- I'll describe Forest's proposal briefly: It's minmax margins (but it's defined as maxmin, with respect to xy - yx), looking at all pairwise comparisons, rather than just at defeats. But, instead of just xy - yx, it's x top or y - yx. As I said in my other posting, it seems to have the same properties as MMT. In other words, FBC, LNHa, 3P, and the (unnecessary) Mono-Add-Plump and the (unnecessary) avoidance of electing C in Kevin's MMPO bad-example. Though Mono-Add-Plump and complying in Kevin's example are unnecessary,they avoid misguided or dishonest criticism by opponents of a reform proposal. As I've said, maybe it's better to ask for a little less than MMPO and MDDTR, in order to avoid the distraction that such criticisms could cause, during an enactment campaign. --especially given that the opponents are likely to have a lot more media money than the proponents. Mike Ossipoff Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] This might be the method we've been looking for:
Chris, you're right that it is very close to MinMax(margins). Let's compare and contrast: In both MinMax versions a matrix M is used to determine the winner in the same way: if the least number in row i is greater than the least number in any other row of the matrix M, then candidate i is elected. [By convention each negative number is less than every positive number, and among several negative numbers the most negative is the least. So -6 -32 5, etc.] In both methods each entry of the matrix M is the difference betwee two numbers (minuend minus subtrahend). The subrtrahend in this difference is exactly the same in both methods. The subtrahend in row i column j of M is the number of ballots on which candidate j is rated or ranked strictly above candidate i. It's in the minuend that the two methods part company: In MinMax(margins) the minuend of the (i, j) entry is the number of ballots on which candidate i is rated or ranked strictly above candidate j. In MinMax(TopTierPairwiseRule) the minuend of the (i, j) entry is the number of ballots on which candidate i is ranked (or rated) strictly above candidate j plus the number of ballots on which candidate i is rated or ranked equal top with candidate j. A third method that I call MinMax(EqualRankPairwiseRule) uses the same subtrahend but defines the minuend as the number of ballots on which candidate i is ranked both above bottom AND above or equal to candidate j. This last method MinMax(ERPR) also satisfies the FBC, and furthermore it nevers gives incentive for insincere order reversal. Both MinMax(ERPR) and MinMax(TTPR) satisfy the mono-add-equal-top criterion: if additional ballots are added with the previous winner ranked top or equal top, the winner is unchanged. Furthermore, suppose that candidate i is the winner under MinMax(ERPR), and that the least number in row i of matrix M is -7. Suppose that this number -7 appears only in columns 3, 9, and 15 of row i. If a new ballot ranks candidate i above or equal to candidates 3, 9, and 15, then the method will still elect candidate i when the new ballot is counted along with the old ones. Note that in the case of MinMax(ERPR) the diagonal entries (i, i) in the matrix M are the respective implicit approvals of the candidates, since ranked candidates are ranked equal to themselves but not above themselves. All three of these MinMax methods are monotone, but fail clone independence in the same sense that MinMax(wv) does. The equal ranking option mitigates this failure. Perhaps further modifications could mitigate it more, if not altogether remove it. For example, incorporating some version of the Cardinal Weighted Pairwise idea might restore clone independence to the same degree enjoyed by Approval and other Cardinal Ratings methods. We can deal with that later. Meanwhile, with a three slot method, clones tend to get equal ranked a lot, so the clone dependence is not much worse than it is in Approval. We need a popular name that can catch on with the public. Any ideas? Forest - Original Message - From: C.Benham Date: Saturday, December 3, 2011 0:24 am Subject: This might be the method we've been looking for: To: em Cc: Forest W Simmons Forest, I don't understand the algorithm's definition. It seems to be saying that it's MinMax(Margins), only computing X's gross pairwise score against Y by giving X 2 points for every ballot on which X is both top-rated and voted strictly above Y, and otherwise giving X 1 point for every ballot on which X is top-rated *or* voted strictly above Y. But from trying that on the first example it's obvious that isn't it. Can someone please explain it to me? Chris Benham Forest Simmons wrote (2 Dec 2011): Here’s a method that seems to have the important properties that we have been worrying about lately: (1) For each ballot beta, construct two matrices M1 and M2: In row X and column Y of matrix M1, enter a one if ballot beta rates X above Y or if beta gives a top rating to X. Otherwise enter a zero. IN row X and column y of matrix M2, enter a 1 if y is rated strictly above x on beta. Otherwise enter a zero. (2) Sum the matrices M1 and M2 over all ballots beta. (3) Let M be the difference of these respective sums . (4) Elect the candidate who has the (algebraically) greatest minimum row value in matrix M. Consider the scenario 49 C 27 AB 24 BA Since there are no equal top ratings, the method elects the same candidate A as minmax margins would. In the case 49 C 27 AB 24 B There are no equal top ratings, so the method gives the same result as minmax margins, namely C wins (by the tie breaking rule based on second lowest row value between B and C). Now for 49 C 27 A=B 24 B In this case B wins, so the A supporters have a way of stopping C from being elected when they know that the B voters
[EM] This might be the method we've been looking for:
Here’s a method that seems to have the important properties that we have been worrying about lately: (1) For each ballot beta, construct two matrices M1 and M2: In row X and column Y of matrix M1, enter a one if ballot beta rates X above Y or if beta gives a top rating to X. Otherwise enter a zero. IN row X and column y of matrix M2, enter a 1 if y is rated strictly above x on beta. Otherwise enter a zero. (2) Sum the matrices M1 and M2 over all ballots beta. (3) Let M be the difference of these respective sums . (4) Elect the candidate who has the (algebraically) greatest minimum row value in matrix M. Consider the scenario 49 C 27 AB 24 BA Since there are no equal top ratings, the method elects the same candidate A as minmax margins would. In the case 49 C 27 AB 24 B There are no equal top ratings, so the method gives the same result as minmax margins, namely C wins (by the tie breaking rule based on second lowest row value between B and C). Now for 49 C 27 A=B 24 B In this case B wins, so the A supporters have a way of stopping C from being elected when they know that the B voters really are indifferent between A and C. The equal top rule for matrix M1 essentially transforms minmax into a method satisfying the FBC. Thoughts? Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] MMPO tiebreakers that don't violate FBC.
Mike, I like MMPO2 because (unlike MMPO1) it takes into account opposition from supporters of eliminated candidates, so is more broad based, and it is easily seen to satisfy the FBC. Also it allows more brad based support than MMPO3 where only the support by top raters is considered in the tie breaking process. Forest Mike Ossipoff wrote ... First of all, no one has found an example in which my most recent MMPO tiebreaker fails FBC. Solve MMPO's ties by MMPO. I'll call that tiebreaker 1. Tiebreaker 2: Whenever there's a tie, the winner is the candidate whose next largest pairwise opposition is the least. (next largest after the pairwise opposition with which s/he is tied with other candidates) Tiebreaker 3: If there is a tie, the winner is the candidate with the most top-ratings. MMPO with tiebreaker 1 is MMPO1 MMPO with tiebreaker 2 is MMPO2 MMPO with tiebreaker 3 is MMPO3. MMPO2 and MMPO3 meet FBC. No one has found an example in which MMPO1 fails FBC. MDDTR meets FBC, CD, and Later-No-Harm (LNHa); and fails Mono-Add-Plump. MMPO2 and MMPO3 meet FBC, CD LNHa, and Mono-Add-Plump; and fails in Kevin's MMPO bad-example: : A 1:A=C 1:B=C : B In this example, MMPO elects C. MCA, MTA, MDDTR, MDDA, ABucklin and MDD,ABucklin return a tie between A B. How bad is this result of MMPO's? Notice that nearly all of the A voters are indifferent between B and C. And the one A voter who isn't indifferent prefers C to B. Likewise nearly all of the B voters are indifferent between A and C. And the one B voter who isn't indiffefrent prefers C to A. So how bad can c be? How bad can MMPO's result be? So, the choice between MMPO and MDDTR is a choice between failing Mono-Add-Plump, vs failing in kevin's MMPO bad-example. Only polling will tell which failure is more likely to appear bad to potential petition-signers and enaction-voters. When we regard the pairwise comparison between C and A, and the pairwise comparison between C and B, in isolation, it's easy to forget that we aren't actually holding those two pairwise elections. We're used to looking at pairwise contests, because we're used to pairwise-count methods. But no one is saying that the one voter who votes C over A is more important than the greater number who vote A over c. I suggest that the strong pairwise defeats against C look more important than they are, because we're used to looking at pairwise count methods, with the result that we actually start believing that there was a a 2-candidate election between A and C, and one between B and C. As I said, MDDTR and MMPO might be controversial, because of failure of Mono-Add-Plump, or failure in Kevin's MMPO bad-example. So, I don't consider them to be the easiest or best public proposals. It would be necessary to talk to some people before being sure that they'd be winnable proposals. A 3-slot version of SFC should be written, to tell of the SFC-like guarantee offered by MDDTR. (Strictly speaking, SFC can only be passed by full ranking methods.) By the way, when people object to random-fill incentive for MDDTR, maybe they're forgetting that MDDTR is a 3-slot method. And, if MMPO were proposed as a 3-slot method, that would avoid the random-fill incentive criticism of it too. Mike Ossipoff Previous message: [EM] This Message didn't post properly, and I'm re-trying it. MDDTR and Mono-Add-Plump reply. Next message: [EM] MMPO tiebreakers that don't violate FBC. Messages sorted by: [ date ] [ thread ] [ subject ] [ author ] More information about the Election-Methods mailing list Election-Methods mailing list - see http://electorama.com/em for list info
[EM] MMPO ideas
There are several ideas that can be used to make variations on MMPO. 1. One is to use a bottom Tier Pairwise rule that counts bottom level candidates on a ballot as being opposed by all other bottom level candidates (analogous to the TTP rule in other methods). Note that this rule doesn't get in the way of the FBC. 2. Use Cardinal Ratings Pairwise opposition. The opposition on a ballot is proportional to how much higher the opposing candidate is rated on the ballot. 3. Instead of Minimizing Max Pairwise Opposition, which is a measure of defensive strength, use a measure of offensive strength: maximize the minimum points scored against the other candidates. To satisfy the FBC use the Top Tier Pairwise rule that gives full points for equal top ratings on a ballot. If X and Y are rated equal top on a ballot then this ballot counts as an offensive point of X against Y, as well as an offensive point of Y against X. 4. Some combination of the defensive and offensive scores in 3. Find the difference or ratio, for example, of the MaxMinPairwisePointsAgainst and the MinMaxPairwiseOpposition. The candidate with the highest ratio or difference is elected. . Election-Methods mailing list - see http://electorama.com/em for list info
[EM] Does High Resolution Range offer a solution to the ABE?
While working with MinMaxCardinalRatingsPairwiseOpposition (MMcrwPO) I got an idea that high resolution Range might have an acceptable solutin to the defection problem that we have been considering: Sincere ballots 49 C x: AB y: BA where x appears to be slightly larger than y in the polls. The A and B factions can agree to enough support such that if they both follow through the one with the larger support will win, but if one defects, C X will win. In this case that level of support is about 96%. If the A voters and the B voters both give 96% to their second choice, then A or B will win, depending on whether or not x is greater than y. If anybody defects from this, then C sill win. Election-Methods mailing list - see http://electorama.com/em for list info
[EM] MMCWPO (minimize maximum cardinal weighted pairwise opposition) satisfies the FBC and solves the ABE problem.
MMCWPO is the method that elects the candidate whose maximal weighted pairwise opposition is minimal. It solves the ABE problem as well as the FBC. I'm being shut down on this computer. More after T day. Forest Election-Methods mailing list - see http://electorama.com/em for list info
[EM] TTPD,TR (an FBC complying ABE solution?)
Chris, your new method includes the statement ... If any candidate X TTP beats any candidate Y, is not in turn TTP beaten by Y and is not TTP beaten by any candidate Z that doesn't also TTP beat Y, then Y is disqualified. In other words, there is no short TTP beatpath from Y to X, where short means fewer than three steps. We could use this condition as the definition for X covers Y in the TTP sense, or more briefly X TTP- covers Y. So your method elects the TTP-uncovered candidate rated top on the greatest number of ballots. If this method fails mono-raise (like most uncovered methods do) we can use one of our covering method ideas to fix it. For example .. Let variable X be the candidate with the highest range score. Then while X is TTP-covered, replace X with the highest range score candidate that TTP-covers it. Elect the last value of X. Forest Election-Methods mailing list - see http://electorama.com/em for list info
[EM] An ABE solution
Mike, thanks for your comments. I'll respond in line below. From: MIKE OSSIPOFF Hi Forest-- Thanks for answering my question about MTA vs MCA. Your argument on that question is convincing, and answers my question about the strategy difference between those two methods. Certainly, electing C in the ABE avoids the ABE problem. I'd been hoping that the election of C can be attained without diverging from Plurality's results enough to upset some people, as MMPO and MDDTR seem to do. So the method that you describe might avoid the public relations (non)problems of methods that elect A. I have a few questions about the method that you describe: 1. What name do you give to it? In this post I'll call it Range till cover-winner or RCW Good Idea. 2. The covering relation doesn't look at pairwise ties? Different variants handle ties differently, but I favor using dominance in the win/loss/tie matrix, in which the entry in row x column y is respectively one, minus one, or zero depending on whether candidate x beats, is beaten by, or ties with candidate y, respectively. We consider a candidate to be tied with itself, so for each x, the diagonal (x, x) entry is zero. Note that this matrix can be gotten by subtracting the transpose of the pairwise matrix from itself, and then replacing each entry by its sign, where sign(t) is 1, -1, or 0, depending on whether t is positive, negativve, or zero. In other words, replace each entry in the pairwise margins matrix with its sign. Row x dominates row y iff the two rows are not identical and every entry in row x is at least as great as the corresponding entry in row y. Candidate x covers candidate y iff row x dominates row y in the above sense. 3. Does the ballot ask the voter for cardinal ratings of the candidates, or is the range score calculated a la Borda? Cardinal ratings are assumed, and for public proposal I contemplate three slots. If the ballots are ordinal, you can transform them (clone free and monotonically) to cardinal ballots via my algorithm under the heading Borda Done Right. 4. How does it do by FBC? And by the criteria that bother some people here about MMPO (Kevin's MMPO bad-example) and MDDTR (Mono-Add-Plump)? I think it satisfies the FBC. In fact, it reduces to Approval in the two slot version or if all voters rate only at the extremes., Approval strategy is probably near near optimal. It satisfies Mono-Add-Plump. Proof: First note that addition of a ballot that truncates all of the candidates doesn't change the winner since it doesn't change the covering relations nor does it change the range score (cardinal rating) order. Now, on such a ballot raise the winner from trunctated status to non-truncated status, and leave the rest of the candidates at the bottom. By the monotonicity of the method, the winner is preserved. I believe it satisfies the Plurality criterion. At least in Kevin's MMPO bad example it ties the two outside candidates. There's much hope that, by electing C instead of A, RCW can avoid those criticisms. I'm also interested in how it does by 1CM and 3P, but I'll look at that, instead of asking you to do everything for me, especially since I'm the one promoting those two criteria. I don't hve those two criteria on the tip of my tongue, but I'll look them up and see if I can help you on that. Mike Forest Here?s my current favorite deterministic proposal: Ballots are Range Style, say three slot for simplicity. When the ballots are collected, the pairwise win/loss/tie relations are determined among the candidates. The covering relations are also determined. Candidate X covers candidate Y if X beats Y as well as every candidate that Y beats. In other words row X of the win/loss/tie matrix dominates row Y. Then starting with the candidates with the lowest Range scores, they are disqualified one by one until one of the remaining candidates X covers any other candidates that might remain. Elect X. For practical purposes this method is the same as Smith//Range. Where they differ, the member of Smith with the highest range score is covered by some other Smith member with a range score not far behind. This method resolves the ABE (approval bad example) in the following way: Suppose that the ballots are 49 C 27 A(top)B(middle) 24 B No candidate covers any other candidate. The range order is CBA. Both A and B are removed before reaching candidate C, which is not covered by any remaining candidate. So the Smith//Range candidate C wins. If the ballots are sincere, then nobody can say that the Range winner was a horrible choice. But more to the point, if the ballots are sincere, the A supporters have a way of rescuing B: just rate hir equal top with A. Suppose, on the other hand that the B supporters like A better than C and the A supporters know this. Then
[EM] An ABE solution.
Here’s my current favorite deterministic proposal: Ballots are Range Style, say three slot for simplicity. When the ballots are collected, the pairwise win/loss/tie relations are determined among the candidates. The covering relations are also determined. Candidate X covers candidate Y if X beats Y as well as every candidate that Y beats. In other words row X of the win/loss/tie matrix dominates row Y. Then starting with the candidates with the lowest Range scores, they are disqualified one by one until one of the remaining candidates X covers any other candidates that might remain. Elect X. For practical purposes this method is the same as Smith//Range. Where they differ, the member of Smith with the highest range score is covered by some other Smith member with a range score not far behind. This method resolves the ABE (approval bad example) in the following way: Suppose that the ballots are 49 C 27 A(top)B(middle) 24 B No candidate covers any other candidate. The range order is CBA. Both A and B are removed before reaching candidate C, which is not covered by any remaining candidate. So the Smith//Range candidate C wins. If the ballots are sincere, then nobody can say that the Range winner was a horrible choice. But more to the point, if the ballots are sincere, the A supporters have a way of rescuing B: just rate hir equal top with A. Suppose, on the other hand that the B supporters like A better than C and the A supporters know this. Then the threat of C being elected will deter B faction defection, and they will rationally vote A in the middle: 49 C 27 A(top)B(middle) 24 B(top)A(middle) Now A covers both other candidates, so no matter the Range score order A wins. This completely resolves the ABE to my satisfaction. The method also allows for easy defense against burial of the CW. In the case 40 AB (sincere ACB) 30 BC 30 CA where C is the sincere CW, the C supporters can defend C's win by truncating A. Then the Nash equilibrium is 40 A 30 BC 30 C in which C is the ballot CW, and so is elected. Now for another topic... MTA vs. MCA I like MTA better than MCA because in the case where they differ (two or more candidates with majorities of top preferences) the MCA decision is made only by the voters whose ballots already had the effect of getting the ”finalists” into the final round, while the MTA decision reaches for broader support. Because of this, in MTA there is less incentive to top rate a lesser evil. If you don’t believe the fake polls about how hot the lesser evil is, you can take a wait and see attitude by voting her in the middle slot. If it turns out that she did end up as a finalist (against the greater evil) then your ballot will give her full support in the final round. Election-Methods mailing list - see http://electorama.com/em for list info
[EM] Descending Acquiescing Coalitions versus Nested Acquiescing Coalitions
DAC (descending acquiescing coalitions) disappointed Woodall because of the
following example:
03: D
14: A
34: AB
36: CB
13: C
The MDT winner is C, but DAC elects B.
DAC elects B even though the set {B} has a DAC score of zero, because the
descending order of
scores includes both the set {C,B} (with a score of 49) and the set {A,B} (with
a score of 48), and the
only candidate common to both sets is B, so B is elected by DAC.
But suppose that we change DAC to NAC (Nested Acquiescing Coalitions) so that
sets in the sequence
of descending scores are not only skipped over when the intersection is empty,
but also skipped over
when the set with the lower score is not a subset of the previously included
sets. Then, in the above
example, C is elected.
