Re: [EM] MMPO(IA>MPO) (was IA/MMPO)
Kevin and Jameson, thanks for the insights and suggestions. It's kind of you to suggest my name, Jameson, but I would rather something more descriptive similar to "the potential approval winner set" of Chris and Kevin or more public relations friendly like the Democratically Acceptable Set. My original motivation (that eventually led to IA/MPO as an approximate solution) was to find a candidate most likely to win two approval elections in a row (going into the second election as front runner) without a change in sincere voter preferences, but with an opportunity to adjust their ballot approval cutoffs. Remember when we were looking at DMC from various points of view? One was to think of the DMC winner as the beats all candidate relative to the set P of candidates that were not doubly defeated, i.e. not defeated both pairwise and in approval by some other candidate. In other words each member of P defeats pairwise every candidate with greater approval. The approval winner, the DMC winner, the Smith\\Approval winner, and the lowest approval candidate that covers all higher approval candidates, etc. are some of the members of P. Also, no member of P has greater MPO than IA (assuming we are talking about Implicit Approval in the definition of P). So every member of P has a fair chance at winning MMPO[IA>=MPO], although the winner is not guaranteed to come from P. The main advantage of MMPO[IA>=MPO] over MMPO(P) is that the former satisfies the FBC while the latter does not. For the record, here's why the entire set P survives step one: Let X be a member of P, and let Y be a candidate whose pairwise opposition against X is maximal, i.e. is MPO(X). If IA(Y) is greater than IA(X), then (by definition of P) X beats Y pairwise, and so X is ranked above Y more than MPO(X), the number of ballots on which Y is ranked over X. In other words, in this case X is ranked on more ballots than the number MPO(X), i.e IA(X)>MPO(X). If IA(Y) is no greater than IA(X), then MPO(X) is no greater than IA(X), since MPO(X) is no greater than IA(Y). Since the cases are exhaustive and in neither case is MPO(X) greater than IA(X) we are done. Personally, I still prefer IA-MPO over MMPO[IA>=MPO] because of the superior participation properties, but I recognize the importance of the Majority Criterion in public proposals. Ironically, in reality Approval satisfies the ballot version of the Majority Criterion, while IA-MPO does not, yet in the face of disinformation or other common sources of uncertainty IA-MPO is at least as likely to elect the actual majority favorite as Appoval is. We need Chris to search for the chinks in the armor of these methods. Where are you Chris? Forest On Sun, Oct 13, 2013 at 10:02 AM, Kevin Venzke wrote: > Hi Forest, > > I read your first message: At first glance I think the new method (elect > the MMPO winner among those candidates whose IA>=MPO) is good. It doesn't > seem to gain SFC, which is actually reassuring, that this might be a > substantially different method from others. It seems like it is mainly an > MMPO tweak (since the MMPO winner usually will not be disqualified) with > corrections for Plurality and SDSC/MD. > > Off the top of my head I can't see that anything is happening that would > break FBC. > > > > De : Forest Simmons > >À : Kevin Venzke > >Cc : em > >Envoyé le : Samedi 12 octobre 2013 13h58 > >Objet : Re: MMPO(IA>MPO) (was IA/MMPO) > > > > > >Kevin, > > > >In the first step of the variant method MMPO[IA >= MPO] (which, as the > name suggests, elects the MMPO candidate from among those having at least > as much Implicit Approval as Max Pairwise Opposition) all candidates with > greater MPO than IA are eliminated. > > > >I have already shown that this step does not eliminate the IA winner. > Now I show that this step does not eliminate the Smith\\IA winner either: > > > >Let X be the Smith candidate with max Implicit Approval, IA(X), and let Y > be a candidate that is ranked above X on MPO(X) ballots. There are two > cases to consider (i) Y is also a member of Smith, and (ii) Y is not a > member of Smith. > > > > > >In both cases we have MPO(X) is no greater than IA(Y), because Y is > ranked on every ballot expressing opposition of Y over X. > > > > > >Additionally in the first case IA(Y) is no greater than IA(X) because X > is the Smith\\IA winner. So in this case MPO(X) is no greater than IA(X) > by the transitive property of "no greater than." > > > > > >In the second case, X beats Y pairwise since X is in Smith but Y is not. > This entails that X is ranked above Y on more ballots than Y is ranked > above X. In other words, X is ranked on more ballots than MPO(X). > Therefore IA