I want to point out that this NAC method also solves the bad approval problem
by electing C, B, and A
respectively, given the respective ballot sets
49 C
27 AB
24 B,
and
49 C
27 A=B
24 B,
and
49 C
27 AB
24 BA .
Which of the good properties of DAC are retained by NAC?
Thanks,
Forest
Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Enhanced DMC
Chris,
It could happen that the lowest approval candidate X that covers all higher
approved candidates is covered by an even lower candidate Y that beats all
higher approved candidates but doesn't cover them all.
In that case X, even though X is the Covering DMC winner, some candidate with
less approval will be the Enhanced DMC winner.
I like Enhanced DMC better than Covering DMC because the latter is more
vulnerable to compromise: a weak favorite which is not covered by compromise can
block compromise's chances by merely getting higher approval. In the former
version favorite has to definitively defeat compromise to ruin her chances.
- Original Message -
From: Chris Benham
I like this. Regarding how approval is inferred, I'm also
happy with Forest's idea of using Range
(aka Score) type ballots (on which voters give their most
preferred candidates the highest numerical
scores) and interpreting any score above zero as approval and
breaking approval ties as any score
above 1 etc. Or any other sort of multi-slot ratings ballot
where all except the bottom-most slot is
interpreted as approval.
Another idea is to enter above-bottom equal-ranking between any
2 candidates in the pairwise matrix
as a whole vote for both candidates, and then take each
candidate X's highest single pairwise score
as X's approval score.
Here are a couple of examples to demonstrate how this method
varies from some other Condorcet
methods.
48: A
01: AD
24: BD
27: CBD
D is the most approved candidate and in the Smith set, and so
Smith//Approval elects D.
Forest's Enhanced DMC or Covering DMC (and your suggested
SARR implementation)
elects B.
B covers D and to me looks like a better winner. This method has
a weaker truncation incentive
than Smith//Approval.
25: AB
27: BC
26: CA
22: C
Approvals: C75, B52, A51.AB 51-49, BC 52-48, CA 75-25
Plain DMC and using MinMax or one of the algorithms that is
equivalent to it when there are three
candidates (such as Schulze and Ranked Pairs and River) and
weighing defeats either by Winning
Votes or Margins all elect B.
If 5 of the 22 C voters change to A those methods all elect C
(a failure of Woodall's mono-sub-delete
criterion).
25: AB
27: BC
26: CA
17: C
05: A (was C)
Approvals: C70, A56, B52.AB 56-49, BC 52-48, CA 70-30.
In both cases our favoured method (like Smith//Approval) elects
C, the positionally dominant candidate. It
seems those other methods are more vulnerable to Push-over strategy.
(To be fair, Woodall has demonstrated that no Condorcet method
can meet mono-raise-delete.)
Chris Benham
From: Ted Stern
To: election-methods@lists.electorama.com
Cc: Forest Simmons ; Chris Benham
Sent: Wednesday, 5 October 2011 8:35 AM
Subject: Re: [EM] Enhanced DMC
After some private email exchanges with Forest and Chris, I'm
proposing a simple way of implementing Enhanced DMC, plus a new name,
Strong Approval Round Robin Voting (SARR Voting).
Ballot:
Ranked Voting, all explicitly ranked candidates considered approved.
Equal ranking allowed. I'm basing this on recommendation from Chris
Benham. I'm open to alternatives, but it seems to be the
easiest way
to do it for now, and the most resistant to burying strategies.
Tallying:
Form the pairwise matrix, using the standard Condorcet
procedure. In
the diagonal entries, save total Approval votes.
For N candidates, the list of candidates in order from highest to
lowest approval is
X_0, X_1, ..., X_k, X_{k+1}, ..., X_{N-1}
Initialize the Strong set to the empty set
Initialize the Weak set to the empty set.
For k = 0 to N-1,
If X_k is already in the Weak set, continue iterating. (X_k is
defeated by a higher approved candidate. This is called being
strongly defeated.)
If X_k loses to a member of the Weak set, continue iterating. (X_k
may defeat all higher approved candidates, but is weakly defeated
by at least one of them.)
If we're still here in the loop, X_k defeats all candidates in the
Strong Set and all candidates in the Weak set. (X_k covers all
previously added members of the Strong set.)
Add X_k to the Strong set and add all of X_k's defeats to the Weak
set.
Set the provisional winner to X_k.
The last provisional winner (the last candidate added to the Strong
set) is the winner of the election.
Note:
The first member of the Strong Set will be X_0.
It is easiest to do this by hand if you first permute the pairwise
array so that it follows the same X_0, ..., X_{N-1} ordering.
As an example election, consider the one on this page:
http://wiki.electorama.com/wiki/Marginal_Ranked_Approval_Voting
Iterating through E, A, C, B, D, we find
E: Strong and Weak Sets are empty, so E has no losses to either.
Strong set = {E}; Weak set = {C, D}
Provisional winner set
Re: [EM] Dodgson and Kemeny done right?
You're right, I forgot that Kemeny only needed the pairwise matrix. And according to Warren Dodgson is summable. I don't see how. - Original Message - From: Kristofer Munsterhjelm Date: Thursday, September 15, 2011 12:14 pm Subject: Re: [EM] Dodgson and Kemeny done right? To: fsimm...@pcc.edu Cc: Warren Smith , election-methods fsimm...@pcc.edu wrote: A fourth common problem with Dodgson and Kemeny that I failed to mention is their common lack of efficient precinct summability. Is that true? My implementation of Kemeny uses a variant of integer program #3 from Improved Bounds for Computing Kemeny Rankings, and this integer program only needs access to the graph itself to find the minimum feedback arc set. In voting terms, that means that the integer program only needs the Condorcet matrix to determine who the winner is. In optimization terms, the problem consists of finding the transitive sequence C_1 beats C_2 beats ... beats C_n so that the sum of [C_1 beats C_2] + [C_2 beats C_3] + ... + [C_(n-1) beats C_n] is maximized. (Equivalently, one could phrase it as minimizing the number of upsets - sum of C_k beats C_(k-1).) Thus, unless I'm wrong, the precinct summability is the same as any other Condorcet method, except to the extent that Kemeny is not summable because it doesn't give the winners in polytime. However, Kemeny can't give the winners in worst case polytime, not even if you have the ballots themselves, unless P = NP. Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Dodgson and Kemeny done right?
Borda done right is detailed here:
http://lists.electorama.com/pipermail/election-methods-electorama.com/2011-July/028043.html
Dodgson done right was sketched here:
http://lists.electorama.com/pipermail/election-methods-electorama.com/2011-July/027888.html
The version of Dodgson I was addressing does not attempt to output a social
order like Kemeny does. (I agree with your treatment of Kemeny below.)
The usual version of Dodgson modifies the ballots minimally to create a
pairwise beats-all candidate without saying who came in second, etc.
My versions of {ranked method} done right also include the clone free
conversion of ordinal ballots to cardinal ballots, as detailed most thoroughly
in the first link above.
A fourth common problem with Dodgson and Kemeny that I failed to mention is
their common lack of efficient precinct summability. In my done right
versions this is taken care of by faction
amalgamation which is easy to do with cardinal ballots, but requires two
rounds in the case of ordinal ballots; the first round sums all of the first
place preferences to make this information available for
the clone free conversion of ordinal ballots to cardinal ballots. Then the
second round can also go forward summably by making use of the cardinalized
ballots.
In the case of Dodgson, once the factions have been amalgamated, if there is no
pairwise beats-all candidate, all interested parties can submit modifications
of these faction scores. The minimal
modification that yields a beats-all candidate determines the winner.
- Original Message -
From: Warren Smith
Date: Tuesday, September 13, 2011 5:29 pm
Subject: Dodgson and Kemeny done right?
To: electionscie...@googlegroups.com, election-methods , fsimm...@pcc.edu
Dodgson and Kemeny done right (F.W.Simmons)
-Warren D. Smith, Sept 2011--
Simmons claims he had posted something called Dodgson done right
which gets around the problem that with Dodgson voting it is NP-hard
to find the winner, and supposedly Kemeny has a similar fix.
I failed to find his post, but reading between the lines am attempting
to try to determine what Simmons probably had in mind by reverse
engineering, and/or the fact I had similar thoughts of my own a long
time back.
DODGSON:
votes are rank-orderings of the N candidates.
Output ordering: the one such that the smallest total
candidate motion (distance moved, summed over all
candidates on all ballots) is required to convert the input orders
into the output order.
KEMENY:
votes are rank-orderings of the N candidates.
Output ordering: the one with the fewest total number of
disagreements with the input orders about candidate-pairs
(summed over
all candidate-pairs on all ballots)
ATTEMPT TO REPAIR/REDEFINE:
Make the ballots instead be range-voting style RATINGS ballots.
Output now also is a ratings-style ballot.
For ratingified Dodgson, output is a ballot with
minimum total candidate motion required to convert all
the input ballots into the output ballot (total distance traveled
along the ratings axis, summed over all candidates and all ballots).
For ratingified Kemeny, output is the ballot with
minimum total 2-candidate travel distance summed over all
candidate-pairs on all input ballots, where that candidate pair
has to
travel along the ratings axis so instead of their original directed
separation, they now have the same separation (in same
direction) as
the output ballot.
THEOREM:
Ratingified Dodgson is no longer NP-hard to find, in fact it is easy,
in fact it is just this: each candidate's output score is the median
of his input scores.
PROOF: Easy.
REMARK:
If instead we were minimizing the sum of the SQUARES of the travel
distances, then ratingified Dodgson would just become average-based
range voting, the output score is the mean of that candidate's input
scores.
THEOREM:
Ratingified Kemeny is by this definition the same thing as
ratingified Dodgson.
--
Warren D. Smith
http://RangeVoting.org -- add your endorsement (by clicking
endorse as 1st step)
Election-Methods mailing list - see http://electorama.com/em for list info
[EM] Kemeny challenge
The problems with Kemeny are the same as the problems with Dodgson: (1) computational intractability (2) clone dependence (3) they require completely ordered ballots (no truncations or equal ranking), so they do not readily adapt to Approval ballots, for example. In my posting several weeks ago under the title Dodgson done right I showed how to overcome these three problems. (The same modifications do the trick for both methods.) However, much of the simplicity of the statements of these two methods (Dodgson and Kemeny) gets lost in the translation. Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Enhanced DMC
Very good Chris.
I tried to build a believable profile of ballots that would yield the approval
order and defeats of this
example without success, but I am sure that it is not impossible.
I think in general that if the approval scores are at all valid I would go for
the enhanced DMC winner over
any of the chain building methods we have considered. I think other
considerations over-ride the
importance of being uncovered.
- Original Message -
From: C.Benham
Date: Sunday, September 11, 2011 10:08 am
Subject: Enhanced DMC
To: election-methods-electorama@electorama.com
Cc: Forest W Simmons
Forest Simmons wrote (15 Aug 2011):
Here's a possible scenario:
Suppose that approval order is alphabetical from most approval
to least A, B, C, D.
Suppose further that pairwise defeats are as follows:
CADBA together with BCD .
Then the set P = {A, B} is the set of candidates neither of
which is pairwise
beaten by anybody with greater approval.
Since the approval winner A is not covered by B, it is not
covered by any
member of P, so the enhanced version of DMC elects A.
But A is covered by C so it cannot be elected by any of the
chain building
methods that elect only from the uncovered set.
Forest,
Is the Approval Chain-Building method the same as simply
electing the
most approved uncovered candidate?
I surmise that the set of candidates not pairwise beaten by a
more
approved candidate (your set P, what I've
been referring to as the Definite Majority set) and the
Uncovered set
don't necessarily overlap.
If forced to choose between electing from the Uncovered set and
electing
from the DM set, I tend towards
the latter.
Since Smith//Approval always elects from the DM set, and your
suggested
enhanced DMC (elect the most
approved member of the DM set that isn't covered by another
member)
doesn't necessarily elect from the Uncovered set;
there doesn't seem to be any obvious philosophical case that
enhanced
DMC is better than Smith//Approval.
(Also I would say that an election where those two methods
produce
different winners would be fantastically unlikely.)
A lot of Condorcet methods are promoted as being able to give
the
winner just from the information contained in the
gross pairwise matrix. I think that the same is true of these
methods
if we take a candidate X's highest gross pairwise
score as X's approval score. Can you see any problem with that?
Chris Benham
- Original Message -
From:
Date: Friday, August 12, 2011 3:12 pm
Subject: Enhanced DMC
To: election-methods at lists.electorama.com,
From: C.Benham
To: election-methods-electorama.com at electorama.com
Subject: [EM] Enhanced DMC
Forest,
The D in DMC used to stand for *Definite*.
Yeah, that's what we finally settled on.
I like (and I think I'm happy to endorse) this Condorcet method
idea, and consider it to be clearly better than regular DMC
Could this method give a different winner from the (Approval
Chain Building ?) method you mentioned in the C//A thread
(on 11
June 2011)?
Yes, I'll give an example when I get more time. But for all
practical
purposes they both pick the highest approval Smith candidate.
Here's a possible scenario:
Suppose that approval order is alphabetical from most approval
to least
A, B, C, D.
Suppose further that pairwise defeats are as follows:
CADBA together with BCD .
Then the set P = {A, B} is the set of candidates neither of
which is
pairwise
beaten by anybody with greater approval.
Since the approval winner A is not covered by B, it is not
covered by any
member of P, so the enhanced version of DMC elects A.
But A is covered by C so it cannot be elected by any of the
chain building
methods that elect only from the uncovered set.
Forest Simmons wrote (12 June 2011):
I think the following complete description is simpler than anything
possible for ranked pairs:
1. Next to each candidate name are the bubbles (4) (2) (1). The
voter rates a candidate on a scale from
zero to seven by darkening the bubbles of the digits that add
up to
the desired rating.
2. We say that candidate Y beats candidate Z pairwise iff Y
is rated
above Z on more ballots than not.
3. We say that candidate Y covers candidate X iff Y pairwise beats
every candidate that X pairwise
beats or ties.
[Note that this definition implies that if Y covers X, then Y
beats X
pairwise, since X ties X pairwise.]
Motivational comment: If a method winner X is covered, then the
supporters of the candidate Y that
covers X have a strong argument that Y should have won instead.