Re: [EM] MMPO(IA>MPO) (was IA/MMPO)
Kevin, In the first step of the variant method MMPO[IA >= MPO] (which, as the name suggests, elects the MMPO candidate from among those having at least as much Implicit Approval as Max Pairwise Opposition) all candidates with greater MPO than IA are eliminated. I have already shown that this step does not eliminate the IA winner. Now I show that this step does not eliminate the Smith\\IA winner either: Let X be the Smith candidate with max Implicit Approval, IA(X), and let Y be a candidate that is ranked above X on MPO(X) ballots. There are two cases to consider (i) Y is also a member of Smith, and (ii) Y is not a member of Smith. In both cases we have MPO(X) is no greater than IA(Y), because Y is ranked on every ballot expressing opposition of Y over X. Additionally in the first case IA(Y) is no greater than IA(X) because X is the Smith\\IA winner. So in this case MPO(X) is no greater than IA(X) by the transitive property of "no greater than." In the second case, X beats Y pairwise since X is in Smith but Y is not. This entails that X is ranked above Y on more ballots than Y is ranked above X. In other words, X is ranked on more ballots than MPO(X). Therefore IA(X) > MPO(X), In sum, in neither case is the Smith\\IA winner X eliminated by the first step in the method MMPO[IA>=MPO]. We see as a corollary that step one never eliminates a (ballot) Condorcet Winner. In particular, it does not eliminate a (ballot) majority winner. And since MMPO always elects a ballot majority unshared first place winner when there is one, and MMPO is the second and final step of the method under consideration, this method satisfies the Majority Criterion. Also worth pointing out is this: since step one eliminates neither the IA winner nor the Smith\\IA winner, if there is only one candidate that survives the first step, then the IA winner is a member of Smith, and the method elects this candidate. Also in view of this result, I suggest a strengthening of the Plurality Criterion as a standard required of any method worthy of public proposal. A method (involving rankings or ratings) satisfies the Minimum Ranking Requirement MRR if it never elects a candidate whose max pairwise opposition is greater than the number of ballots on which it is rated above MinRange or (in the case of ordinal ballots) ranked above at least one other candidate. What do you think? Also we need a nice name for the set of candidates that is not eliminated by step one. Any suggestions? Forest Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] IA/MPO
Kevin, Let's call the method that elects the candidate with the max (non-negative) difference between IA and MPO, "IA-MPO." This method satisfies FBC, Plurality, Monotonicity, and the Mono-Add-Top version of Participation. In addition it has the property that if a ballot that does not truncate the current winner is added, the new winner (if any) will be someone not truncated on the added ballot. IA-MPO fails the Majority Criterion, but the following variant satisfies the MC and therefore should be more proposable: 1. Eliminate all candidates that have greater MPO than IA. 2. Elect from among the remaining candidates one with the least MPO (not recalculating MPO's). Here's a short proof that step one does not eliminate the Implicit Approval winner: Let A be the candidate with max IA, and suppose that this max value is IA(A). Let MPO(A) be the max pairwise opposition against A, and let B be a candidate that is ranked above A on MPO(A) ballots. Then B's IA is at least MPO(A), which cannot be greater than the approval winner's IA. Therefore MPO(A) is no greater than IA(A). In other words the Implicit Approval winner is never eliminated by step one above. If I am not mistaken, this MC compliant method still satisfies the FBC, Plurality, etc. and even satisfies Mono-Add-Plump, but not Mom-Add-Top. What do you think? Forest On Fri, Oct 11, 2013 at 6:53 AM, Kevin Venzke wrote: > Hi Forest, > > > > De : Forest Simmons > > > >On Thu, Oct 10, 2013 at 9:23 AM, Kevin Venzke wrote: > > > >Hi Forest, > > > () > Well, if the elimination in step 1 recalculates MPO for step 2, you > probably lose FBC. > > Important reason to not re-calculate the MPO's > Hrm. MDDA's approach (i.e. for satisfying Majority Favorite, and SFC more > broadly) is that if your MPO >.5 then you mostly can't win. MAMPO's > approach is that if your IA is >.5 then only your MPO is considered, not > your IA. I wonder if there are any other options. Both of these approaches > are kind of drastic, and I don't think a method "needs" to completely > satisfy SFC. > > Kevin Venzke > Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] IA/MPO