Now that we have the basic concepts that we need, and
assuming that
the ballots have been marked
and collected, here's the method of picking the winner:
4. Initialize the variable X with (the name of) the
Re: [EM] Sincere Zero Info Range
One afterthought: Of all the cardinal ratings methods for vvarious values of p, the only one that satisfies the Favorite Betrayal Criterion (FBC) is the case of p=infinity, i.e. where the max absolute rating is limited, or equivalently, the scores are limited to some finite range, i.e. the case with which we are most familiar. - Original Message - From: Date: Friday, September 2, 2011 5:48 pm Subject: Sincere Zero Info Range To: election-methods@lists.electorama.com, Range voting is cardinal ratings with certain constraints on the possible ratings, namely that they have to fall within a certain interval or range of values, and usually limited to whole number values. Ignoring the whole number requirement, we could specify a constraint for an equivalent method by simply limiting the maximum of the absolute values of the ballot scores. Call this method infinity. We could get another (non-equivalent) system by limiting the sum of the absolute values of the scores. Call this method one. Yet another system is obtained by limiting the sum of the squared values of the scores. Call this method two. Other methods are obtained by limiting the sum of the p powers of the absolute values of the scores. In thise scheme method two corresponds to p=2, and methods infinity and one, respectively, are the limits of method p as p approaches infinity or one. Suppose that there are three candidates. Then graphically the constraints for the three respective methods corresponding to p equal to infinity, one, and two, turn out to be a cube, an octahedron, and a ball with a perfectly spherical boundary, respectively.The optimal strategies for methods infinity and one generically involve ballots represented by corners of the cube and octahedron, respectively. In the case of method infinity, this means that all scores on a strategically voted ballot will be at the extremes of the allowed range, i.e. method infinity is strategically equivalent to Approval. In the case of method one, the corners represent the ballots that concentrate the entire max sum value in one candidate, and since negative scores are allowed, this method is strategically equivalent to the method that allows you to vote for one candidate or against one candidate but not both. I don't think anybody has studied this method (Kevin has studied a different method that allows you to vote for one candidate and against another.), but in the case of only three candidates it is the same as Approval. The unit ball for method two has no corners or bulges (which all other values of p involve), so the strategy is not so obvious. But if Samuel Merrill is right, then in the zero information case, the optimum strategy for method two is to vote appropriately normalized sincere utilities. The appropriate normalization is accomplished by subtracting the mean sincere utility from the other utilities, and then dividing all of them by their standard deviation. In practice, the subtraction part is not necessary, because adding the same constant to all of the ratings on the same ballot makes no difference in the final outcome of a cardinal ratings election. Note that this fact is the basis of one way of seeing why methods infinity and one are strategically equivalent in the case of only three candidates. Election-Methods mailing list - see http://electorama.com/em for list info
[EM] Sincere Zero Info Range
Range voting is cardinal ratings with certain constraints on the possible ratings, namely that they have to fall within a certain interval or range of values, and usually limited to whole number values. Ignoring the whole number requirement, we could specify a constraint for an equivalent method by simply limiting the maximum of the absolute values of the ballot scores. Call this method infinity. We could get another (non-equivalent) system by limiting the sum of the absolute values of the scores. Call this method one. Yet another system is obtained by limiting the sum of the squared values of the scores. Call this method two. Other methods are obtained by limiting the sum of the p powers of the absolute values of the scores. In thise scheme method two corresponds to p=2, and methods infinity and one, respectively, are the limits of method p as p approaches infinity or one. Suppose that there are three candidates. Then graphically the constraints for the three respective methods corresponding to p equal to infinity, one, and two, turn out to be a cube, an octahedron, and a ball with a perfectly spherical boundary, respectively. The optimal strategies for methods infinity and one generically involve ballots represented by corners of the cube and octahedron, respectively. In the case of method infinity, this means that all scores on a strategically voted ballot will be at the extremes of the allowed range, i.e. method infinity is strategically equivalent to Approval. In the case of method one, the corners represent the ballots that concentrate the entire max sum value in one candidate, and since negative scores are allowed, this method is strategically equivalent to the method that allows you to vote for one candidate or against one candidate but not both. I don't think anybody has studied this method (Kevin has studied a different method that allows you to vote for one candidate and against another.), but in the case of only three candidates it is the same as Approval. The unit ball for method two has no corners or bulges (which all other values of p involve), so the strategy is not so obvious. But if Samuel Merrill is right, then in the zero information case, the optimum strategy for method two is to vote appropriately normalized sincere utilities. The appropriate normalization is accomplished by subtracting the mean sincere utility from the other utilities, and then dividing all of them by their standard deviation. In practice, the subtraction part is not necessary, because adding the same constant to all of the ratings on the same ballot makes no difference in the final outcome of a cardinal ratings election. Note that this fact is the basis of one way of seeing why methods infinity and one are strategically equivalent in the case of only three candidates. Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] the meaning of a vote (or lack thereof)
An example, due to Samuel Merrill (of Brams, Fishburn, and Merrill fame), simply normalizes the scores on each range ballot the same way that we convert a garden variety normal random variable into a standard one: i.e. on each ballot subtract the mean (of scores on that ballot) and divide by the standard deviation (of scores on that ballot). Once each ballot has been normalized in this way, elect the candidate with the greatest total of normalized scores (over all ballots). Let's call the above version of range voting Merrill's method. As I mentioned before it is strategy free in the zero information case. For the partial or complete info case we can make a double range version (as Warren calls it) using Merrill's method as method X, or more simply ... (1) Have the voters fill out two range ballots. (2) From the first set of range ballots (the potentially strategic ones) extract a candidate A using Merrill's method. (3) Also from the first set, find the Smith set, and the random ballot Smith probabilities. (4) Use the second set of range ballots to decide between the random ballot smith lottery and candidate A. (5) Elect A if more voters prefer A over random ballot Smith than vice versa. (6) Else elect the Smith candidate rated highest on a random ballot (from the first set). This method has the advantage of sincerity on both ballot sets under zero info conditions, and sincerity on the second set under any conditions. Furthermore it always elects from the Smith set when not electing the sincere range winner. It is monotone, clone free, satisfies the Condorcet Criterion, etc. Yes, it relies on chance to a small degree, but doesn't actually pick the winner by chance unless there is no Condorcet winner, and even then only when the expected utility of random Smith is greater than the utility of the (potentially strategic) range winner, which would be rare. Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] the meaning of a vote (or lack thereof)
After Kevin's and Kristopher's comments, which I agree with, I am hesitant to beat a dead horse, but I have two more things for the record that should not be overlooked: First, just as there are deterministic voting methods that elicit sincere ordinal ballots under zero information conditions, there are deterministic methods that elicit precise sincere utilities under zero information conditions. An example, due to Samuel Merrill (of Brams, Fishburn, and Merrill fame), simply normalizes the scores on each range ballot the same way that we convert a garden variety normal random variable into a standard one: i.e. on each ballot subtract the mean (of scores on that ballot) and divide by the standard deviation (of scores on that ballot). Once each ballot has been normalized in this way, elect the candidate with the greatest total of normalized scores (over all ballots). Second, I want to get at the heart of the incommensurability complaint: in most elections some voters will have a much greater stake in the outcome than others. For some it may be a life or death issue; if X is elected your friend's death sentence is commuted, if Y is elected he goes to the chair. Other voters may have only a mild interest in the outcome. How can this problem of incommensurability of stakes be addressed by election methods? Answer: it cannot be addressed by any method that satisfies the basic requirements of neutrality, anonymity, secret ballot, one-person-one-vote, etc. So this failure to provide for stark differences in stakes is not unique to Range. It applies to all decent voting methods. Having said that, Range has an option that is better than most methods that are based on ordinal ballots: give top rating to all candidates that might pardon or commute your friend's death sentence, and give bottom rating to all recent former governors of Texas and their ilk. - Original Message - From: Kristofer Munsterhjelm Date: Thursday, August 25, 2011 7:38 am Subject: Re: [EM] the meaning of a vote (or lack thereof) To: fsimm...@pcc.edu Cc: election-methods@lists.electorama.com fsimm...@pcc.edu wrote: Here's a link to Jobst's definitive posting on individual and social utility: http://lists.electorama.com/htdig.cgi/election-methods- electorama.com/2007-February/019631.html Also, I would like to make another comment in support of Warren's thesis that cardinal range scores are as meaningful or more so than ordinal rankings: Consider that Borda is a method based on rankings. Do the rankings in Borda have the same meaning to the voter as the rankings in IRV do? From Arrow's point of view they do; the ballots are identical in format, and in either case (for a sincere vote) you simply rank A ahead of B if you prefer A over B. But now let's compare Borda with Range; Suppose that there are ten candidates and that the Range ballots ask you to rate them on a scale of zero to nine. On the Borda ballot you are asked to rank them from one to 10. Borda elects the candidate with the highest average rank (i.e. the lowest average rank number). Range elects the candidate with the highest average range score. Now, tell me why Arrow worries about the supposed incommensurable ratings on a scale of zero to 9, but sees no problem with the one to ten ranking scale? Doesn't that confuse the meaning of ranking (versus rating) in itself with the meaning of ranking, as interpreted by the system? I could make a ranked ballot system like IRV that would produce non-monotone results given the ranked ballots that are input to it -- but I could also make a rated ballot system, say the winner is the candidate with the greatest mode, that would also give non-monotone results (since if X is the candidate with greatest mode, rating X higher may lower his mode). Thus, if ratings and rankings are to have meaning, it would seem that this meaning would be independent of the system in question. Otherwise, the meaning would have to be considered with respect to the space of possible voting methods that could use the ballot type in question, and there would be very many outright weird voting methods on both ballot types. If, then, meaning is independent of the method, then Borda's internal workings (where it assigns a score to each ranking) doesn't mean that Borda makes use of a rated ballot, but simply that Borda acts *as if* the ranked ballot is a rated ballot. Because of this, it may produce counterintuitive outcomes (e.g. failing the majority criterion). For that matter, we know that every ranked ballot method can produce a counterintuitive outcome (if we consider determinism, unanimity, non-dictatorship, and IIA intuitive). However, in the independent-of-method point of view, that doesn't make the ranked ballot itself ill-defined. To use an analogy, say you
Re: [EM] the meaning of a vote (or lack thereof)
Here's a link to Jobst's definitive posting on individual and social utility: http://lists.electorama.com/htdig.cgi/election-methods-electorama.com/2007-February/019631.html Also, I would like to make another comment in support of Warren's thesis that cardinal range scores are as meaningful or more so than ordinal rankings: Consider that Borda is a method based on rankings. Do the rankings in Borda have the same meaning to the voter as the rankings in IRV do? From Arrow's point of view they do; the ballots are identical in format, and in either case (for a sincere vote) you simply rank A ahead of B if you prefer A over B. But now let's compare Borda with Range; Suppose that there are ten candidates and that the Range ballots ask you to rate them on a scale of zero to nine. On the Borda ballot you are asked to rank them from one to 10. Borda elects the candidate with the highest average rank (i.e. the lowest average rank number). Range elects the candidate with the highest average range score. Now, tell me why Arrow worries about the supposed incommensurable ratings on a scale of zero to 9, but sees no problem with the one to ten ranking scale? Note that in this case a scoring challenged voter could rank the candidates, and then subtract their respective ranks from 10 to get evenly spaced range scores on the required scale. Thus 1 , 2, 3, 4, ... 9, 10 transform to 9, 8, 7, 6, ... 1, 0, respectively. [When Borda is counted, this transformation is part of the counting process; Borda elects the candidate with the largest Borda score.] If the scoring challenged voter doesn't like the evenly spaced aspect, there is nothing she can do about it in the ranking context, but in the range context she can adjust some of the ratings to reflect bigger and smaller gaps in preference. It seems to me that Arrow must want a unique generic meaning that people can relate to independent of the voting system. Perhaps he is right that ordinal information fits that criterion slightly better than cardinal information, but as Warren says, what really matters is the operational meaning. But back to a possible generic meaning of a score or cardinal rating: if you think that candidate X would vote like you on a random issue with probability p percent, then you could give candidate X a score that is p percent of the way between the lowest and highest possible range values. Note that this meaning is commensurable across the electorate. Furthermore, with regard to commensurability of range scores, think of the example that Warren gave in which the optimum strategy is sincere range strategy; in that example it makes no difference (except for ease of counting) whether or not each voter uses a different range; some could use zero to 100, some negative 64 to positive 64, etc. A ballot will distinguish among the two finalist lotteries in the same way after any affine transformation of the scores. A few years ago Jobst gave a rather definitive discussion of this issue. His investigation led to the result that ideally the scores should allow infinitesimals of various orders along with the standard real values that we are used to. Jobst is skeptical about generic objective meaning for utilities, but in the context of voting, especially lottery methods, he can give you a precise objective meaning of the scores. For example, if you have a choice between alternative X or a coin toss to decide between Y and Z, and you don't care one whit whether or not X is chosen or the the coin toss decides between Y and Z, then (for you)objectively X has a utility value half way between Y and Z. A sequence of questions of this nature can help you rationally assign scores to a set of alternatives. I'll see if I can locate Jobst's results in the archives. Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] the meaning of a vote (or lack thereof)
It seems to me that Arrow must want a unique generic meaning that people can relate to independent of the voting system. Perhaps he is right that ordinal information fits that criterion slightly better than cardinal information, but as Warren says, what really matters is the operational meaning. But back to a possible generic meaning of a score or cardinal rating: if you think that candidate X would vote like you on a random issue with probability p percent, then you could give candidate X a score that is p percent of the way between the lowest and highest possible range values. Note that this meaning is commensurable across the electorate. Furthermore, with regard to commensurability of range scores, think of the example that Warren gave in which the optimum strategy is sincere range strategy; in that example it makes no difference (except for ease of counting) whether or not each voter uses a different range; some could use zero to 100, some negative 64 to positive 64, etc. A ballot will distinguish among the two finalist lotteries in the same way after any affine transformation of the scores. A few years ago Jobst gave a rather definitive discussion of this issue. His investigation led to the result that ideally the scores should allow infinitesimals of various orders along with the standard real values that we are used to. Jobst is skeptical about generic objective meaning for utilities, but in the context of voting, especially lottery methods, he can give you a precise objective meaning of the scores. For example, if you have a choice between alternative X or a coin toss to decide between Y and Z, and you don't care one whit whether or not X is chosen or the the coin toss decides between Y and Z, then (for you)objectively X has a utility value half way between Y and Z. A sequence of questions of this nature can help you rationally assign scores to a set of alternatives. I'll see if I can locate Jobst's results in the archives. Election-Methods mailing list - see http://electorama.com/em for list info
[EM] Voting Reform Statement
The study of voting systems has made significant progress over the last decade, and our understanding is even farther beyond what it was 20 years ago. One important place where that has happened is on the election methods mailing list. This mailing list is likely to include the largest and most diverse group of voting systems theorists in the world. It is a place where opinions vary and debate is vigorous. Thus, we think that the broad, though imperfect, consensus on the following ideas is worth paying attention to. We believe that the voting systems currently used in most of the English-speaking world, including single-round plurality voting (also termed First Past the Post, FPTP) and single-member districts (aka seats, ridings, or electorates), represent some of the worst voting systems known. We believe that reforming these systems would provide important societal benefits, and that there are clearly not corresponding reasons to oppose such reform from the perspective of the public interest. We may disagreeabout which specific reforms might provide the absolutely optimum results, but we can nevertheless agree that there are a number of options which would represent worthwhile improvements. *Single-winner reform* There are various criteria, both formally-defined and informal, by which one can judge a voting system. These criteria can be divided into several classes: 1. Honest-results-oriented criteria. These include such measures as Bayesian regret (that is, simulated societal satisfaction), the majority criterion,and the Condorcet criterion, which focus on whether the correct candidate, according to some definition, is elected. Although these criteria in some cases can favor different candidates as being correct, in most practicalcases they agree. 2. Strategy-resistance criteria. Voting is a complex process, and inevitably there are some cases where some group could get an advantage by changingtheir votes. It is desirable to keep such cases to a minimum. For one thing, it's fairer not to reward such strategic voting behavior. But it's not just that. Perhaps more importantly, a voting system which gives too much of an incentive to strategic voters, can lead to widespread strategywhich systematically distorts the results. 3. Process-oriented criteria. These include such measures as simplicity of the ballot, simplicity of the ballot-counting process, and feasibility of auditing or other fraud-prevention measures. 4. Candidate-incentive criteria. Systems which encourage or discourage clone candidates; give too much power to parties, as opposed to voters; have problems here. These criteria also include less strictly-defined concerns about the type of candidates and campaign strategies a system encourages; for instance, systems which effectively reduce the field to 2 major candidates could encourage negative advertising. There is a broad consensus among researchers plurality voting is among the worst systems for honest results, for strategy-resistance, and for candidate incentives. Honest voting can split votes among similar candidates, spoiling the election and leading opposing candidates to win. Voters respond by strategically choosing the lesser evil among the two major candidates, which can lead to complacent candidates because even corrupt,widely-disliked candidates can win. The system discourages candidates from entering the race, and encourages negative advertising. Although pluralityhas good simplicity and fraud-resistance, this is not enough to recommend its use. A number of proposed single-winner replacements for plurality exist. Although theorists can not find consensus about which of these systems is best, we can agree that many of them are clearly head-and- shoulders above plurality. Systems advanced as as best by some of us, and accepted as good by all of us, include (in categorical and alphabetical order): Put Approval Voting here in alphabetical order, and mention that each of the following methods is a generalization of Approval in a slightly different direction. In other words all of the most highly esteemed methods on the EM list turn out to be generalizations of Approval I know that you made this point in a slightly different way, but it could easily be passed over without registering mentally if we are not careful. - Various *Bucklin* or median-based systems such as *Majority Judgment* - Various *Condorcet* systems, including *Condorcet//Approval, various Condorcet//IRV hybrids, Ranked Pairs, *and* Schulze*. - *Range Voting* (aka Score Voting) - *SODA voting* Notably absent from the above list is IRV (aka Alternative Vote, Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Enhanced DMC
- Original Message -
From:
Date: Friday, August 12, 2011 3:12 pm
Subject: Enhanced DMC
To: election-methods@lists.electorama.com,
From: C.Benham
To: election-methods-electorama@electorama.com
Subject: [EM] Enhanced DMC
Forest,
The D in DMC used to stand for *Definite*.
Yeah, that's what we finally settled on.
I like (and I think I'm happy to endorse) this Condorcet
method
idea,
and consider it to be clearly better than regular DMC
Could this method give a different winner from the (Approval
Chain
Building ?) method you mentioned in the C//A thread (on 11
June 2011)?
Yes, I'll give an example when I get more time
Here's a possible scenario:
Suppose that approval order is alphabetical from most approval to least A, B,
C, D.
Suppose further that pairwise defeats are as follows:
CADBA together with BCD .
Then the set P = {A, B} is the set of candidates neither of which is pairwise
beaten by anybody with greater approval.
Since the approval winner A is not covered by B, it is not covered by any
member of P, so the enhanced version of DMC elects A.
But A is covered by C so it cannot be elected by any of the chain building
methods that elect only from the uncovered set.
Election-Methods mailing list - see http://electorama.com/em for list info
[EM] Enhanced DMC
From: C.Benham To: election-methods-electorama@electorama.com Subject: [EM] Enhanced DMC Forest, The D in DMC used to stand for *Definite*. Yeah, that's what we finally settled on. I like (and I think I'm happy to endorse) this Condorcet method idea, and consider it to be clearly better than regular DMC Could this method give a different winner from the (Approval Chain Building ?) method you mentioned in the C//A thread (on 11 June 2011)? Yes, I'll give an example when I get more time. But for all practical purposes they both pick the highest approval Smith candidate. Initialize a variable X to be the candidate with the most approval. While X is covered, let the new value of X be the highest approval candidate that covers the old X. Elect the final value of X. For all practical purposes this is just a seamless version of C//A, i.e. it avoids the apparent abandonment of Condorcet in favor of Approval after testing for a CW. Assuming cardinal ballots, candidate A covers candidate B, iff whenever B is rated above C on more ballots than not, the same is true for A, and (additionally) A beats (in this same pairwise sense) some candidate that B does not. Your newer suggestion (enhanced DMC) seems to have an easier-to-explain and justify motivation. Chris Benham Forest Simmons wrote (12 July 2011): One of the main approaches to Democratic Majority Choice was through the idea that if X beats Y and also has greater approval than Y, then Y should not win. If we disqualify all that are beaten pairwise by someone with greater approval, then the remaining set P is totally ordered by approval in one direction, and by pairwise defeats in the other direction. DMC solves this quandry by giving pairwise defeat precedence over approval score; the member of P that beats all of the others pairwise is the DMC winner. The trouble with this solution is that the DMC winner is always the member off P with the least approval score. Is there some reasonable way of choosing from P that could potentially elect any of its members? My idea is based on the following observation: There is always at least one member of P, namely the DMC winner, i.e. the lowest approval member of P, that is not covered by any member of P. So why not elect the highest approval member of P that is not covered by any member of P? By this rule, if the approval winner is uncovered it will win. If there are five members of P and the upper two are covered by members of the lower three, but the third one is covered only by candidates outside of P (if any), then this middle member of P is elected. What if this middle member X is covered by some candidate Y outside of P? How would X respond to the complaint of Y, when Y says, I beat you pairwise, as well as everybody that you beat pairwise, so how come you win instead of me? Candidate X can answer, That's all well and good, but I had greater approval than you, and one of my buddies Z from P beat you both pairwise and in approval. If Z beat me in approval, then I beat Z pairwise, and somebody in P covers Z. If you were elected, both Z and the member of P that covers Z would have a much greater case against you than you have against me. Election-Methods mailing list - see http://electorama.com/em for list info
[EM] A variant of DSC
Thanks for the thorough analysis, Chris.
It seems to me that the crux of the matter is the same as the open vs. closed
primaries dilemma.
If you vote sincerely in a closed primary, you may be supporting a candidate
that will not be competitive
in the larger competition. On the other hand an open primary might be
highjacked, among other
problems.
Woodall's DSC is like a closed primary. It narrows down on a clone set, and
the ballots that rank
members of that set highly are the ones that determine the winner.
My variant is more like an open primary; it allows all ballots equal voice in
determining whom to elect
from that clone set.
It seems that this dilemma is one of the most fundamental of our subject.
From: C.Benham
To: election-methods-electorama@electorama.com
Subject: [EM] A variant of DSC
Forest,
Your suggested variant of DSC doesn't address DSC's bad failures
of
Mutual Dominant Third and Minimal Defense.
49: A
48: B
03: CB
The biggest solid coalition is {A}49, so both DSC and your
suggestion
elect A. But MD says not A and MDT says B.
As near as I can tell, my version still has all of the
advantages of DSC, including later-no-harm, clone
independence, monotonicity, etc.
Your version fails Later-no-Harm:
49: A
27: BA
24: CB
It (like DSC) elects A, but if the 49 A voters change to AB
your
version eliminates C and then elects B.
DSC (like DAC, DHSC and SC-DC) meets Participation.
31: ACB
33: BAC
36: C
Your version (like DSC) elects C, but if we add 6 CAB ballots
the
winner changes to A (a failure of both Participation and Mono-
add-Top).
31: ACB
33: BAC
36: C
06: CAB
And then if 2 of those CAB ballots changes to ABC the winner
changes
back to C, failing Mono-raise.
31: ACB
33: BAC
36: C
04: CAB
02: ABC
the only advantage of DSC over DAC is that DAC does not
satisfy later-no-harm.
DSC meets Independence from Irrelevant Ballots, but DAC badly
fails it,
as shown from this old example from
Michael Harman (aka Auros):
03: D
14: A
34: AB
36: CB
13: C
B wins, but if the 3D ballots are removed then C wins.
(Also B is an absurd-looking unjustified winner.)
I regard DSC as FPP elegantly fixed up to meet Clone-Winner and
Majority for Solid Coalitions, but it's shortcomings
help to show that its set of of criterion compliances isn't
sufficient
(and that Participation is 'expensive').