Kevin, good work! On Thu, Oct 10, 2013 at 9:23 AM, Kevin Venzke wrote: > Hi Forest, > ... > > Unfortunately, I realized that an SFC problem is possibly egregious: > > 51 A>B > 49 C>B > > B would win easily, contrary to SFC (which disallows both B and C). But > more alarmingly it's a majority favorite problem. > So it is non-majoritarian in the same sense that Approval is. In this case the count is too close for approval voters to drop their second preferences, so B will be the Approval winner. Of course with perfect information, they would bullet, and A would win. Philosophically, in this situation I sympathize with electing the candidate broader support (the "consensus candidate") over the mere majority favorite, which is why Approval's failure of (one version of) the Majority Criterion has never bothered me. Jobst and I have gone to a lot of trouble to contrive methods that make B the game theoretic winner in the face of such preferences. I'm sure you remember his challenge to find a method that makes B the perfect information game theoretic winner when utilities are given by (say) 60 A(100), B(70) 40 C(100), B(50) It seems that only lottery methods can solve this challenge in a satisfactory way. We co-authored a paper with the double entendre title of "Some Chances for Consensus" on this topic for the benefit of people who take the "tyranny of the majority" problem seriously. In sum, my non-majoritarian leanings allow me to bid adieu to SFC without too much regret. Now here is a proof of the fact that you observed ... that not all candidates can have greater MPO than IA: Suppose to the contrary that (for some ballot set) every candidate has a greater MPO than IA. Let C(1), C(2), C(3), ... be a sequence of candidates such that C(n+1) gives max opposition to C(n). Since the MPO for C(n) is greater than the IA for C(n), and the IA for C(n+1) is at least as great as the opposition of C(n+1) to C(n), we can conclude that the IA for C(n+1) is greater than the IA for C(n). Therefore,IA increases along the sequence C(1), C(2), C(3), ... so the sequence cannot cycle. therefore there must be an infinite number of candidates. This impossibility shows that the existence of the posited ballot set.was a false assumption. In light of this fact I propose the following variation on our method: 1. Eliminate all candidates that have higher MPO than IA. 2. Elect the remaining candidate with the greatest difference between its IA and its MPO. I like differences better than ratios in this context, but I used ratios in IA/MPO because I worried about people who couldn't easily agree that (25 - 30) > (72 - 90) , for example. But now that we know eliminating all of the negative differences is possible without eliminating all of the candidates, let's switch to differences. Forest Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] IA/MPO
Kevin, thanks for working on the property compliances. I agree that this method does satisfy the FBC, is monotone, and is at least marginally clone independent, like Score and ratings based Bucklin and MMPO. I am not as expert as you in the various defense criteria. My main focus so far is that the method seems to remedy some of the problems of Approval and some of the problems of MMPO. Approval has a problem with this (true preferences) scenario: 30 A 3 A>C 15 C>A 4 C 15 C>B 3 B>C 30 C Of course, (under Approval voting) the two 15 member factions should, and would bullet C, if they were sure of the numbers, but it is more likely that due to disinformation from the A and B parties (and other sources of uncertainty) they would not truncate their second preferences, so A and B would be tied for most approval. However, IA/MPO robustly elects C. Our friend MMPO has a problem with 19 A>B>C 18 B>C>A 18 C>A>B 15 D>A>B 15 D>B>C 15 D>C>A electing the Condorcet Loser D, (unless some preferences are strategically collapsed). But IA/MPO elects the "right winner" A, with no need to collapse preferences among members of the ABC clone set.. Can you think of any other examples where one or the other of IA or MMPO is by itself inadequate? Does IA/MPO always improve the outcome in such cases? My Best, Forest On Tue, Oct 8, 2013 at 9:06 PM, Kevin Venzke wrote: > Hi Forest, > > > > > De : Forest Simmons > >À : EM > >Envoyé le : Mardi 8 octobre 2013 16h59 > >Objet : [EM] IA/MPO > > > >Kevin, > > > >I'm afraid that IA/MPO does fail Plurality: > > > >33 A > >17:A=C > >17:B=C > >33 B > > > >The IA/MPO ratio for both A and B is 50/50 = 1, while the ratio for C is > 34/33, which is greater than 1. > > > >But this is about the worst violation posssible, and it doesn't seem too > bad to me. > > > >If equal top ranking were not allowed, then Plurality would not be > violated. Or (in other words) the method satisfies a weaker version of > Plurality that says if C is ranked on fewer ballots than X is ranked top > but not equal to) C, then C cannot win. > > > > > >I don't know if that is helpful. > > Actually, we are OK here because Plurality only counts strict first > preferences. This aspect is useful when trying to make proofs about it. In > this particular case, I say that if Plurality disqualifies some candidate X > to due another candidate Y, I know that pairwise opposition to X exceeds > X's approval, so X's score is below 100%. (And the same sentence is true if > you swap in SDSC/MD for Plurality.) Since we know somebody will have >=100% > as a score, X won't win. > > I think the question for methods like this is how far away you can get > from the ideal strategy resembling approval strategy. I feel optimistic > because the role given to MPO is large. In MDDA and MAMPO majority > threshold rules are hard-coded and key to seeing any ranking sensitivity. > They satisfy SFC (basically a weak LNHarm) but I think IA/MPO is awfully > close to satisfying that as well. > > > Basically: > Let a be the approval of candidate X > Let b be the approval of candidate Y and also Y's opposition to X > Let c be the maximum opposition to Y > > Then IA/MPO violates SFC when a/b > b/c and a > b > 0.5 > c. Possible to > do, but it would hardly ever happen, I think. > > Kevin Venzke > > Election-Methods mailing list - see http://electorama.com/em for list info
[EM] IA/MPO
Kevin, I'm afraid that IA/MPO does fail Plurality: 33 A 17:A=C 17:B=C 33 B The IA/MPO ratio for both A and B is 50/50 = 1, while the ratio for C is 34/33, which is greater than 1. But this is about the worst violation posssible, and it doesn't seem too bad to me. If equal top ranking were not allowed, then Plurality would not be violated. Or (in other words) the method satisfies a weaker version of Plurality that says if C is ranked on fewer ballots than X is ranked top but not equal to) C, then C cannot win. I don't know if that is helpful. When i get more time, I'll show you why I think that IA/MPO is a good method when the true preferences are given by something like 30 A 3 A>C 15 C>A 4 C 15 C>B 3 B>C 30 B All of our favorite methods, including IA/MPO, say that C should win. But disinformation when A and B are the two big party candidates, may easily result in voted ballots of 30 A 3 A>C 15 C=A 4 C 15 C=B 3 B>C 30 B Candidate C still wins under IA/MPO, even though this is a violation of Plurality. Message: 1 > Date: Sun, 6 Oct 2013 15:21:35 -0700 > From: Forest Simmons > To: EM > Subject: [EM] Try this method on your favorite election scenario > Message-ID: > < > cap29onfvlcopwcx7c0i-hqz_2qujtc332cjtyux2yhurhnr...@mail.gmail.com> > Content-Type: text/plain; charset="iso-8859-1" > > Ballots are ranked or rated. If ranked, then equal ranking and truncation > are allowed. > > Let IA stand for Implicit Approval, which for any candidate X is the number > of ballots on which X is ranked or rated above bottom, i.e. neither > truncated nor rated at zero. > > Let MPO stand for maximum pairwise opposition, which (for candidate X) is > the maximum (as Y varies over the other candidates) of the number of > ballots on which a strict preference of Y over X is indicated. > > The winner of this method (IA/MPO) is the candidate with the highest ratio > of IA to MPO. > > Example > > 45 A>B > 35 B>C > 20 C > > For A IA is 45 and MPO is 55, so IA/MPO is 45/55 or 9/11. > For B IA is 80 and MPO is 45, so IA/MPO is 80/45 or 16/9. > For C IA is 55 and MPO is 80, so IA/MPO is 55/80 or 11/16. > > The IA/MPO winner is B. > > If, instead, the A faction votes 45 A, then the ratios become ... > > For A (the same) 9/11. > For B IA is 35 and MPO is still 45, so the ratio is 7/9. > For C IA is still 55 and MPO is 45, so the ratio is 11/9. > > This time C wins. > -- next part -- > An HTML attachment was scrubbed... > URL: < > http://lists.electorama.com/pipermail/election-methods-electorama.com/attachments/20131006/30dcb66d/attachment.html > > > > -- > > Message: 2 > Date: Mon, 7 Oct 2013 17:51:37 +0100 (BST) > From: Kevin Venzke > To: em > Subject: Re: [EM] Try this method on your favorite election scenario > Message-ID: > <1381164697.61992.yahoomail...