I still think IRV (Alternative Vote, no above-bottom equal-
ranking,
voters can strictly rank from the top as many or few
candidates as they like) is the best of the single-winner
methods that
meets Later-no-Harm.
Chris Benham
Forest Simmons wrote (Sun 7 Aug 2011):
That Q in the previous subject heading was a typo.
Here's an example that illustrates the difference in Woodall's
DSC and my modified version:
25 A1A2
35 A2A1
20 BA1
20 CA1
Woodall's DSC assigns 60 points to {A1, A2} and then the only
other positive point coalitions that have
non-empty intersections with this set are {A2}, {A1}, {A1, B},
and {A1, C}, with respective points of 35,
25, 20 and 20. The 35 point set {A2} decides the result: A2 wins.
In my version, the 60 point coalition is the highest point
proper coalition {A1, A2}, so candidates B and C
are struck from the ballots and we are left with
25 A1A2
35 A2A1
40 A1
This time A1 wins.
As near as I can tell, my version still has all of the
advantages of DSC, including later-no-harm, clone
independence, monotonicity, etc.
Note that Woodall and I get the same result for
25 A1A2
35 A2A1
40 DA1
namely, that A1 wins. But if you split the D faction in half,
you get the original scenario above. It seems
to me that A1 should continue to win, but classical DSC switches
to A2 without any good reason. In
other words, it lacks a certain kind of consistency that our
modified version has.
Jameson,
the only advantage of DSC over DAC is that DAC does not satisfy
later-no-harm. In the context of
chicken this would keep the bluffer from truncating, but to no
avail; the plurality winner (with 48 points)
would win, since (singleton) it would form the highest point
solid coalition all by itself.
Under DAC the bluffer would truncate but would still form an
assenting coalition with the guy who did not
truncate her, but not a solid coalition. An even bigger
assenting coalition would be the plurality winner
together with the bluffer. Of these two, only the bluffer would
be in the second largest coalition, so the
bluffer would win under DAC.
- Original Message -
From:
Date: Saturday, August 6, 2011 3:13 pm
Subject: AQ variant of DSC
To: election-methods at lists.electorama.com
electorama.com,
/ One way of looking at Woodall's DSC method is that it is
// designed to elect from the clone set that
// extends up to the top rank on the greatest number
Re: [EM] A variant of DSC
Last night I realized that my example below shows that my variant of DSC fails
later-no-harm.
Here's an example that illustrates the difference in Woodall's
DSC and my modified version:
25 A1A2
35 A2A1
20 BA1
20 CA1
In my modification of
DSC A1 wins. If the A2 faction truncates A1, then the solid coalition with the
greatest support is just
{A2}, so A2, wins. This means that later does indeed harm A2. Therefore,
like DAC, my modification
of DSC fails later-no-harm.
Does my modification of DAC fail later-no-help?
Election-Methods mailing list - see http://electorama.com/em for list info
[EM] Another method based ratings of zero to three.
I know that Kevin is using four levels (zero through three) to test various methods, so here's an idea: 1. Find the number of votes at each level for each candidate. 2. If any candidates have scores of one on more than fifty percent of the ballots, convert the surplus ones to twos. 3 If the number of candidates with a score of zero on less than fifty percent of the ballots is zero or one, then elect the candidate with the greatest number of positive scores. 4. Else elect (from among the candidates with satisfying the quota of positive scores) the one with the greatest number of twos and threes. 5. If there is a tie after this stage, elect the tied candidate with the greatest number of ballots having a score of three. Election-Methods mailing list - see http://electorama.com/em for list info
[EM] What kind of monotonicity whould we exspect from a PR method?
It seems that if a PR method chose slate {X, Y} for a two winner election, and
only X or Y received
increased support in the rankings or ratings, then {X, Y} should still be
chosen by the method.
But consider the following approval profile (for a two winner election):
3 X
1 XY
2 Y
2 Z
It seems pretty clear that the slate {X, Y} should be elected, and that is the
PAV decision.
Now suppose that X gets additional approval on some ballots but the Y and Z
approvals stay the same:
2 X
3 XY
2 Z
Now PAV elects {X, Z}, and this seems like the right choice, because this slate
completely covers the
electorate, unlike any other pair. Candidate Y has more approvals than Z, but
everybody that approves
Y also approves X, so given that X is part of the slate, Y would only
contribute half a satisfaction point
per ballot, while Z adds a full point per ballot. Since 21.5, Z wins over Y
for the remaining position on
the slate.
This violates the strong monotonicity ideal of the first paragraph, but does
not violate a weaker version
that says if only one candidate X of the winning slate gets additional support
on some ballots (and all
other candidates have the same or less support as before on all ballots) then
that one candidate X
should be a part of the new slate.
Now let's look at this example from the point of view of the Ultimate Lottery:
In the before scenario, the Ultimate Lottery probabilities x, y, and z for the
respective candidates X, Y,
and Z are obtained by maximizing the product
x^3*(x+y)*y^2*z^2 subject to the constraint x+y+z=1
The solution is exactly (x,y,z)=(45%, 30%, 25%) .
After the increase in support for x the Ultimate Lottery probabilities are
obtained by maximizing the
product
x^3*(x+y)^3*z^3 subject to the same constraint x+y+z=1.
The solution is precisely (x, y, z) = (75%, 0, 25%) .
Note that (in keeping with the strong ideal expressed at the beginning of this
message) the only
candidate to increase in probability was X, the one that received increased
support. It did so at the
expense of Y whose probability decreased to zero. So Z passed up Y without any
change in its
probability. That's basically why Z took Y's place on the slate without any
increased support on the
ballots.
So this helps us understand (in the PR election) why the weaker member of the
two winner slate
changes from Y to Z, and why we cannot expect the strong monotone property for
a finite winner PR
election; the discretization in going from the ideal proportion of the Ultimate
Lottery to a finite slate
allows only a crude approximation to the ideal proportion.
In other words, it is just one of the classical apportionment problems in
disguise.
How do other PR methods stack up with regard to monotonicity?
Since IRV is non-montone, automatically STV fails even the weak montonicity
sartisfied by PAV.
How about the other common methods?
Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] : Chicken problem (was: SODA and the Condorcet
To sum up my point of view suppose that the candidates publicly announce the respective preferences (with levels of support shown): 48 A 27 CB 25 B We cannot tell from these ballots alone if B is bluffing or if B really despises A and C equally. If the decision is made only on the basis of these ballots, then the right decision for the case when B is bluffing will be the wrong decision for the case when B is not bluffing, so no method that relies on the ballots alone will solve the problem. But if C is allowed to play before B, and C strongly believes that B is bluffing, then C can bullet. If C is right, then B will approve C also, and C will win. If C is wrong, then A will win. Under actual conditions it is very unlikely that C is going to guess wrongly as to whether or not B is bluffing. SODA allows C to play before B, so the problem is basically solved, as long as B is allowed to approve someone that she did not rank on her ballot, or else as long as there is a very strong incentive for B to rank significant preferences. I've been thinking that perhaps we should allow candidtes to approve candidates that they did not rank ahead of time, as long as they also approve all candidates that they did rank in that case. This would allow a candidate to back down when their bluff was called. Would the candidates then just rank themselves in the pre-election public rankings so that they would have free reign when it came to approval designations? I don't thnk so, because there are other dynamics that make it advantageous for them to commit to ranking their significant preferences ahead of time, especially when there is no chicken standoff, but even in that case as well. Am I misjudging this orI over-looking a worse problem? - Original Message - From: Jameson Quinn Date: Saturday, August 6, 2011 4:04 pm Subject: Re: [EM] : Chicken problem (was: SODA and the Condorcet To: fsimm...@pcc.edu Cc: election-methods@lists.electorama.com 2011/8/6 Jan, IRV elects C like all of the other methods if the B faction doesn't truncate. But IRV elects A when the B faction truncates. Of course, with this knowledge, the B faction isn't likely to truncate, and as you say C will be elected. The trouble with IRV is that in the other scenario when the B faction truncates sincerely because of detesting both A and C, IRV still elects A instead of B. Also, if the A faction votes AB, then B clearly should win, but does not under IRV. So yes, IRV solves the chicken dilemma, but in so doing causes other problems. (This same argument, as it happens, works against tree-based methods.) I still claim that SODA is the only system I know of that can solve the chicken dilemma without over-solving it and making other problems. Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] A variant of DSC
That Q in the previous subject heading was a typo.
Here's an example that illustrates the difference in Woodall's DSC and my
modified version:
25 A1A2
35 A2A1
20 BA1
20 CA1
Woodall's DSC assigns 60 points to {A1, A2} and then the only other positive
point coalitions that have
non-empty intersections with this set are {A2}, {A1}, {A1, B}, and {A1, C},
with respective points of 35,
25, 20 and 20. The 35 point set {A2} decides the result: A2 wins.
In my version, the 60 point coalition is the highest point proper coalition
{A1, A2}, so candidates B and C
are struck from the ballots and we are left with
25 A1A2
35 A2A1
40 A1
This time A1 wins.
As near as I can tell, my version still has all of the advantages of DSC,
including later-no-harm, clone
independence, monotonicity, etc.
Note that Woodall and I get the same result for
25 A1A2
35 A2A1
40 DA1
namely, that A1 wins. But if you split the D faction in half, you get the
original scenario above. It seems
to me that A1 should continue to win, but classical DSC switches to A2 without
any good reason. In
other words, it lacks a certain kind of consistency that our modified version
has.
Jameson,
the only advantage of DSC over DAC is that DAC does not satisfy later-no-harm.
In the context of
chicken this would keep the bluffer from truncating, but to no avail; the
plurality winner (with 48 points)
would win, since (singleton) it would form the highest point solid coalition
all by itself.
Under DAC the bluffer would truncate but would still form an assenting
coalition with the guy who did not
truncate her, but not a solid coalition. An even bigger assenting coalition
would be the plurality winner
together with the bluffer. Of these two, only the bluffer would be in the
second largest coalition, so the
bluffer would win under DAC.
- Original Message -
From:
Date: Saturday, August 6, 2011 3:13 pm
Subject: AQ variant of DSC
To: election-methods@lists.electorama.com,
One way of looking at Woodall's DSC method is that it is
designed to elect from the clone set that
extends up to the top rank on the greatest number of ballots,
i.e. kind of the plurality winner among
clone sets.
There are two ways in which this description is not precise, but
maybe we would get a better method if
we follwed this description more closely.
(1) The solid coalitions look like clone sets on the ballots
that reach up to the top, but they don't have to
look like clone sets on the other ballots.
(2) This description doesn't tell how DSC narrows down after
finding the plurality winner solid coalition.
In fact the entire set of candidates is automatically the solid
coalition that extends to the top rank on
100% of the ballots, so for starter we need to narrow down to a
proper sub-coalition.
With regard to (1), imagine a one dimensional issue space with
the candidates distributed as follows:
A..B1..B2..B3...C..D1..D2...E
The set {B1, B2, B3} and the set {D1, D2} will be solid
coalitions that extend to the top rank on the
ballots of the voters that have a favorite among them, and they
will appear as clone sets on all of the
ballots that do not rank C first. But voters near C may well
intermingle the B's and the D's like
C B3D1B2D2B1EA
This shows that a geometrical clone doesn't have to end up as a
classical ballot clone except on the
ballots of the voters that are situated in the middle of the
clone set, in which case they will appear as
solid (or assenting) coalitions that extend to the top rank.
So Woodal had the right idea for making his method clone independent.
If I uderstand correctly, Woodall invented DSC to prove a point,
viz. that a method can satisfy later no
harm, be clone free, and montone. He didn't invent the method
as a serious proposal. So I don't think
his feelings will be hurt if we suggest an improvement.
My suggestion is that once we have found the proper subset solid
coalition that extends to the top rank
on the greatest number of ballots, strike from the ballots the
candidates that are not in that coalition, and
iterate until there is only one candidate left. Elect the sole
remaining candidate.
For incomplete rankings we can modify DAC in the same way, by
replacing the term solid with the
term assenting.
Election-Methods mailing list - see http://electorama.com/em for list info
[EM] : Chicken problem (was: SODA and the Condorcet
Jan, IRV elects C like all of the other methods if the B faction doesn't truncate. But IRV elects A when the B faction truncates. Of course, with this knowledge, the B faction isn't likely to truncate, and as you say C will be elected. The trouble with IRV is that in the other scenario when the B faction truncates sincerely because of detesting both A and C, IRV still elects A instead of B. Date: Sat, 6 Aug 2011 11:46:12 -0600 From: Jan Kok To: Jameson Quinn , Election Methods Mailing List Subject: Re: [EM] Chicken problem (was: SODA and the Condorcet criterion) To review for other readers, we're talking about the scenario 48 A 27 CB 25 BC Candidates B and C form a clone set that pairwise beats A, and in fact C is the Condorcet Winner, but under many Condorcet methods, as well as for Range and Approval, there is a large temptation for the 25 B faction to threaten to truncate C, and thereby steal the election from C. ?Of course C can counter the threat to truncate B, but then A wins. ?So it is a classical game of chicken. Some methods like IRV cop out by giving the win to A right off the bat, so there is no game of chicken. Wait a minute! IRV elects C in this scenario, if that is how the voters actually vote, and those are the sincere preferences (A voters have no preference between B and C). Much as I hate to say it, IRV works OK in that scenario. On the other hand, if the A voters prefer B over C, (as in the 2009 Burlington, VT mayoral election, http://scorevoting.net/Burlington.html) IRV ignores the preference and still elects C, which seems to be the wrong choice. -- ___ Election-Methods mailing list Election-Methods@lists.electorama.com http://lists.electorama.com/listinfo.cgi/election-methods- electorama.com End of Election-Methods Digest, Vol 86, Issue 18 Election-Methods mailing list - see http://electorama.com/em for list info
[EM] AQ variant of DSC
One way of looking at Woodall's DSC method is that it is designed to elect from
the clone set that
extends up to the top rank on the greatest number of ballots, i.e. kind of the
plurality winner among
clone sets.
There are two ways in which this description is not precise, but maybe we would
get a better method if
we follwed this description more closely.
(1) The solid coalitions look like clone sets on the ballots that reach up to
the top, but they don't have to
look like clone sets on the other ballots.
(2) This description doesn't tell how DSC narrows down after finding the
plurality winner solid coalition.
In fact the entire set of candidates is automatically the solid coalition that
extends to the top rank on
100% of the ballots, so for starter we need to narrow down to a proper
sub-coalition.
With regard to (1), imagine a one dimensional issue space with the candidates
distributed as follows:
A..B1..B2..B3...C..D1..D2...E
The set {B1, B2, B3} and the set {D1, D2} will be solid coalitions that
extend to the top rank on the
ballots of the voters that have a favorite among them, and they will appear as
clone sets on all of the
ballots that do not rank C first. But voters near C may well intermingle the
B's and the D's like
C B3D1B2D2B1EA
This shows that a geometrical clone doesn't have to end up as a classical
ballot clone except on the
ballots of the voters that are situated in the middle of the clone set, in
which case they will appear as
solid (or assenting) coalitions that extend to the top rank.
So Woodal had the right idea for making his method clone independent.
If I uderstand correctly, Woodall invented DSC to prove a point, viz. that a
method can satisfy later no
harm, be clone free, and montone. He didn't invent the method as a serious
proposal. So I don't think
his feelings will be hurt if we suggest an improvement.
My suggestion is that once we have found the proper subset solid coalition that
extends to the top rank
on the greatest number of ballots, strike from the ballots the candidates that
are not in that coalition, and
iterate until there is only one candidate left. Elect the sole remaining
candidate.
For incomplete rankings we can modify DAC in the same way, by replacing the
term solid with the
term assenting.