@web171501.mail.ir2.yahoo.com> > Content-Type: text/plain; charset=iso-8859-1 > > Hi Forest, > > > > > > De?: Forest Simmons > >??: EM > >Envoy? le : Dimanche 6 octobre 2013 17h21 > >Objet?: [EM] Try this method on your favorite election scenario > > > >Ballots are ranked or rated.? If ranked, then equal ranking and > truncation are allowed. > >? > >Let IA stand for Implicit Approval, which for any candidate X?is the > number of ballots on which?X is?ranked or rated above bottom, i.e. neither > truncated nor rated at zero. > >? > >Let MPO stand for maximum pairwise opposition, which (for candidate X)?is > the maximum (as?Y varies over the other candidates) of the number of > ballots on which a strict preference of Y over?X is indicated. > >? > >The winner of this method (IA/MPO) is the candidate with the highest > ratio of IA to MPO. > >? > >Example > >? > >45 A>B > >35 B>C > >20 C > >? > >For A? IA is 45 and MPO is 55, so IA/MPO is 45/55 or 9/11. > >For B IA is 80 and MPO is 45, so IA/MPO is 80/45 or 16/9. > >For C IA is 55 and MPO is 80, so IA/MPO is 55/80 or 11/16. > >? > >The IA/MPO winner is B. > >? > >If, instead, the A faction votes 45 A, then the ratios become ... > >? > >For A? (the same) 9/11. > >For B? IA is 35 and MPO is still 45, so the ratio is 7/9. > >For C IA is still 55 and MPO is 45, so the ratio is 11/9. > >? > >This time C wins. > > IA/MPO seems like a pretty good method. It seems to be guaranteed that at > least one candidate will have a score >= 100%. That's elegant. With that > assumption it seems easy to demonstrate that the method satisfies Plu
[EM] Try this method on your favorite election scenario
Ballots are ranked or rated. If ranked, then equal ranking and truncation are allowed. Let IA stand for Implicit Approval, which for any candidate X is the number of ballots on which X is ranked or rated above bottom, i.e. neither truncated nor rated at zero. Let MPO stand for maximum pairwise opposition, which (for candidate X) is the maximum (as Y varies over the other candidates) of the number of ballots on which a strict preference of Y over X is indicated. The winner of this method (IA/MPO) is the candidate with the highest ratio of IA to MPO. Example 45 A>B 35 B>C 20 C For A IA is 45 and MPO is 55, so IA/MPO is 45/55 or 9/11. For B IA is 80 and MPO is 45, so IA/MPO is 80/45 or 16/9. For C IA is 55 and MPO is 80, so IA/MPO is 55/80 or 11/16. The IA/MPO winner is B. If, instead, the A faction votes 45 A, then the ratios become ... For A (the same) 9/11. For B IA is 35 and MPO is still 45, so the ratio is 7/9. For C IA is still 55 and MPO is 45, so the ratio is 11/9. This time C wins. Election-Methods mailing list - see http://electorama.com/em for list info
[EM] A safe elimination method
This method requires complete preference ballots: no equal ranking nor truncation allowed. However it could be adapted to other types of ballots including range ballots in various ways. Method: Let N be the number of candidates. On every ballot insert approval cutoffs in the N-1 spaces between the ranks. Find the approval winners for each of the cutoffs. Eliminate every candidate that is not an approval winner for any of the N-1 cutoffs. Find the new ranks on each ballot for the remaining candidates. Repeat the process, until only one candidate remains. I call this method "safe elimination" because, unlike superficial IRV, it sounds the depths of the ranks at every stage of the elimination. I am confident that the Yee diagrams of this method will exhibit none of the pathologies that the Yee diagrams of IRV reveal. And unlike Borda, it does satisfy the Majority Criterion. However, I do not believe this complete rankings version to be clone free or burial resistant, let alone monotonic. So at this stage it is just an interesting idea meant to stimulate the imagination more than anything else. Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] How to find the voters' honest preferences
On Mon, Sep 9, 2013 at 3:34 PM, Kristofer Munsterhjelm wrote: > On 09/09/2013 11:50 PM, Forest Simmons wrote: > >> Kristofer, >> >> Thanks for your insights and considerations regarding the method and my >> questions. >> >> It is true as you noted that there is no game theoretic incentive to >> vote insincerely on the second ballot,(unless this election is >> considered as part of an ongoing game including future elections as >> well), it is also true, as you point out, that there is no incentive to >> vote complete preferences on the second ballot. >> >> You then say ... >> >> "If you just want to find a winner, then an ordinary runoff might work >> as well: select the finalists as above, then have a majority-rule >> election in the second round." >> >> The trouble with this ordinary runoff idea is that the runoff stage >> (potentially) over-rides the strategic pairwise preferences implicit in >> the three slot ballots. In other words it throws out our burial >> disincentive. >> > > Isn't that easily fixed? Consider the runoff part of your method > description: > > > The runoff between them is decided