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[EM] SODA and the Condorcet criterion
Jameson, as you say, it seems that SODA will always elect a candidate that beats every other candidate majority pairwise. If rankings are complete, then all pairwise wins will be by majority. So at least to the degree that rankings are complete, SODA satisfies the Condorcet Criterion. Also, as I mentioned briefly in my last message under this subject heading, SODA seems to completely demolish the chicken problem. To review for other readers, we're talking about the scenario 48 A 27 CB 25 BC Candidates B and C form a clone set that pairwise beats A, and in fact C is the Condorcet Winner, but under many Condorcet methods, as well as for Range and Approval, there is a large temptation for the 25 B faction to threaten to truncate C, and thereby steal the election from C. Of course C can counter the threat to truncate B, but then A wins. So it is a classical game of chicken. Some methods like IRV cop out by giving the win to A right off the bat, so there is no game of chicken. But is there a way of really facing up to the problem? i.e. a way that elects from the majority clone set by somehow diffusing the game of chicken? The problem is that in most methods both factions must decide more or less simultaneously. However, if the decisions can be made sequentially, then the faction that plays first can safely forestall the chicken threat of the other. That's one reason that it makes sense for SODA to have the candidates play sequentially, and to have the strongest candidate of a clone (or near clone) set go before the other candidate or candidates in the clone set. Since DAC is designed to pick out the strongest candidate in the plurality winner clone set, it is a natural for setting the order of play (in the sophisticated version of SODA). Another way to solve the chicken problem is to not allow truncations. But in SODA it seems essential to allow the candidates to truncate. However there is a pressure for the candidates to not truncate too high up in the rankings; if they do, they lose credibility with their supporters, so fewer of them will delegate their approval decisions to them. Since having complete rankings helps both in chicken and with regard to the Condorcet Criterion, it might be worth using the implicit order in the approval ballots of the supporters of candidate X to complete X's rankings by using that implicit order to rank the candidates truncated by X (or otherwise ranked equal by X). This would discourage X from too much truncation, and would make it more likely that the true CW was elected in the (usual?) case where there is one. Forest From: Jameson Quinn To: EM Subject: [EM] SODA and the Condorcet criterion Here's the new text on the SODA page Delegated_Approval#Criteria_Compliancerelatingto the Condorcet criterion: It fails the Condorcet criterion, although the majority Condorcet winner over the ranking- augmented ballots is the unique strong, subgame-perfect equilibrium winner. That is to say that, the method would in fact pass the *majority* Condorcet winner criterion,assuming the following: - *Candidates are honest* in their pre-election rankings. This could be because they are innately unwilling to be dishonest, because they are unable to calculate a useful dishonest strategy, or, most likely, because they fear dishonesty would lose them delegated votes. That is, voters who disagreed with the dishonest rankings might vote approval-style instead of delegating, and voters who perceived the rankings as dishonest might thereby value the candidate less. - *Candidates are rationally strategic* in assigning their delegated vote. Since the assignments are sequential, game theory states that there is always a subgame-perfect Nash equilibrium, which is always unique except in some cases of tied preferences. - *Voters* are able to use the system to *express all relevant preferences*. That is to say, all voters fall into one of two groups: those who agree with their favored candidate's declared preference order and thus can fully express that by delegating their vote; or those who disagree with their favored candidate's preferences, but are aware of who the Condorcet winner is, and are able to use the approval-style ballot to express their preference between the CW and all second-place candidates. Second place means the Smith set if the Condorcet winner were removed from the election; thus, for this assumption to hold, each voter must prefer the CW to all members of this second-place Smith set or vice versa. That's obviously always true if there is a single second-place CW. The three assumptions above would probably not strictly hold true in a real-life election, but they usually would be close enough to ensure that the system does elect the CW. SODA does even better than this if there are only 3 candidates, or if the Condorcet winner goes
Re: [EM] A DSV method inspired by SODA
Of course DSC and DAC are the same when rankings are complete. I was only going to use it to determine the first player, and with amalgamated factions (almost surely) the rankings would be complete. Of course there are many variations of this DSV idea [e.g. we could use chiastic approval to pick the first player], but the main contribution of SODA is the idea of sequential determination of the approval cutoffs. That eliminates the need for mixed (i.e. probabilistic) strategies. In other words, it makes the DSV method deterministic instead of stochastic. I think a deterministic DSV method is easier to sell than a stochastic one, even though personally I would be happy with strategy A applied to the ballots one by one in some random order. In other words, the approval cutoff on the current ballot is next to the current approval winner on the side of the approval runnerup. If there is no CW, then the winner depends on the random order of the ballot processing. The public might have a hard time with that fact. - Original Message - From: Jameson Quinn Date: Thursday, August 4, 2011 7:41 am Subject: Re: [EM] A DSV method inspired by SODA To: fsimm...@pcc.edu Cc: election-methods@lists.electorama.com I suspect that SODA would be Condorcet compliant (over ballots) if the first player was, not the DSC winner, but the DAC winner (re-ordering between each delegated assignment). I'll see if I can work up a proof on this. JQ 2011/7/30 One of the features of SODA is a step where the candidates decide what their approval cutoffs will be.on behalf of themselves and the voters for whom they are acting as proxies. One of the many novel features is that instead of making these decisions simultaneously, the candidates make them sequentially with full knowledge of the decisions of the candidates preceding them in the sequence. I wonder if anybody has ever tried a DSV (designated strategy voting) method based on these ideas. Here's one way it could go: Voters submit range ballots. Factions are amalgamated via weighted averages, so that each candidate ends up with one faction that counts according to its total weight. For large electorates, these faction scores will almost surely yield complete rankings of the candidates. From this point on, only these rankings will be used. The ratings were only needed for the purpose of amalgamating the factions. If we had started with rankings, we could have converted them to ratings via the method of my recent post under the subject Borda Done Right. In either case, once we have the rankings from the amalgamated factions we proceed as follows: Based on these rankings the DSC (descending solid coalitions) winner D is found. The D faction ranking determines the sequential order of play. When it is candidate X's turn in the order of play, X's approval cutoff decision is made automatically as follows: For each of the possible cutoffs, the winner is determined recursively (by running through the rest of the DSV tentatively). The cutoff that yields the best (i.e. highest ranked) candidate according to X's faction's ranking, is the cutoff that is applied to X's faction. After all of the cutoffs have been applied, the approval winner (based on those cutoffs) is elected. It would be too good to be true if this method turned out to be monotone. For that to be true moving up one position in the sequence of play could not hurt the winner. Although I think that this is probably usually true, I don't think that it is always true. Anybody know any different? Election-Methods mailing list - see http://electorama.com/em for list info Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Amalgamation details, hijacking, and free-riding
- Original Message - From: Jameson Quinn Date: Wednesday, August 3, 2011 4:10 pm Subject: Re: Amalgamation details, hijacking, and free-riding To: fsimm...@pcc.edu Cc: election-methods@lists.electorama.com 2011/8/3 So if the true preferences are 20 AB 45 C? 35 (something else), the C supporters could spare 21 voters to vote AC so that the amalgamated factions would become 41 AC 24 C? 35 (something else) . I can see where it is possible for such a move to payoff, but it seems fairly innocuos compared to other strategy problems like burial, compromising, chicken, etc. Not to me. I would be livid to find out my vote had been hijacked. All the other strategies you mention at least use a voter's own vote. Highjacking sounds bad, but it is just one form of over-riding votes. At least it doesn't over-ride your first place preference like the compromising incentive twists your arm to do. Every method eventually over-rides various preferences at some point in the process. Compromising is a form of extortion that blackmails you into expressing a false preference. That's the most egregious form. In other words, compromising forces you to either lie or lose. If somebody else highjacks, they lie to take advantage of you, but with much more risk than the liar who buries to take advantage of the CW supporters. For this kind of highjacking to work, the highjacking faction would have to have more than three times the support of the highjacked faction, as can be seen from the above example (which lacking that much support in the hijacking faction gives an obvious first place advantage to A). That kind of superiority is more than enough to over-ride pairwise wins in ranked pairs, river, beatpath, etc. Election-Methods mailing list - see http://electorama.com/em for list info
[EM] SODA and the Condorcet criterion
I want to thank Jameson for taking the ball and running with it on SODA. I really appreciate his talented and energetic work on elaborating, explaining, and selling the method. It's exciting to me to see the possibilities. Here's more evidence of monotonicity: With a three candidate cycle x ABC y BCA z CAB if xyz, then A plays first, but B wins the election. If the B faction increases at the expense of the x faction so that yxz, then B goes first, and still wins! (because ACB is opposite the cyclic order of the beat cycle) The other nice thing about SODA and strong first play order is that it makes the game of chicken go away. Date: Thu, 4 Aug 2011 08:01:30 -0500 From: Jameson Quinn To: EM Subject: [EM] SODA and the Condorcet criterion Message-ID: Content-Type: text/plain; charset=iso-8859-1 Here's the new text on the SODA page Delegated_Approval#Criteria_Compliancerelatingto the Condorcet criterion: It fails the Condorcet criterion, although the majority Condorcet winner over the ranking- augmented ballots is the unique strong, subgame-perfect equilibrium winner. That is to say that, the method would in fact pass the *majority* Condorcet winner criterion,assuming the following: - *Candidates are honest* in their pre-election rankings. This could be because they are innately unwilling to be dishonest, because they are unable to calculate a useful dishonest strategy, or, most likely, because they fear dishonesty would lose them delegated votes. That is, voters who disagreed with the dishonest rankings might vote approval-style instead of delegating, and voters who perceived the rankings as dishonest might thereby value the candidate less. - *Candidates are rationally strategic* in assigning their delegated vote. Since the assignments are sequential, game theory states that there is always a subgame-perfect Nash equilibrium, which is always unique except in some cases of tied preferences. - *Voters* are able to use the system to *express all relevant preferences*. That is to say, all voters fall into one of two groups: those who agree with their favored candidate's declared preference order and thus can fully express that by delegating their vote; or those who disagree with their favored candidate's preferences, but are aware of who the Condorcet winner is, and are able to use the approval-style ballot to express their preference between the CW and all second-place candidates. Second place means the Smith set if the Condorcet winner were removed from the election; thus, for this assumption to hold, each voter must prefer the CW to all members of this second-place Smith set or vice versa. That's obviously always true if there is a single second-place CW. The three assumptions above would probably not strictly hold true in a real-life election, but they usually would be close enough to ensure that the system does elect the CW. SODA does even better than this if there are only 3 candidates, or if the Condorcet winner goes first in the delegation assignment order, or if there are 4 candidates and the CW goes second. In any of those circumstances,under the assumptions above, it passes the *Condorcet* criterion, not just the majority Condorcet criterion. The important difference between the Condorcet criterion (beats all others pairwise) and the majority Condorcetcriterion (beats all others pairwise by a strict majority) is that the former is clone-proof while the latter is not. Thus, with few enough strong candidates, SODA also passes the independence of clones criterion . Note that, although the circumstances where SODA passes the Condorcet criterion are hemmed in by assumptions, when it does pass, it does so in a perfectly strategy-proof sense. That is *not* true of any actual Condorcetsystem (that is, any system which universally passes the Condorcet criterion). Therefore, for rationally-strategic voters who believe that the above assumptions are likely to hold, *SODA may in fact pass the Condorcetcriterion more often than a Condorcet system*. -- next part -- An HTML attachment was scrubbed... URL: electorama.com/attachments/20110804/d8f85fc2/attachment-0001.htm Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Amalgamation details, hijacking, and free-riding
So if the true preferences are
20 AB
45 C?
35 (something else),
the C supporters could spare 21 voters to vote AC so that the amalgamated
factions would become
41 AC
24 C?
35 (something else) .
I can see where it is possible for such a move to payoff, but it seems fairly
innocuos compared to other
strategy problems like burial, compromising, chicken, etc.
In any case, it can only be a problem in methods that forget the ratings after
the amalgamation and use
only the rankings (like DSC), because when two candidates are rated closely a
small hijacking effort
could tip the balance and reverse the ranking of the two candidates in question.
On the free rider problem of some PR methods, what do you think about the
following?
Because of its free riding problem Plurality is a fairly decent PR method in
a perfect information
setting, as long as voters agree to randomize in order to take advantage of the
free riding effect. For
example in a three winner election where the voter preferences are
60 A1A2
25 B
15 C
If the A supporters agreed to toss coins and vote A! or A2 in the case of heads
or tails, respectively,
then the winning slate would be {A1, A2, B}, the best possible outcome in this
case.
So, in at least one PR method, the free-riding possibilities are essential
for the fairness of the method.
In fact, that is the basic principle of Asset voting (for PR); the candidates
share their assets so that
none will be wasted unnecessarily. Whether the voters or the candidates do the
redistribution doesn't
natter in the perfect info case.
In the zero info case, free-riding doesn't work, so it can neither harm nor
help.
So, I don't worry too much about it.
From: Jameson Quinn
OK, that's what I thought. So, candidate hijacking does not work
for any
amalgamated ballot blind method, that is, a method which
forgets which
rating came from which ballot. However, on a non-ballot-blind system,
including the ranking-based DSC which was the next step in your
SODA-inspired sequential play method, it can work. Basically,
it involves
finding a faction a bit smaller than yours, and ranking its favorite
candidate first. Since your faction is larger, you will be able
to set the
ranking of the remaining candidates, and you will gain the
ballot weight of
the smaller faction. Of course, you must be sure that the false flag
candidate does not win. This is similar to Wodall free riding in PR.
JQ
2011/8/1
To amalgamate factions so that there is at most one faction
per candidate X
(in the context of range
style ballots) take a weighted average of all of the ballots
that give X
top rating, where each ballot has
weight equal to one over the number of candidates rated equal
top on that
ballot. The total weight of the
resulting faction rating vector for candidate X is the sum
of the weights
that that were used for the
weighted average.
Note that these faction rating vectors are efficiently
summable. A running
sum (together with its weight)
is kept for each candidate. Any single ballot is incorporated
by taking a
weighted average of the running
sum and the ballot, where the respective weights are those
mentioned above.
For the running sum it is
the running sum weight. For the ballot it is zero if the
candidate is not
rated top, and 1/k if it is rated top
with (k-1) other candidates..
To combine two running sums for the same candidate take a
weighted average
of the two using the
running sum weights, and then add these weights together to
get the
combined running sum weight.
If you multiply each faction rating vector by its weight and
add up all
such products, you get the vector of
range totals.
Of course Range as a method is summable more efficiently without
amalgamating factions, but other
non-summable methods, when willing to accept amalgamated
factions, thereby
become summable.
So, for example, we can make a summable form of Dodgson:
(1) Use ratings instead of rankings.
(2) amalgamate the factions.
(3) let each candidate (with help from advisors) propose a
modification of
the ballots that will created a
Condorcet Winner.
(4) the CW that is created with the least total modification
is the winner.
Modifications are measured by how much they change the ratings
on how many
ballots.
For example if you change X's rating by .27 on 10 of the 537
ballots of one
faction, and by .32 on 15
ballots from another faction, then the total modification is
2.7 + 4.8 =
7.5
The reason for the competition is that otherwise the method
would be
NP-complete.
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[EM] Record activity on the EM list?
Towards the end of July, I noticed that I had to scroll down a long ways in the archive to get to the most recent messages. I wonder if we set some kind of record. If we were approaching or receding from a major election, it would be more understandable. Maybe all of the feisty guys are getting too tame, so nothing gets censored. Maybe Rob is getting lax on filtering out the stuff that doesn't have Nobel prize potential :) Election-Methods mailing list - see http://electorama.com/em for list info
[EM] Amalgamation details
To amalgamate factions so that there is at most one faction per candidate X (in the context of range style ballots) take a weighted average of all of the ballots that give X top rating, where each ballot has weight equal to one over the number of candidates rated equal top on that ballot. The total weight of the resulting faction rating vector for candidate X is the sum of the weights that that were used for the weighted average. Note that these faction rating vectors are efficiently summable. A running sum (together with its weight) is kept for each candidate. Any single ballot is incorporated by taking a weighted average of the running sum and the ballot, where the respective weights are those mentioned above. For the running sum it is the running sum weight. For the ballot it is zero if the candidate is not rated top, and 1/k if it is rated top with (k-1) other candidates.. To combine two running sums for the same candidate take a weighted average of the two using the running sum weights, and then add these weights together to get the combined running sum weight. If you multiply each faction rating vector by its weight and add up all such products, you get the vector of range totals. Of course Range as a method is summable more efficiently without amalgamating factions, but other non-summable methods, when willing to accept amalgamated factions, thereby become summable. So, for example, we can make a summable form of Dodgson: (1) Use ratings instead of rankings. (2) amalgamate the factions. (3) let each candidate (with help from advisors) propose a modification of the ballots that will created a Condorcet Winner. (4) the CW that is created with the least total modification is the winner. Modifications are measured by how much they change the ratings on how many ballots. For example if you change X's rating by .27 on 10 of the 537 ballots of one faction, and by .32 on 15 ballots from another faction, then the total modification is 2.7 + 4.8 = 7.5 The reason for the competition is that otherwise the method would be NP-complete. Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] A DSV method inspired by SODA
Jameson, for my benefit could you elaborate on what you mean by hijacking strategy, especially in the context of amalgamation of factions. Is ordinary Range susceptible to hijacking? If not, then neither is amalgamation of factions per se, since Range scores are identical with or without amalgamation of factions. Forest - Original Message - From: Jameson Quinn Date: Saturday, July 30, 2011 4:35 pm Subject: Re: [EM] A DSV method inspired by SODA To: fsimm...@pcc.edu Cc: election-methods@lists.electorama.com 2011/7/30 One of the features of SODA is a step where the candidates decide what their approval cutoffs will be.on behalf of themselves and the voters for whom they are acting as proxies. One of the many novel features is that instead of making these decisions simultaneously, the candidates make them sequentially with full knowledge of the decisions of the candidates preceding them in the sequence. I wonder if anybody has ever tried a DSV (designated strategy voting) method based on these ideas. Here's one way it could go: Voters submit range ballots. Factions are amalgamated via weighted averages, so that each candidate ends up with one faction that counts according to its total weight. For large electorates, these faction scores will almost surely yield complete rankings of the candidates. From this point on, only these rankings will be used. The ratings were only needed for the purpose of amalgamating the factions. If we had started with rankings, we could have converted them to ratings via the method of my recent post under the subject Borda Done Right. In either case, once we have the rankings from the amalgamated factions we proceed as follows: Based on these rankings the DSC (descending solid coalitions) winner D is found. The D faction ranking determines the sequential order of play. When it is candidate X's turn in the order of play, X's approval cutoff decision is made automatically as follows: For each of the possible cutoffs, the winner is determined recursively (by running through the rest of the DSV tentatively). The cutoff that yields the best (i.e. highest ranked) candidate according to X's faction's ranking, is the cutoff that is applied to X's faction. After all of the cutoffs have been applied, the approval winner (based on those cutoffs) is elected. It would be too good to be true if this method turned out to be monotone. For that to be true moving up one position in the sequence of play could not hurt the winner. Although I think that this is probably usually true, I don't think that it is always true. Anybody know any different? I'm pretty certain that even if a method like this could be monotone, the amalgamation in the first step breaks it, because of a candidate hijacking strategy. I have no opinion if some other way to do this step would give monotonicity.I'd like to think so, but I wouldn't bet on it. JQ Election-Methods mailing list - see http://electorama.com/em for list info Election-Methods mailing list - see http://electorama.com/em for list info
[EM] Andy's Question
I think that Andy's question about who the PR winners should be in the three
winner (approval) scenario
20 AC
20 AD
20 AE
20 BC
20 BD
20 BE
needs more consideration.
As was pointed out {C, D. E} seems the best, even though PAV would say the
slates
{A,B,C}, {A,B,D}, and {A,B,E} are tied for best.
For those that lean towards {C, D, E}, would you go so far as to say it is the
best solution for the
scenario
40 ABC
40 ABD
40 ABE ?
If not, then how do we decide? If so, then how about
40 CA1A2A3(at 90%)(all others)
40 DA2A3A1(at 90%)(all others)
40 EA3A1A2(at 90%)(all others)
Should {A1, A2, A3} win? or should we continue with {C, D, E} ?
If I understand it, STV would elect {C, D, E}, while RRV (sequential or not)
would elect {A1, A2, A3}.
How would Warren's three district connection solve this problem?
I'm not saying that these scenarios are likely, but I think we need a clearer
idea of what we want in these
extreme cases when we are designing and evaluating practical methods. The
exceptional cases test
the rule, which is the original meaning of the aphorism, The exception proves
the rule.
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[EM] A DSV method inspired by SODA
One of the features of SODA is a step where the candidates decide what their approval cutoffs will be.on behalf of themselves and the voters for whom they are acting as proxies. One of the many novel features is that instead of making these decisions simultaneously, the candidates make them sequentially with full knowledge of the decisions of the candidates preceding them in the sequence. I wonder if anybody has ever tried a DSV (designated strategy voting) method based on these ideas. Here's one way it could go: Voters submit range ballots. Factions are amalgamated via weighted averages, so that each candidate ends up with one faction that counts according to its total weight. For large electorates, these faction scores will almost surely yield complete rankings of the candidates. From this point on, only these rankings will be used. The ratings were only needed for the purpose of amalgamating the factions. If we had started with rankings, we could have converted them to ratings via the method of my recent post under the subject Borda Done Right. In either case, once we have the rankings from the amalgamated factions we proceed as follows: Based on these rankings the DSC (descending solid coalitions) winner D is found. The D faction ranking determines the sequential order of play. When it is candidate X's turn in the order of play, X's approval cutoff decision is made automatically as follows: For each of the possible cutoffs, the winner is determined recursively (by running through the rest of the DSV tentatively). The cutoff that yields the best (i.e. highest ranked) candidate according to X's faction's ranking, is the cutoff that is applied to X's faction. After all of the cutoffs have been applied, the approval winner (based on those cutoffs) is elected. It would be too good to be true if this method turned out to be monotone. For that to be true moving up one position in the sequence of play could not hurt the winner. Although I think that this is probably usually true, I don't think that it is always true. Anybody know any different? Election-Methods mailing list - see http://electorama.com/em for list info
[EM] HBH
In HBH a pecking order is established on the basis of implicit approval or some other monotonic, clone consistent order like chiastic approval that has no incentive for order reversals and minimal incentive for collapsing (i.e. merging) of ratings. A monotone, clone consistent measure of distance or proximity of candidtes is also established. Then, after initializing the variable X as the lowest candidate in the pecking order ... While there remain two or more uneliminated candidates compare X pairwise with the candidate Y with least proximity to X. If Y beats X convincingly, then replace X with Y and discard X. Else discard Y. EndWhile By beats X convincingly I mean either Y covers X, or beats X by a strict majority, or is higher than X on the pecking order AND beats X pairwise. The importance of convincingly is so that this method will satisfy the plurality criterion. Like MMPO the method has a strong center seeking dynamic, and a danger with all such methods is failure of the Plurality criterion, i.e. we don't want a winner to be rated above zero (or ranked above truncation) on fewer ballots than some loser has top ratings or rankings. One possibility for the proximity measure is the sum (over all ballots b) of products of the form b(X)*b(Y), where b(Z) means the rating of Z on on ballot b. Other suitable proximity measures are (1) the average rating of X on all ballots that rate Y tops. (2) the average rating of Y on all ballots that rate X tops. (3) the smaller of (1) and (2). (4) a weighted average of (1) and (2), where the weights are the numbers of the respective ballots of each kind. I call the above suggestions direct or explicit proximity measures, because they reflect a direct interpretation of scores on range style ballots. There are also indirect or implicit measures of proximity. The simplest of these that is clone consistent goes as follows: First average together all of the ballots that rate X tops. Then average together all of the ballots that rate Y tops. Call these average ballots ratings f and g, respectively. The find the sum of all expressions of the form |f(Z)-g(Z)|*p(Z), where p(Z) is the proportion of the original ballots that rated Z at the top. This factor p(Z) is what makes the method clone consistent. Any other clone free lottery distribution p would do just as well. This sum is an implicit or indirect measure of the distance between X and Y. If the sum is zero, then the functions f and g are identical, which means that the supporters of X and Y have (on average) identical ratings for all of the candidates. The measure is indirect because the candidates' distances from each other are judged by how their supporters differ in their ratings of (all of) the candidates. If this measure of distance were used in HBH, I'm afraid that it would destroy the monotonicty of the method, because increasing p(X) makes |f(X)-g(X)| more important in the distance calculation, which in turn, could make X and Y either closer together or further apart in this metric, depending on how this absolute difference compares with the other absolute differences in the calculation. However, this indirect measure could be used in an election post mortem to detect insincere voting. If the implicit distances are not roughly consistent with the explicit proximities used in the method, the you have evidence of insincere ratings for the purpose of manipulating the election results. Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Borda Done Right (with proof of clone consistency and monotonicity)
A modification I am considering:
If all of the candidates are ranked on a ballot, then on that ballot keep the
raw range scores without
normalization, so the lowest ranked candidate Z will be ranked at p(Z) which
may or may not be zero.
But on ballots with one or more truncations do the normalization to ensure that
the truncated candidates
have a final rating of zero.
Also, the method is not summable because the probability distribution values
are computed from the
whole ballot set. In the case of the benchmark lottery distribution, the first
place votes would need to be
tallied before the ranked ballots could be converted to range form. The method
could be carried out in
two separate summable steps .. first the top vote talley, and then, given the
results of that talley, each
precinct could encode the rest of the information in a summable way.
When someone pointed out to Borda that his method led to
strategic order reversals, he replied that he
only intended it for honest voters. Unfortunately, that's only
half the problem; Borda is highly sensitive to
cloning:
Assume honest votes:
80 AB
20 BA
Candidate A wins by Borda and any other decent method.
Now clone B:
80 AB1B2B3B4B5B6
20 B1B2B3B4B5B6A
B1 wins with a Borda score of 5*80+6*20=520 compared with A's
score of 6*80=480 .
Range, which awards the winner to the candidate with the highest
average rating instead of the highest
average ranking, doesn't suffer from this problem, since ratings
are not constrained to spread out like
rankings.
In short, Range is the cardinal ratings analog of Borda, without
the drastic clone problem. There is still
an incentive to exagerate sincere ratings to the extent of
collapsing to the extremes, but not to the
extent of order reversals. Honest voting with Range would give
perfectly satisfactory results, unlike the
case with Borda.
But can we find a Borda Done Right method based on Rankings
instead of ratings?
Yes. We just need a natural way of converting rankings to
ratings that automatically takes clone sets
into account, rating their members near each other.
One way to do that is (for each candidate X) let p(X) be the
percentage of ballots that rank X in first
place. If X is replaced with a clone set {X1, X2, ...} then the
sum p(X1)+p(X2)+ ... will be the same as p
(X) was before the replacement. Furthermore, if X is moved up
in the rankings relative to Y (but no other
relative move) then p(X) will not decrease, and p(Z) will not
increase for any other candidate Z.
These two properties (clone consistency and monotonicity) of the
ballot favorite lottery p are the only
ones needed for the following construction and discussion. So
the result will apply for any other lottery
distribution p that is both clone consistent and monotone.
We do the transformation from rankings to ratings in two steps:
first a conversion to raw ratings, and
then a normalization. Since the normalization will preserve the
monotonicity and clone consistency, we
will concentrate our attention mostly on the raw ratings.
But just for the record, to normalize a raw ratings ballot,
subtract the lowest rating from each of the other
ratings and then divide them all by the highest resulting
rating. For example if (on some ballot) the raw
ratings for the respective candidates are 1, .8, .5, .3,
and .2, first subtract the lowest rating .2 fromo
each of the other numbers to get .8, .6, .3, .1, and 0, and
then divide by the largest of these, namely .8
to get
1, .75, .375, .125, and 0. This is the affine transformation
that normalizes the ratings to a scale of zero
to one.
The more interesting part is the conversion of rankings to raw
range scores by use of the lottery
distribution p. For a given ballot b and an arbitrary candidate
X, the raw score of X is the sum over all Z
ranked (on ballot b) equal to or behind (i.e. lower than) X, of
the values p(Z). In other words the raw
score of X is
p(X)+p(Z1)+p(Z2)+ ... where the sum is over all Z ranked below
or equal to X on ballot b.
The way to visualize this is the candidates (or their names)
stacked up on top of each other with the
highest ranked candidate at the top of the stack, where the
spacing between the candidates Z1 and Z2
is given by the value of p(Z1) where Z1 is the higher of the two
candidates. The total height of the
candidate X in this stack of names is the raw score of X. Since
the probabilities add up to unity, the
candidates ranked equal top will all have raw scores of unity.
Now suppose that X is replaced with a clone set {X1, X2, ...},
then in the new stack of candidates the
clone set will precisely fill up the space p(X)=p(X1)+p(X2)+...
that separated X from the candidate ranked
immediately below X. This is what we mean when we say that the
conversion is clone consistent.
Now suppose that X moves
[EM] Minimal Zero Info Range Strategy
Here's a minimal range strategy that anybody with an adding machine could carry out: First rate all of the candidates sincerely (whatever that means). Then add up all of the scores to get the number S. Divide S by the maxRange value to get a whole number quotient Q and remainder R.less than the divisor maxRange. On your ballot rate your Q most favored candidates at maxRange. Rate the next one in line with a score of R. Rate all of the rest with zero. Election-Methods mailing list - see http://electorama.com/em for list info
[EM] Single Contest Method
Andy's chiastic method is a way of utilizing range ballots that has a much more mild incentive than Range itself to inflate ratings. He locates the method in a class of methods each of which is based on a different increasing function f from the interval [0,1 ] into the same interval: Elect the candidate with the highest fraction q such that at least the fraction f(q) of the ballots rate the candidate at fraction q of the maxRange value (assuming that minRange is zero). Just as the median is more stable than the mean, so also these methods are more resistant to rating inflation. In any case there is no incentive for strategically rating X above Y unless sincere ratings also put X above Y. It seems to me that the incentive for rating inflation is so mild in these methods, that if the rankings induced by the ratings are later used to decide between two finalists, that fact by itself is enough to strongly discourage the extreme inflation or collapsing to the top that optimal range strategy requires. With that in mind, here is my proposal for a Single Contest method: Elect the pairwise winner of the contest between the chiastic winners for the cases where f(q)=q/2, and f(q)=(q+1)/2, respectively. If two candidates have range scores symmetrically distributed about the mean range value, the second winner will be the one with the smaller standard deviation of ratings, i.e. less controversial, while the first one could have about half of its ratings at maxRange and the other half at minRange. In general, if both graphs have approximate symmetry centered at the point (1/2, 1/2) it's about even odds as to which one would win the pairwise contest. I hope that Kevin will put this one in the mix as Chiastic Single Contest. Election-Methods mailing list - see http://electorama.com/em for list info
[EM] Borda Done Right (with proof of clone consistency and monotonicity)
When someone pointed out to Borda that his method led to strategic order
reversals, he replied that he
only intended it for honest voters. Unfortunately, that's only half the
problem; Borda is highly sensitive to
cloning:
Assume honest votes:
80 AB
20 BA
Candidate A wins by Borda and any other decent method.
Now clone B:
80 AB1B2B3B4B5B6
20 B1B2B3B4B5B6A
B1 wins with a Borda score of 5*80+6*20=520 compared with A's score of 6*80=480
.
Range, which awards the winner to the candidate with the highest average rating
instead of the highest
average ranking, doesn't suffer from this problem, since ratings are not
constrained to spread out like
rankings.
In short, Range is the cardinal ratings analog of Borda, without the drastic
clone problem. There is still
an incentive to exagerate sincere ratings to the extent of collapsing to the
extremes, but not to the
extent of order reversals. Honest voting with Range would give perfectly
satisfactory results, unlike the
case with Borda.
But can we find a Borda Done Right method based on Rankings instead of
ratings?
Yes. We just need a natural way of converting rankings to ratings that
automatically takes clone sets
into account, rating their members near each other.
One way to do that is (for each candidate X) let p(X) be the percentage of
ballots that rank X in first
place. If X is replaced with a clone set {X1, X2, ...} then the sum
p(X1)+p(X2)+ ... will be the same as p
(X) was before the replacement. Furthermore, if X is moved up in the rankings
relative to Y (but no other
relative move) then p(X) will not decrease, and p(Z) will not increase for any
other candidate Z.
These two properties (clone consistency and monotonicity) of the ballot
favorite lottery p are the only
ones needed for the following construction and discussion. So the result will
apply for any other lottery
distribution p that is both clone consistent and monotone.
We do the transformation from rankings to ratings in two steps: first a
conversion to raw ratings, and
then a normalization. Since the normalization will preserve the monotonicity
and clone consistency, we
will concentrate our attention mostly on the raw ratings.
But just for the record, to normalize a raw ratings ballot, subtract the lowest
rating from each of the other
ratings and then divide them all by the highest resulting rating. For example
if (on some ballot) the raw
ratings for the respective candidates are 1, .8, .5, .3, and .2, first
subtract the lowest rating .2 fromo
each of the other numbers to get .8, .6, .3, .1, and 0, and then divide by
the largest of these, namely .8
to get
1, .75, .375, .125, and 0. This is the affine transformation that normalizes
the ratings to a scale of zero
to one.
The more interesting part is the conversion of rankings to raw range scores by
use of the lottery
distribution p. For a given ballot b and an arbitrary candidate X, the raw
score of X is the sum over all Z
ranked (on ballot b) equal to or behind (i.e. lower than) X, of the values
p(Z). In other words the raw
score of X is
p(X)+p(Z1)+p(Z2)+ ... where the sum is over all Z ranked below or equal to X on
ballot b.
The way to visualize this is the candidates (or their names) stacked up on top
of each other with the
highest ranked candidate at the top of the stack, where the spacing between the
candidates Z1 and Z2
is given by the value of p(Z1) where Z1 is the higher of the two candidates.
The total height of the
candidate X in this stack of names is the raw score of X. Since the
probabilities add up to unity, the
candidates ranked equal top will all have raw scores of unity.
Now suppose that X is replaced with a clone set {X1, X2, ...}, then in the new
stack of candidates the
clone set will precisely fill up the space p(X)=p(X1)+p(X2)+... that separated
X from the candidate ranked
immediately below X. This is what we mean when we say that the conversion is
clone consistent.
Now suppose that X moves up in the ranking one place by moving X up relative to
the other candidates
on some of the ballots. If the distribution p changes, then p(X) is the only
value that increases.
First let's consider the effect on the ballots where no swap was made: If all
of the candidates that lost
probability are ranked below X, then the raw score of X stays the same, because
whatever is subtracted
from the ones under X is added to the space immediately below X. In this
subcase some of the other
candidates' raw scores decrease, but none increase.
On the other hand if some of the candidates above X lose probability, then X
may well push some of the
other candidates upward in raw score, but only by the same amount that X's raw
score increases at
most. In either of these subcases, no other candidate's total raw range score
(over all such ballots) will
increase more than X's range score increases.
On the ballots where X moves up in the
[EM] Automated Approval Winner
Here's an example of a monotone method for converting ranked ballots into approval ballots automatically: x: ABC y: BCA z: CAB First we convert to range ballots using first place numbers cumulatively: x: A(x+y+z), B(y+z),C(z) y: B(x+y+z), C(x+z), A(x) z: C(x+y+z), A(x+y), B(y) Now we normalize the ratings: x: A(1), B(y/(x+y)), C(0) y: B(1), C(z/(y+z)), A(0) z: C(1), A(x/(x+z)), B(0) For fun we calculate the range totals for the respective candidates: x +z*x/(x+z), y+x*y/(x+y), and z+y*z/(y+z), respectively. These values simplify to x*(x+2*z)/(x+z), y*(y+2*z)/(y+x), and z*(z+2*y)/(z+y) , respectively. The largest of these values determines the range winner. Now we begin the conversion of range to approval. First we find the range total for each ballot: For each of the ballots from the first faction the range total is (x+2*y)/(x+y). For each of the ballots from the next faction the range total is (y+2*z)/(y+z). For each of the ballots from the last faction the range total is (z+2*x)/(z+x). These totals tell how many candidates each of the respective ballots should approve. Since these numbers are all greater than one, each ballot should approve at least one candidate. The remaining fractions y/(x+y), z/(y+z), and x/(z+x), respectively, tell us what fraction of each faction should approve two candidates. So the approvals are x*y/(x+y) : AB, x*x/(x+y): A only y*z/(y+z): BC, y*y/(y+z) : B only x*z/(x+z): CA, z*z/(x+z): C only The respective approval totals are x*(x+2*z)/(x+z), y*(y+2*z)/(y+x), and z*(z+2*y)/(z+y) , which are identical to the respective range totals (in this example of a three faction Condorcet cycle, but not in general). I'll give the proof of monotonicity in another message. Forest Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Automated Approval methods (was Single Contest)
This kind of approach has been experimented with for a long time by Rob LeGrand, and there doesn't seem to be any good way to make it monotone. Here's a very conservative and simple approach that may have some value in some context, if not this one: For each rating ballot b approve the top N candidates where N is the (rounded) sum of the ballot b ratings of all of the candidates divided by the maxRange value Let S be the sum over candidates X of the ballot ratings b(X) . Then N is S divided by maxRange, rounded to the nearest whole number (or rounded to even when exactly halfway between floor and ceiling of S/maxRange). The N highest rated candidates on ballot b are approved. If these approvals are used to elect an approval winner, the method is montone and as clone free as possible for automated approval. (It can split clone sets at the approval boundary on a ballot). Here is a possible heuristic for the method: If the ballot b ratings are normalized (by dividing by maxRange) and taken to represent probabilities, so that b(X) is the probability that candiadte X would correctly represent the ballot b voter on a random question, then the sum S is the expected number of candidates that would agree with this voter on a random question. So why not approve the top S voters, since they are the most likely to be the ones that would agree with the voter? Note that this is a zero information strategy, and for all I know, it could well be zero-info-optimal by some criterion or other. The usual zero info strategy is to assume that all of the candidates are equally likely to win, and to approve above expectation on that basis, but the insertion of lots of clones can radically change those probabilities. This kind of reminds me of the rule that Kristofer suggested for how many winners there should be in a PR election when that number hasn't been decided ahead of time. Date: Sun, 24 Jul 2011 20:01:48 +0100 (BST) From: Kevin Venzke Hi Kristofer, --- En date de?: Dim 24.7.11, Kristofer Munsterhjelm a ?crit?: I also tried implementing the most obvious (I suppose) method: Take the ratings and conduct simulated approval polling, either for some determined or semi-random number of iterations, or until someone wins twice in a row. This doesn't test as well as I thought it would though. What Approval strategy do you use? I always use better than expectation when it is allowed to assume the voters know the method is approval. (Which is just to say that the main sim, when during pure Approval, can't use better than expectation.) I put a tiny amount of average utility of all candidates into the expectation just to try to avoid the situation where your favorite won all the polls so therefore you don't approve him. Kevin Venzke Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] How to make a summable version of STV
Kristopfer. Look at it this way, the process of amalgamating the factions is a low pass filter that gets rid of some fo the noise. So why not consider the resulting ballots as the true ballots, and the associated weights tell how many of them there are of each kinsd. STV can be done with these new true ballots, Droop quotas and all. - Original Message - From: Kristofer Munsterhjelm Date: Saturday, July 23, 2011 1:53 am Subject: Re: [EM] How to make a summable version of STV To: fsimm...@pcc.edu Cc: election-methods@lists.electorama.com fsimm...@pcc.edu wrote: This is to illustrate a point that Warren has recorded on his website somewhere (I don't remember exactly where); namely that lack of summability is not insurmountable. We start with the assumption that the voters have range style ballots on a scale of zero to six. [Seven levels are about optimal according to the psychometrics experts.] I thought five was, not seven. Do you have any papers? At each precinct the ballots are sorted into n piles, one for each candidate. The ballots in each pile are averaged together to get a rating vector for each candidate. [At this first stage if a candidate shares (with k-1 other candiates) top rating on a ballot, then a copy of that ballot is sent to each of those candidate's piles, along with a weight of 1/k .] The precincts send the n candidate vectors, together with their respective totalweights to the counting center. For each candidate a weighted average of the vectors for that candidate from all of the precincts is computed, and the total weight is taken as the size of that candidate's faction. The STV computation is then based on these n almagamated factions. That would fail the Droop proportionality criterion. Just take your favorite example where Range fails it, then stick an universal favorite candidate X in front of every voter's vote. Now, there's only one rating vector - X's - and the averaging will smooth out any structure beyond X. This is an extreme example, but the averaging could hide detail in more realistic ballot sets, too. Election-Methods mailing list - see http://electorama.com/em for list info