by the voters' pairwise >>>> >>> >>> preferences as expressed on the three slot ballots (when these > >>> finalists are not rated equally thereon), or (otherwise) on the > >>> ordinal ballots when the three slot ballot makes no distinction > >>> between them. > > Couldn't you adapt this to a manual runoff? Something like: > > 1. Determine the virtual runoff candidates X and Y. > 2. If there's a pairwise difference between the two on the three slot > ballots, then whoever it favors, wins. > 3. If there's no difference between the two, then go to a manual tiebreak > runoff. > > Or would that mean runoffs would happen so rarely that the electorate > would fail to get used to them? > It could be done that way, but my intention was to take into account that some three slots ballots might not distinguish the two finalists even though others would. The ones that do distinguish them are used for that purpose, but the ones that do not distinguish them make use of the higher resolution preferences. Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] How to find the voters' honest preferences (Kevin Venzke)
> (Well, both methods have strategy there anyway, since the ballots either > must conform to each other or are interpreted as conforming to each other. > I don't think you stated it was required for the second ballot to not > reverse preferences of the first ballot.) > The second ballot, being non-strategic will perforce sometimes reverse preferences that are implicit in the three slot ballot. However, the second ballot is only used where the first ballot shows no preference. This usage gives no incentive for insincere order (even though it does not always require sincere order). So even though the second ballot will be sinere if the voter has any impulse for honesty, no matter how small, the winner between finalists may not be the true preference of the two, since some of the preferences in that pairwise contest may come from the strategic ballots. That's the price we pay in order to discourage burial. > > > If I understand this correctly then this means that if you rate one > candidate 100%, one candidate 99%, and then a dozen candidates at 0%, you > will approve only the candidate at 100%. Can that be right? > When the last cup fills only part way, you have to round to either zero or 100%, which ever is closer. In the three slot case it would work something like this: Suppose the original cardinal ratings are on a scale of zero to two, but with high resolution: A(2), B(1.9), C(1.6), D(1.3), E(1.2), F(1), G(0.7), H(0.2 ), I(0.1), J(0), etc. Then in B through E where the ratings are strictly between 1 and 2, we add the fractional parts to get .9 +.6 +.3 +.2 = 2. This tells us that exactly two of these four ratings should be rounded up: B(2), C(2),D(1),E)(1). Then in G, H, and I which have ratings strictly between zero and one, the fractional parts add up to one, so exactly one of these gets rounded up. Putting all of this together we get A(2), B(2), C(2), D(1), E(1), F(1), G(1), H(0), I(0), J(0), etc. In this example, the fractional parts that were added together have whole sums. When that does not happen, we round to the nearest whole (with round to even tie breaker), etc as in the following example: Example two: A(2), B(1.3), C(1.25), D(1.2), E(1.15), F(0), G(0), etc. The fractional parts sum to .9, which rounds to one, so the modified ratings are A(2), B(2), C(1), D(1), E(1), F(0), G(0), etc. I'll try to give the rationale for this in my remaining time on this public computer: Think of it this way: suppose I use Bernoulli random variables x1, x2, x3, x4 with expected vales equal to the respective fractional parts .3, .25, .2, and .15 : B(1+x1), C(1+x2), D(1+x3), and E(1+x4). Use a random number generator to get values for x1, x2, x3, and x4. [A Bernoulli random variable takes on a value of one with probability equal to its expectation, and zero otherwise.] If we use these values and all voters do the same kind of thing, the total cardinal ratings will turn out the same statistically in the case of large numbers of ballots. But wait, what if x3 and x4 turn out to be one's, while x1 and x2 turn out to be zeros? Wouldn't it be better to trade those one's into the the preferred positions? Now we ask ourselves, what is the expected number of one's? The answer is the sum of the individual expectations, namely .3 + .25 +.2 +.15 = .9 . Instead of using the Bernoulli experiments, just use this expected number of one's as the appropriate number of one's. We expect approximately one bernoulli variable to be a one. So we round up only the first rating (of those with ratings strictly between one and two). Thanks, Forest > > -- > > ___ > 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 111, Issue 10 > * > Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] How to find the voters' honest preferences
Kristofer, Thanks for your insights and considerations regarding the method and my questions. It is true as you noted that there is no game theoretic incentive to vote insincerely on the second ballot,(unless this election is considered as part of an ongoing game including future elections as well), it is also true, as you point out, that there is no incentive to vote complete preferences on the second ballot. You then say ... "If you just want to find a winner, then an ordinary runoff might work as well: select the finalists as above, then have a majority-rule election in the second round." The trouble with this ordinary runoff idea is that the runoff stage (potentially) over-rides the strategic pairwise preferences implicit in the three slot ballots. In other words it throws out our burial disincentive. Concerning the burden of two ballots: In practice, voters that consider their three slot ballots to be sincere could opt out of the second ballot, or they could opt to replicate their favorite's preference order, etc. My Best, Forest On Sun, Sep 8, 2013 at 3:27 AM, Kristofer Munsterhjelm wrote: > On 09/08/2013 02:50 AM, Forest Simmons wrote: > >> The following method makes use of two ballots for each voter. The first >> ballot is a three slot ballot with allowed ratings of 0, 1, and 2. The >> second ballot is an ordinal preference ballot that allows equal rankings >> and truncations as options. >> >> The three slot ballot is used to select two finalists: one of them is >> the candidate rated at two on the greatest number of ballots. The other >> one is the candidate rated zero on the fewest ballots. >> >> The runoff between them is decided by the voters' pairwise preferences >> as expressed on the three slot ballots (when these finalists are not >> rated equally thereon), or (otherwise) on the ordinal ballots when the >> three slot ballot makes no distinction between them. >> >> [Giving priority to the three slot pairwise preference over the ordinal >> ballot preferences is necessary to remove the burial incentive.] >> >> Note that there is no strategic advantage for insincere rankings on the >> ordinal ballots. >> > > This sounds like an automated runoff. You use the first ballot to perform > an initial election, then you use the second ballot to determine the > outcome of a runoff between the two you picked from the initial election. > Because majority rule is strategy-proof with n=2, there is no incentive to > be insincere on the second ballot. > > But some might say that in certain situations there's no incentive to > fully rank the second ballot either. Say that you're pretty sure the runoff > will come to X vs Y. Then you only need to fill in X vs Y. > > Now, if the voters are basically honest and strategy concerns only come in > second place (overriding their honesty if the pressure towards strategy is > strong enough), then the voters will submit a full honest ranking anyway. > But if they're not, then they might not give you all their preferences. > (This question is related to other discussion as well. I've sometimes > argued that if you have a criterion X that says "there is no incentive to > be insincere in way Y unless enough people do it", that will keep the > voters from doing so, because their primary concern is honesty; while > others think that the voters will be insincere in way Y anyway because > there's no disincentive either. The difference is whether we need "there > must be an incentive to being honest" or whether "there is no incentive to > being dishonest" is enough.) > > Anyway, to get back on topic, the method you mention seems to work in a > game-theoretical sense. The ordinal ballots will be sincere. However, I > think real world voters would be confused by it. "Why do I have to submit > two ballots?", and so on. I suspect that strategic voters will be strategic > on both ballots, while honest voters will be honest on both unless they > feel like they have to use strategy on the first. > > If you just want to find a winner, then an ordinary runoff might work as > well: select the finalists as above, then have a majority-rule election in > the second round. If the purpose is determining the honest preferences, > then your method would have an advantage since it requires the voters to > state their preferences ahead of time (before they know who the finalists > are going to be) and so will have to submit more information than in a > runoff. > > > Questions. >> >> (1) What are some near optimal strategies for voters to convert their >> complete cardinal ratings into three slot ratings in this context? >>