[EM] Single Contest
If one of the finalists is chosen by a method that satisfies the majority criterion, then you can skip step one, and the method becomes smoother. Here are some possibilities for the method that satisfies the majority criterion: DSC, Bucklin, and the following range ballot based method: Elect the candidate X with the greatest value of p such that more than p/2 percent of the ballots rate X at least p percent of the maxRange value. That method is similar to the one that Andy Jennings suggested recently, and which I think could be the method to choose the other finalist: Elect the candidate Y with the greatest value of p such that at least p percent of the ballots rate Y at p percent of the maxRange value or higher. If these last two methods are used to choose the finalists, X and Y, then a strict majority top rated candidate will automatically win. The voters don't have to agonize over approval cutoffs, they can just grade the candidates on a scale of zero to maxRange. In fact that's what Andy had in mind ... an approval-like method that sets the cutoff level (in the sense that Bucklin can be thought of as a method for setting the approval cutoff level), but in a more robust way than Bucklin. In addition the composite method is monotone, and at least marginally clone independent (i.e. in the same way that Range is).. Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Single Contest
From: Jameson Quinn To be clear: if X and Y are the same, there's no need for a runoff? That's right. I hope that isn't be too anticlimatic! 2011/7/23 If one of the finalists is chosen by a method that satisfies the majority criterion, then you can skip step one, and the method becomes smoother. Here are some possibilities for the method that satisfies the majority criterion: DSC, Bucklin, and the following range ballot based method: Elect the candidate X with the greatest value of p such that more than p/2 percent of the ballots rate X at least p percent of the maxRange value. That method is similar to the one that Andy Jennings suggested recently, and which I think could be the method to choose the other finalist: Elect the candidate Y with the greatest value of p such that at least p percent of the ballots rate Y at p percent of the maxRange value or higher. If these last two methods are used to choose the finalists, X and Y, then a strict majority top rated candidate will automatically win. The voters don't have to agonize over approval cutoffs, they can just grade the candidates on a scale of zero to maxRange. In fact that's what Andy had in mind ... an approval-like method that sets the cutoff level (in the sense that Bucklin can be thought of as a method for setting the approval cutoff level), but in a more robust way than Bucklin. In addition the composite method is monotone, and at least marginally clone independent (i.e. in the same way that Range is).. Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] HBH
Toby, it is much easier to get a clone independent measure of distance or of proximity with range style ballots than with voter rankings, i.e. cardinal ratings are better than ordinal rankings in this context. Once you have a way of measuring distance (or alternatively proximity) between candidates, then this can be extended by the Hausdorff measure of distance (or proximity) between sets of candidates as follows: Form a pairwiwe distance (or else proximity) matrix whose entry in row X and column Y is the distance (or else proximity) between candidate X from the first set and candidate Y from the second set. To get the Hausdorff distance, form the set m consisting of all of the row and column minima (of the distance matrix). The maximum element of this set is the Hausforff distance between the two sets. To get the Hausdorff proximity, form the set consisting of all the row and column maxima (of the proximity matrix). The smallest element of this set is the Hausdorf proximity. From: Toby Pereira I'm not sure I've followed everything about how to determine the pecking order and how to calculate the distance between candidates, and why it's good to base the challenge order on this, but I'll go along with it! Could we work on a similar distance basis for STV? There are S candidates to be elected, so we could start with the bottom S in the pecking order. For the challenger (single candidate), can distance be measured between a set of candidates and a single candidate? I'm sure it can, so we could calculate the most distant candidate from the current champion set. Then we need to see which of the S+1 candidates is to be eliminated. We now need to determine the order of comparison here as well. Essentially there's two orders of comparison we need to worry about. The first, as already described, is which candidate is to be pitted against the current champion set. Then, once this is determined, we need to determine the order of comparison of the S+1 sets that contain all members of the champion set and the challenger candidate. Each set here will have just one of the S+1 candidates missing so in terms of starting at the bottom of the pecking order, we probably want to start off with the set that excludes the candidate at the top of the pecking order (the top out of the S+1 that are being considered here). Then on the basis that we can probably calculate distance between a set and a candidate, we can also probably calculate the distance between two sets. So we take our champion set containing S of the S+1 candidates, and compare it against the most distant other remaining set that still remains in this group. When only one of these sets remains, this becomes the new champion set, and we find the most distant challenger and continue as before. Also, the winner in the set comparisons can be done as in Schulze STV (or another one if you prefer) as Schulze STV compares sets that differ by only one candidate anyway, so it seems fit for purpose. So, Forest, is this of any potential use? From: Toby Pereira To: fsimm...@pcc.edu Cc: election-methods@lists.electorama.com Sent: Thu, 21 July, 2011 12:19:36 Subject: Re: [EM] HBH Excellent - if you think it's a good idea it must be! I don't think it would be as simple as checking one possible result against one other that differs by one candidate. How would we decide which of the current champion set to remove to put the new candidate in for our comparison? We'd have to check the merits of each possible result. There wouldn't be that many though. So if S candidates are to be elected, the current champion set would obviously have S members. Then we introduce a challenger candidate giving us S+1 candidates. There will then only be S+1 possible sets of S members involving these candidates (one for each candidate's absence). Instead of comparing each of these sets against each other, we can stick to the principle of HBH and its winner-stays-on nature. So when we determine the winning set of S out of the S+1 candidates, we eliminate the candidate not in the winning set. Then we pick another challenger and so on, one at a time, until S candidates remain. These S are elected. As for deciding the order of comparison, I'm not sure I'm as well qualified as you! From: fsimm...@pcc.edu To: Toby Pereira Cc: election-methods@lists.electorama.com Sent: Thu, 21 July, 2011 2:22:03 Subject: Re: [EM] HBH Good idea. Let's play with it. - Original Message - From: Toby Pereira Date: Wednesday, July 20, 2011 4:44 pm Subject: Re: [EM] HBH To: fsimm...@pcc.edu Cc: election-methods@lists.electorama.com I was thinking - Schulze STV compares every result against every other result that
Re: [EM] SODA
I like it! - Original Message - From: Jameson Quinn Date: Thursday, July 21, 2011 4:11 am Subject: Re: [EM] SODA To: fsimm...@pcc.edu Cc: election-methods@lists.electorama.com For generic SODA, the current rule is: candidates exercise their ballots in descending order of current approval score. By current approval score do you mean the non-delegated scores? If so, what do we do when everbody delegates? No, I mean total votes - including non-delegated approval from voters,delegated bullet votes from voters, and assigned delegated votes from other candidates. Yes, these totals increase as the game is played, so descending order is a bit of a misnomer; perhaps I should say the next player is always the candidate with the highest current votes. JQ Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Stats on HBH and a few others
Kevin, Thanks for running these! This is valuable information. From: Kevin Venzke Hi Forest, I ran some small batches of simulations under a handful of scenarios (1D and also aspectral) to try to get a sense of general trends. Then I averaged the numbers. Hopefully I didn't implement anything incorrectly. Definitions: HBH3 and HBH4 are three- and four-slot HBH. ELMDP is the eliminate loser of most distant pair method, 3-slot. ELLDP is the eliminate loser of least distant pair method, 3-slot. Appr and WV are what you'd guess. MinAvt is the Condorcet method that picks the outcome that minimizes the number of voters that could and would avert it. (Average of two versions' scores, but they are quite close) SC is the currently best version of my Single Contest method that I won't define just yet. (If I can still improve it I want to wait.) Finally, MAIRO or Majority Approval//Instant Runoff is an irritating method that tests well but has obvious clone concerns. It's a rank ballot with explicit cutoff. If zero or one candidate has maj approval,the AW wins. Otherwise, take the pairwise comparison between the top two approval candidates. (I wish I had included the Approval-Weighted Pairwise methods as well. They are usually stiff competition.) For percentage of polls won by the candidate who won in the fewest polls (in a given scenario) aka method that comes closest to Random Candidate, the ranking goes: Appr 0.80%, ELLDP 0.47%, HBH3 0.44%, ELMDP 0.34%, HBH4 0.34%, MinAvt 0.23%, SC 0.22%, WV 0.22%, MAIRO 0.05%. Average % of top ratings/rankings of the candidate who had the fewest:Appr 30.5%, ELMDP 24.8%, WV 23.1%, HBH3 20.1%, SC 18.7%, HBH4 17.9%, MinAvt 17.6%, MAIRO 16.7%, ELLDP 15.1%. Voters compromising: ELLDP 7.5%, HBH4 5.6%, MAIRO 4.6%, SC 4.1%, MinAvt 3.7%, HBH3 2.2%, WV 1.5%, ELMDP 0.3%, Appr 0.0%. Voters compressing: Appr 34.3%, ELMDP 19.4%, WV 14.8%, HBH3 6.9%, MinAvt 3.5%, HBH4 3.0%, ELLDP 0.5%, SC 0.04%, MAIRO 0.0%. Voters bullet-voting: Appr 65.7%, HBH3 39.3%, HBH4 29.6%, ELLDP 15.9%, SC 1.2%, MAIRO 0.44%, ELMDP 0.25%, WV 0.19%, MinAvt 0.10%. Voters burying: WV 10.1%, ELLDP 6.3%, ELMDP 5.6%, MinAvt 5.4%, HBH4 5.3%, MAIRO 4.4%, HBH3 2.9%, SC 0.4%, Appr 0.0%. Voters ranking worst first: ELMDP 1.8%, ELLDP 0.3%, SC 0.1%, HBH4 0.05%, HBH3 0.004%, WV MinAvt MAIRO Appr = 0.0%. Overall sincerity: SC 94.2%, MAIRO 90.5%, MinAvt 87.2%, ELMDP 76.3%, WV 73.4%, ELLDP 70.0%, HBH4 56.5%, HBH3 48.7%, Appr N/A. These numbers above are why I am interested in SC and MinAvt... Plurality failures: detected under ELLDP only. Sincere Condorcet efficiency: MAIRO 93.6%, HBH3 92.1%, HBH4 91.0%, SC 89.7%, MinAvt 89.5%, WV 88.8%, ELLDP 88.7%, ELMDP 88.0%, Appr 87.4%. Sincere Condorcet *Loser* efficiency (i.e. a bad thing): ELMDP 1.3%, MinAvt 0.6%, Appr 0.4%, WV 0.4%, HBH3 0.4%, HBH4 0.3%, ELLDP 0.3%, MAIRO and SC = 0.0%. Utility maximizer efficiency: The range was 71.4% to 74.7%. Best to worst: MAIRO, HBH3, HBH4, MinAvt,ELMDP, Appr, ELLDP, SC, WV. Hopefully you or others find this interesting to look over. That's it for now. Kevin Venzke Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Correspondences between PR and lottery methods (was Centrist vs. non-Centrists, etc.)
From: Andy Jennings
On Mon, Jul 18, 2011 at 6:00 PM, wrote:
Andy and I were thinking mostly of Party Lists via RRV. His
question was
that if we used RRV, either
sequential or not, would we get the same result as the
Ultimate Lottery
Maximization. I was able to
show to our satisfaction, that at least in the non-sequential
RRV version,
the results would be the
same. It seems like the initial differences between
sequential and
non-sequential RRV would disappear
in the limit as the number of candidates to be seated
approached infinity.
Would that imply P=NP? In other words, sequential RRV might
be an
efficient method of
approximating a solution (for large n) of non-sequential RRV
(which is
undoubtedly NP hard). What
would be analogous in the Traveling Salesman Problem? Don't
hold your
breath, but it would be
interesting to sort out the analogy, if possible.
I am still hopeful that sequential RRV with a large number of
seats, leaving
each candidate in as if they were their own party, would be a
good and
tractable way to choose legislators and give them each a
different amount of
voting power. I'm hoping it would be possible to calculate the
proportions in the limit as n goes to infinity.
But sequential RRV is completely ignorant about how many seats
need to be
filled, so it's not really going to find the globally optimum N-winner
representative body like ULM and non-sequential RRV aim to do. This
infinite sequential RRV might be good when there is no pre-
determinednumber of seats to fill but instead we want the method
to choose the number
of winners. For real elections, however, I suspect that it will
give some
voting power to every candidate, so maybe it's not that good for
choosing a
representative body.
Good points!
Here's an example, on the other hand, where this method chooses
too few
winners:
10 voters approve A and C
10 voters approve A and D
10 voters approve A and E
10 voters approve B and C
10 voters approve B and D
10 voters approve B and E
If you're choosing two winners, I think the obvious winners are
A and B.
But if you want to choose three winners, I think the obvious
choice is C,
D, and E. Only a method that knows how many winners you're
going to choose
can make the correct decision here. In this case, RRV will
choose A and B.
If A and B are left in (pretending they are parties even if
they are
candidates) then RRV will continue to alternate between A and B.
In the
limit, it will give half of the voting power to A and half to B.
This is
just not helpful if you wanted to choose three winners.
I would like to point out that if the election were single winner, this RRV
result of 50%A+50%B would be
the closest thing to a consensus lottery. In other words, your method of using
RRV with repetition is a
great way of generating proportional lotteries. So it seems that good
lotteries are easier to generate
than good PR results.
ULM and non-sequential RRV evaluate each possible combination of
winners and
can do the right thing in the three winner case.
If I am thinking straight, PAV would give the same max score of
20*(1+1/2)+40
to three slates of three candidates {A, B, C}, {A, B, D}, and {A, B, E}.
If higher resolution range values were available, non-sequential PAV would
probably favor one of these
three.
This reminds me of envy free versus merely fair division. The {C, D, E}
slate would be envy free as
well as fair, even though {A, B, C} would give more total satisfaction, while
still being fair in the sense
that each voter got at least the satisfaction guaranteed by PR. In the case of
{A, B, C} the D and E
voters might well envy the extra satisfaction of the C voters.
Andy
Election-Methods mailing list - see http://electorama.com/em for list info
[EM] SODA
In our SODA development we came to something of an impasse for determining the
order of play for
the candidates casting their approval cutoffs.
Here's a suggestion:
Let the DSC winner go first, because the DSC winner is easily calculated,
satisfies Later-No-Harm (so
does not unduly encourage truncation), and can be thought of as the minimal
acceptable modification of
plurality, namely de-cloning it without destroying its montonicity. In a way,
DSC elegantly
accomplishes what IRV attempts but botches.
[In the context of SODA where there is only one faction for each of the n
candidates, the DSC method
has to score at most n*(n-1) subsets, and it takes no more than the order of
n^2 steps to determine the
DSC score of each of these subsets. So the whole thing can be done in the
order of n^4 steps at worst.]
From then on the next player in the sequence is the candidate that ranked the
previous player X the
highest. If there is a tie, say Y1, Y2, and Y3 each ranks X equally high (and
higher than anybody else
does) then the member of {Y1, Y2, Y3} ranked highest by X is the next player.
This order is clone consistent, i.e. if Y is replaced by a clone set, then the
entire clone set will be
intercalated into the order in place of Y.
This order discourages burial, because if X is first in the order, and Y buries
X, then Y will not follow X,
unless all of the other candidates bury X, too, in which case X could not have
been first.
Note that we could reverse the roles of X and Y in determining the order and
breaking ties: The
remaining candidate Y that X ranks the highest is next, and if X ranks no
remaining candidate, then the
candidate that ranks X the highest is next. My intuition is that this order
might not be quite as burial
resistant, but it would be better at discouraging what we could call fawning,
namely ranking the
presumed DSC winner artificially high for the sole purpose of getting into the
order earlier.
Another option would be to use the DSC winner's rankings for all of the rest of
the players, and passing
to the second player's rankings to resolve any equal rankings made by the DSC
winner, etc.
We need to experiment to see if any of these is adequate, and if so, which is
best.
What are some good scenarios to test?
Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] SODA
Sounds good. - Original Message - From: Jameson Quinn I would like to keep generic SODA as simple as possible, to make it easier to promote for practical use. However, I am still interested in figuring out the best possible SODA+ method, using DSC or whatever. For generic SODA, the current rule is: candidates exercise their ballots in descending order of current approval score. By current approval score do you mean the non-delegated scores? If so, what do we do when everbody delegates? This will correctly get the CW in all 3-candidate scenarios (including non-delegable votes), and I suspect in all 4-candidate scenarios without delegable votes and full ranking. (I can get a very fragile non-CW scenario for 5 candidates, all-delegable votes, and full ranking, in a (1+*2*)v(2) clone scenario, where the CW is one of the starred 2.) Note that, simple as it is, this rule tends to follow clone sets down, as Forest's proposed rules do, because if A delegates to B, then B is almost certain to go next. Thus, this rule is highly clone resistant, for reasonable numbers of clones, although not perfectly clone-proof. It is also clone-proof (ie, IIA) for 1D scenarios. I'd like to make some Yee diagrams for SODA with this rule. Does anybodyknow what algorithms I could use? It would be pretty easy if you assumed that all voters gave a delegable vote; but I think that a more realisticrule would be that voters give a delegable vote iff they agree with their preferred candidate's first delegation, and approves the top max(2 candidates or 1/3 of all candidates) if not. That, plus the delegation order, is getting a bit hairy for calculating Yee diagrams from, so I'd appreciate any tips on algorithmic short cuts. (eg, is there some way to prove that this voting rule in a 2-dimensional space always gives a ballot-CW under SODA, and that therefore I can avoid dealing with delegation order?). If the diagrams can be calculated relatively quickly, I'd be interested in doing this using a real-time web tool (as an excercise in programming in go with GAE); otherwise, I could do it in whatever language, offline. As for Forest's DSC rule for SODA+: how about starting with the DSC winner, then proceeding to the DSC winner among those who got votes from the last player, or the DSC winner among the remainder if the last player did not assign votes? It's in effect similar to the higest ranking from the previous player rule, but closer to the generic SODA rule. JQ ps. I vaguely know how DSC works, but I'd appreciate a refresher. Here's a link: http://wiki.electorama.com/wiki/Descending_Solid_Coalitions It's quite fun to play with. Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] HBH
Good idea. Let's play with it. - Original Message - From: Toby Pereira Date: Wednesday, July 20, 2011 4:44 pm Subject: Re: [EM] HBH To: fsimm...@pcc.edu Cc: election-methods@lists.electorama.com I was thinking - Schulze STV compares every result against every other result that differs by just one candidate, which could be a lot of work for a computer! So could your HBH system be used for STV elections? Determine the order of comparison and compare two results that differ by one candidate and the losing candidate is eliminated. So each pairwise comparison eliminates a candidate and it's all done much more quickly. From: fsimm...@pcc.edu To: fsimm...@pcc.edu Cc: election-methods@lists.electorama.com Sent: Mon, 18 July, 2011 19:25:01 Subject: [EM] HBH HBH stands for Hog Belly Honey, the name of an inerrant nullifier invented by a couple of R.A. Lafferty characters. The HBH is the only known nullifier that can posit moral and ethical judgments, set up and enforce categories, discern and make full philosophical pronouncements, in other words eliminate the garbage and keep what's valuable. The main character, the flat footed genius, Joe Spade, picks the name Hog Belly Honey, for it on account it's so sweet. The whole idea of HBH is just starting at the bottom of a pecking order and pitting (for elimination) the current champ against the most distant challenger. I hope you will keep that in mind as we introduce the necessary technical details. HBH is based on range style ballots that allow the voters to rate each alternative on a range of zero to some maximum value M. [Keep this M in mind; we will make explicit use of it presently.] Once the ballots are voted and submitted, the first order of business is to set up a pecking order for the purpose of resolving ties, etc. Alternative X is higher in the pecking order than alternative Y if alternative X is rated above zero on more ballots than Y is rated above zero. If both have the same number of positive ratings, then the alternative with the most ratings greater than one is higher in the pecking order. If that doesn't resolve the tie, then the alternative with the greatest number of ratings above two is higher, etc. In the practically impossible case that two alternatives have exactly the same number of ratings at each level, ties should be broken randomly. The next order of business is to establish a proximity relation between alternatives. For our purposes closeness or proximity between two alternatives X and Y is given by the number Sum over all ballots b, min( M*(M-1), b(X)*b(Y) ). [The minimization with M*(M-1) clinches the method's resistance to compromise, as explained below.] This proximity value is a useful measure of a certain kind of closeness of the two alternatives: the larger the proximity number the closer the alternatives in this limited sense, while the smaller the number the more distant the alternatives from each other (again, in this limited sense). For the purposes of this method, if two alternatives Y and Z have equal proximity to X, then the one that is higher in the pecking order is considered to be closer than the other. In other words, the pecking order is used to break proximity ties. Next we compute the majority pairwise victories among the alternatives. Alternative X beats alternative Y majority-pairwise if X is rated above Y on more than half of the ballots. For the purposes of this method, the victor of a pair of alternatives is the one that beats the other majority pairwise, or in the case where neither beats the other majority-pairwise it is the one that is higher in the pecking order. Of the two, the non-victor alternative is called the loser. In other words, the pecking order decides pairwise victors and losers when there is no majority defeat. [This convention on victor and loser is what makes the method plurality compliant, as explained below.] Next we initialize an alphanumeric variable V with the name of the lowest alternative in the pecking order, and execute the following loop: While there remain two or more discarded alternatives discard the loser between V and the alternative most distant from V, and replace V with the name of the victor of the two. EndWhile Finally, elect the alternative represented by the final value of V. This HBH method is clone free, monotone, Plurality compliant, compromise resistant, and burial resistant. Furthermore, it is obviously the case that if some alternative beats each of the other alternatives majority pairwise, then that alternative will be elected. Let's see why the method is plurality compliant: If there is
[EM] covering in the context of range style ballots
It recently struck me that in range we can strengthen the covering relation if we include the range levels as virtual candidates: An alternative beats level L pairwise iff it is rated above L on more ballots than it is rated below L. Then for an alternative to cover Y, it has to beat Y pairwise, as well as all of the alternatives that Y beats pairwise, including each virtual candidate (i.e. level) that Y beats pairwise. This makes the relation stronger, so that we should have fewer qualms about using covering to over-ride other kinds of defeats, In particular, if X covers Y in this strong sense, then Y should not win, even if Y is the approval winner (unless the method is just plain vanilla Approval). In particular, I'm thinking of the method that goes like this: Initialize variable X as the highest approval candidate. Then while X is covered (in the strong sense) replace X with the highest (non-virtual) approval candidate that covers X (in the strong sense) EndWhile Elect the final value of X. It would be interesting to see how this changes things, in particular when the approval winner is covered in the old weak sense, but uncovered in this new strong sense. In most realistic cases, the old version is just Smith//Approval. But what should we mean by Smith when we have all of these virtual candidates? If it is included, then the top level virtual candidate automatically pairwise beats (and covers) all of the other candidates, both real and virtual. Also notice that an ordinary CW covers (in the strong sense) the Bucklin winner, iff the CW has its median rating at the same level as the Bucklin winner. Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] HBH
From: Kevin Venzke Hi Forest, --- En date de?: Lun 18.7.11, fsimm...@pcc.edu a ?crit?: The pecking order is the Range order. Assume no ties. I suppose that you could use the range order for the pecking order, but as you mention below that could lead to some strategic distortions. The pecking order I had in mind is more like implicit approval; if implicit approval (number of above zero ratings) has no ties, that's all it is. Oops. I understood this when I read it but didn't retain it. I understand the Plurality argument. Still wrapping my head around the monotonicity and compromise ones. The key to montonicity is that from the first time the eventual winner becomes the defending champ, she has to defeat every challenger, i.e. all of the remaining candidates.? So if a winner W gets barely enough increased support to change the proximity order by just one pair, say W is now closer to V, than W' instead of vice-versa, then W' challenges V before W.? Whether or not W' or the current V weathers this challenge doesn't matter, because W beat the current V before, and then beat W' and all of the other remaining candidates later. If the winner gets lots of increased support, just gradually add that support so that it makes this minimal change in the order, and note that each time this kind of minimal change is made, the winner stays the same.? If we wanted to get fancy, we would call this a homotopy argument. Ok. I'll just trust it for now. If W is the only one whose rating is raised, then the sequence of eliminations is the same as before right up to the point where W entered the fray before (at which point she was the beats all alternative for all of the remaining alternatives). After the change, the order of elimination is different from that point on, but W is still the beats all candidate for the remaining alternatives, so the change in order doesn't matter, no matter where she enters into the fray she is faced only with a subset of the alternatives that she beat before. I was thinking a bit while trying to code it. In the 3-candidate case, if there is a Condorcet winner in the HBH sense then he wins. Otherwisewe have one of two cycles as always. I notice that no matter which cycle it is, the winner is going to be whichever candidate is deemed further from the approval loser. My first thought is that that sounds pretty bad. It sounds like the approval loser is set up to spoil the outcome. On the other hand, in order to have a cycle, everybody must have majority approval, which makes it very strange to think of there being weak noise candidates as such... A candidate lacking majority approval (who isn't the AW) has no effect. So it's hard to imagine appalling outcomes... I'm pretty sure it can overrule a lone majority contest, but it won't even violate minimal defense when it does... Kevin Venzke Kevin, I've been thinking of two other possible proximity measures, both based on the idea of amalgamation of factions, (where each candidate's amalgamateed faction is a weighted average of the factions that rated her tops). One possibility is to say that the most distant alternative from alternative V is the one that rates V the lowest in its amalgamated rating. The other basic possibility is to say that the most distant alternative from V is the one that V rates the lowest in its amalgamated ratings. If one were adopted, the other could be used to break ties. I lean towards the first mentioned possibility because it strongly discourages burial. The only problem I can see with it is that it might encourage top heavy ballots; everybody wants to rate everbody highly to delay being drawn into the fray as long as possible. What do you think? The proximity measure that we have been using up to now is more of a mutual measure; both have to rate each other highly to make the measure judge them to be close. Also it takes into consideration (to some degree) what the other factions think about them. Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Correspondences between PR and lottery methods (was Centrist vs. non-Centrists, etc.)
It sounds like you guys are straightening out the confusion, and exploring some
good ideas.
- Original Message -
From: Toby Pereira
Date: Tuesday, July 19, 2011 7:47 am
Subject: Re: [EM] Correspondences between PR and lottery methods (was Centrist
vs. non-Centrists, etc.)
To: Kristofer Munsterhjelm
Cc: fsimm...@pcc.edu, election-methods@lists.electorama.com
OK, thanks for the information. But what I meant regarding a
result (group of
winners) having a score itself is that this score is just the
total satisfaction
score for a particular result, and then it is this number that
is proportional
to the probability of that set of candidates being elected. So
rather than
looking at each candidate's chances in the lottery individually,
you could look
at whole results and the candidates are elected as one. I was
thinking that this
might be an analogue to random ballot in the single winner case.
From: Kristofer Munsterhjelm
To: Toby Pereira
Cc: fsimm...@pcc.edu; election-methods@lists.electorama.com
Sent: Tue, 19 July, 2011 15:15:15
Subject: Re: [EM] Correspondences between PR and lottery methods
(was Centrist
vs. non-Centrists, etc.)
Toby Pereira wrote:
For proportional range or approval voting, if each result has
a score, you
could make it so that the probability of that result being the
winning result is
proportional to that score. Would that work?
For a lottery derived from PAV or PRV, each winner has a single
score, which is
the probability that the winner would be selected in that
lottery. However, an
entire assembly (group of winners) does not have a single score
as such.
That is, you get an output of the sort that {A: 0.15, B: 0.37,
C: 0.20, D: 0.17,
E: 0.11}, which means that in this lottery, A would win 15% of
the time. It's
relatively easy to turn this into a party list method - if party
A wins 15% of
the time, that just means that party A should get 15% of the
seats. You could
also use it in a system where each candidate has a weight, but
to my knowledge
that isn't done anywhere.
However, if A can only occupy one seat in the assembly, it's
less obvious
whether or not A should win (or how often, if it's a
nondeterministic system) in
a two-winner election. In his reply to my question, Forest gave
some ideas on
how to figure that out.
Also, how is non-sequential RRV done? Forest pointed me to
this a while back -
http://lists.electorama.com/pipermail/election-methods-
electorama.com/2010-May/026425.html
- the bit at the bottom seems the relevant bit. Is that what
we're talking
about?
Very broadly, you have a function that depends on a prospective
assembly (list
of winners) and on the ballots. Then you try every possible
prospective assembly
and you pick the one that gives the best score.
In proportional approval voting, each voter gets one
satisfaction point if one
of the candidates he approved is in the outcome, one plus a half
if two
candidates, one plus a half plus a third if three candidates,
and so on. The
winning assembly composition is the one that maximizes the sum
of satisfaction
points. It's also possible to make a Sainte-Laguë version where
the point
increments are 1, 1/3, 1/5... instead of 1, 1/2, 1/3 etc.
Proportional range voting is based on the idea that you can
consider the
satisfaction function (how many points each voter gets depending
on how many
candidates in the outcome is also approved by him) is a curve
that has f(0) = 0,
f(1) = 1, f(2) = 1/2 and so on. Then you can consider ratings
other than maximum
equal to a fractional approval, so that, for instance, a voter
who rated one
candidate in the outcome at 80%, one at 100%, and another at
30%, would have a
total satisfaction of 1 + 0.8 + 0.3 = 2.1.
All that remains to generalize is then to pick an appropriate
continuous curve,
because the proportional approval voting function is only
defined on integer
number of approvals (1 candidate in the outcome, 2 candidates, 3
candidates).
That's what Forest's post is about.
(Mathematically speaking, the D'Hondt satisfaction function f(x)
is simply the
xth harmonic number. Then one can see that f(x) = integral from
0 to 1 of (1 -
x^n)/(1-x) dx. This can be approximated by a logarithm, or
calculated by use of
the digamma function. Forest gives an integral for the
corresponding
Sainte-Laguë satisfaction function in the post you linked to,
and I give an
expression in terms of the harmonic function in reply:
http://lists.electorama.com/pipermail/election-methods-
electorama.com/2010-May/026437.html
)
Election-Methods mailing list - see http://electorama.com/em for list info
[EM] HBH
HBH stands for Hog Belly Honey, the name of an inerrant nullifier invented by a couple of R.A. Lafferty characters. The HBH is the only known nullifier that can posit moral and ethical judgments, set up and enforce categories, discern and make full philosophical pronouncements, in other words eliminate the garbage and keep what's valuable. The main character, the flat footed genius, Joe Spade, picks the name Hog Belly Honey, for it on account it's so sweet. The whole idea of HBH is just starting at the bottom of a pecking order and pitting (for elimination) the current champ against the most distant challenger. I hope you will keep that in mind as we introduce the necessary technical details. HBH is based on range style ballots that allow the voters to rate each alternative on a range of zero to some maximum value M. [Keep this M in mind; we will make explicit use of it presently.] Once the ballots are voted and submitted, the first order of business is to set up a pecking order for the purpose of resolving ties, etc. Alternative X is higher in the pecking order than alternative Y if alternative X is rated above zero on more ballots than Y is rated above zero. If both have the same number of positive ratings, then the alternative with the most ratings greater than one is higher in the pecking order. If that doesn't resolve the tie, then the alternative with the greatest number of ratings above two is higher, etc. In the practically impossible case that two alternatives have exactly the same number of ratings at each level, ties should be broken randomly. The next order of business is to establish a proximity relation between alternatives. For our purposes closeness or proximity between two alternatives X and Y is given by the number Sum over all ballots b, min( M*(M-1), b(X)*b(Y) ). [The minimization with M*(M-1) clinches the method's resistance to compromise, as explained below.] This proximity value is a useful measure of a certain kind of closeness of the two alternatives: the larger the proximity number the closer the alternatives in this limited sense, while the smaller the number the more distant the alternatives from each other (again, in this limited sense). For the purposes of this method, if two alternatives Y and Z have equal proximity to X, then the one that is higher in the pecking order is considered to be closer than the other. In other words, the pecking order is used to break proximity ties. Next we compute the majority pairwise victories among the alternatives. Alternative X beats alternative Y majority-pairwise if X is rated above Y on more than half of the ballots. For the purposes of this method, the victor of a pair of alternatives is the one that beats the other majority pairwise, or in the case where neither beats the other majority-pairwise it is the one that is higher in the pecking order. Of the two, the non-victor alternative is called the loser. In other words, the pecking order decides pairwise victors and losers when there is no majority defeat. [This convention on victor and loser is what makes the method plurality compliant, as explained below.] Next we initialize an alphanumeric variable V with the name of the lowest alternative in the pecking order, and execute the following loop: While there remain two or more discarded alternatives discard the loser between V and the alternative most distant from V, and replace V with the name of the victor of the two. EndWhile Finally, elect the alternative represented by the final value of V. This HBH method is clone free, monotone, Plurality compliant, compromise resistant, and burial resistant. Furthermore, it is obviously the case that if some alternative beats each of the other alternatives majority pairwise, then that alternative will be elected. Let's see why the method is plurality compliant: If there is even one majority defeat in the sequence of eliminations, every value of V after that will be the name of an alternative that is rated positively on more than half of the ballots. If none of the victories are by majority defeat, then the winner is the alternative highest on the pecking order, i.e. the one with the greatest number of positive ratings. Let's see why the method is monotone: Suppose that the winner is moved up in the ratings. Then its defeat strengths will only be increased, and any proximity change can only delay its introduction into the fray, so it will only face alternatives that lost to it before. Let's see why it is compromise resistant: Since Favorite and Compromise are apt to be in relatively close proximity, and pairwise contests are always between distant alternatives, if Compromise gets eliminated, it will almost certainly be by someone besides Favorite, so there can hardly be any incentive for rating Favorite below Compromise. Furthermore, there is no likely advantage of rating Compromise
Re: [EM] HBH (typo correction)
- HBH stands for Hog Belly Honey, the name of an inerrant nullifier invented by a couple of R.A. Lafferty characters. The HBH is the only known nullifier that can posit moral and ethical judgments, set up and enforce categories, discern and make full philosophical pronouncements, in other words eliminate the garbage and keep what's valuable. The main character, the flat footed genius, Joe Spade, picks the name Hog Belly Honey, for it on account it's so sweet. The whole idea of HBH is just starting at the bottom of a pecking order and pitting (for elimination) the current champ against the most distant challenger. I hope you will keep that in mind as we introduce the necessary technical details. HBH is based on range style ballots that allow the voters to rate each alternative on a range of zero to some maximum value M. [Keep this M in mind; we will make explicit use of it presently.] Once the ballots are voted and submitted, the first order of business is to set up a pecking order for the purpose of resolving ties, etc. Alternative X is higher in the pecking order than alternative Y if alternative X is rated above zero on more ballots than Y is rated above zero. If both have the same number of positive ratings, then the alternative with the most ratings greater than one is higher in the pecking order. If that doesn't resolve the tie, then the alternative with the greatest number of ratings above two is higher, etc. In the practically impossible case that two alternatives have exactly the same number of ratings at each level, ties should be broken randomly. The next order of business is to establish a proximity relation between alternatives. For our purposes closeness or proximity between two alternatives X and Y is given by the number Sum over all ballots b, min( M*(M-1), b(X)*b(Y) ). [The minimization with M*(M-1) clinches the method's resistance to compromise, as explained below.] This proximity value is a useful measure of a certain kind of closeness of the two alternatives: the larger the proximity number the closer the alternatives in this limited sense, while the smaller the number the more distant the alternatives from each other (again, in this limited sense). For the purposes of this method, if two alternatives Y and Z have equal proximity to X, then the one that is higher in the pecking order is considered to be closer than the other. In other words, the pecking order is used to break proximity ties. Next we compute the majority pairwise victories among the alternatives. Alternative X beats alternative Y majority-pairwise if X is rated above Y on more than half of the ballots. For the purposes of this method, the victor of a pair of alternatives is the one that beats the other majority pairwise, or in the case where neither beats the other majority-pairwise it is the one that is higher in the pecking order. Of the two, the non-victor alternative is called the loser. In other words, the pecking order decides pairwise victors and losers when there is no majority defeat. [This convention on victor and loser is what makes the method plurality compliant, as explained below.] Next we initialize an alphanumeric variable V with the name of the lowest alternative in the pecking order, and execute the following loop: While there remain two or more discarded alternatives This should say while there are two or more undiscarded ... discard the loser between V and the alternative most distant from V, and replace V with the name of the victor of the two. EndWhile Finally, elect the alternative represented by the final value of V. This HBH method is clone free, monotone, Plurality compliant, compromise resistant, and burial resistant. Furthermore, it is obviously the case that if some alternative beats each of the other alternatives majority pairwise, then that alternative will be elected. Let's see why the method is plurality compliant: If there is even one majority defeat in the sequence of eliminations, every value of V after that will be the name of an alternative that is rated positively on more than half of the ballots. If none of the victories are by majority defeat, then the winner is the alternative highest on the pecking order, i.e. the one with the greatest number of positive ratings. Let's see why the method is monotone: Suppose that the winner is moved up in the ratings. Then its defeat strengths will only be increased, and any proximity change can only delay its introduction into the fray, so it will only face alternatives that lost to it before. Let's see why it is compromise resistant: Since Favorite and Compromise are apt to be in relatively close proximity, and pairwise contests are always between distant alternatives, if Compromise gets eliminated, it will almost certainly be by someone besides Favorite, so there can hardly be any incentive for rating Favorite below
Re: [EM] HBH
From: Kevin Venzke Hi Forest, So here's my summary using a 4-slot ballot and 3 candidates let's say. The pecking order is the Range order. Assume no ties. I suppose that you could use the range order for the pecking order, but as you mention below that could lead to some strategic distortions. The pecking order I had in mind is more like implicit approval; if implicit approval (number of above zero ratings) has no ties, that's all it is. The proximity between two candidates is the smaller of 6 (3*2), and the product of the two candidates' ratings, summed from each ballot. When finding the nearest candidate to another specific candidate, break ties such that higher pecking order placement means one is closer.(Assuming, I suppose, that a higher-order candidate wants to go later.) Pairwise contests are determined by majority wins if available. Otherwise the pecking order winner wins. You start the method with V = the Range loser generally. Call him C. Find the most distant candidate from C, call him B. Eliminate the loser of the pairwise contest (according to the above paragraph), which leaves you with either B or C to compare to A, who is by default the furthest candidate from whoever is left. Right! I understand the Plurality argument. Still wrapping my head around the monotonicity and compromise ones. The key to montonicity is that from the first time the eventual winner becomes the defending champ, she has to defeat every challenger, i.e. all of the remaining candidates. So if a winner W gets barely enough increased support to change the proximity order by just one pair, say W is now closer to V, than W' instead of vice-versa, then W' challenges V before W. Whether or not W' or the current V weathers this challenge doesn't matter, because W beat the current V before, and then beat W' and all of the other remaining candidates later. If the winner gets lots of increased support, just gradually add that support so that it makes this minimal change in the order, and note that each time this kind of minimal change is made, the winner stays the same. If we wanted to get fancy, we would call this a homotopy argument. I get what the latter mechanic is doing, I just wonder if it can really be that easy? Well, it could be that voter Joe is the only one that thinks that his compromise is a good compromise for his favorite, so that they don't end up with high proximity after all. If he doesn't realize this, he won't have any incentive to compromise. If he does realize it, then he may well have significant incentive to compromise. But most voters are not going to be confused about who the natural compromise candidates are for their favorites, and this agreement will ensure that they have high proximity, i.e. most voters will not find their compromise pitted against their favorite unless they both survive until very late in the game. Strategy-wise I am a bit concerned that your raw Range score is going to be important more often. I'll implement it when I get a moment. Kevin Election-Methods mailing list - see http://electorama.com/em for list info