Re: [EM] How to find the voters' honest preferences (Kevin Venzke)
Kevin, thanks for your thoughtful response. See some inline responses below: > Message: 3 > Date: Sun, 8 Sep 2013 19:51:03 +0100 (BST) > From: Kevin Venzke > To: em > Subject: Re: [EM] How to find the voters' honest preferences > Message-ID: > <1378666263.87766.yahoomail...@web171502.mail.ir2.yahoo.com> > Content-Type: text/plain; charset=iso-8859-1 > > . > > Except for the way you determine the runoff finalists, it reminds me of > what I named Single Contest. > After considering many single contest possibilities I settled on this one because it seemed like the simplest monotonic one: If I improve the winner W's position on the three slot ballot, then she still wins if the two finalists, W and L, remain the same. W will still be a finalist, but what about L? Since W is the only candidate with a score change, if L changes to L', the L' = W, i.e. now W is the only finalist, and therefore she still wins. I'll have to continue in another message, since my time is gone. Forest Election-Methods mailing list - see http://electorama.com/em for list info
[EM] How to find the voters' honest preferences
The following method makes use of two ballots for each voter. The first ballot is a three slot ballot with allowed ratings of 0, 1, and 2. The second ballot is an ordinal preference ballot that allows equal rankings and truncations as options. The three slot ballot is used to select two finalists: one of them is the candidate rated at two on the greatest number of ballots. The other one is the candidate rated zero on the fewest ballots. The runoff between them is decided by the voters' pairwise preferences as expressed on the three slot ballots (when these finalists are not rated equally thereon), or (otherwise) on the ordinal ballots when the three slot ballot makes no distinction between them. [Giving priority to the three slot pairwise preference over the ordinal ballot preferences is necessary to remove the burial incentive.] Note that there is no strategic advantage for insincere rankings on the ordinal ballots. Questions. (1) What are some near optimal strategies for voters to convert their complete cardinal ratings into three slot ratings in this context? (2) We have a "sincere approval" method of converting cardinal ratings into two slot ballots. What is the analogous "sincere three slot" method? [Sincere approval works by topping off the upper ratings with the lower ratings; think of the ratings as full or partially full cups of rating fluid next to each candidate's name. If you rate a candidate at 35%, then that candidate's cup is 35% full of rating fluid. Empty all of the rating fluid into one big pitcher and use it to completely fill as many cups as possible from highest rated candidate down. Approve the candidates that end up with full cups. This is called "sincere approval" because generically (and statistically) the total approval (over all voters) for each candidate turns out to be the same as the total rating would have been.] Thanks, Forest Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Instead of Top 2
On Sun, Apr 21, 2013 at 1:27 AM, Kristofer Munsterhjelm < km_el...@lavabit.com> wrote: > On 04/20/2013 12:09 AM, Forest Simmons wrote: > >> Suppose the two methods were IRV and Approval, and that each voter could >> choose which of the two methods to vote on their strategic ballots, and >> then rank the candidates non-strategically as well for the choice >> between the two method winners. >> >> We would learn something about the popularity of the two methods, which >> one chose the final winner the most often, which one elicited the most >> order reversals, etc. >> >> The same experiment could be done with any two methods. >> > > For that matter, the experiment could be done with ordinary runoff to > check if the voters change their minds between the rounds of the runoff. > > The experiment would go like this: first round, the voters vote using the > two methods in question, and also give a honest preference ordering for a > virtual runoff. Second round, they vote in the actual runoff between the > winner candidates (or some complex tiebreak if the winner is the same for > both methods). If the winner is the same for both methods, we're done. In that case we have to wait for another election to have a runoff. But if one of the methods results in a tie, the other method is the natural tie breaker. Method A is the tie breaker for method B and vice versa. > Then one compares the preference orderings with the runoff results. If the > runoff is A vs B, A won, but the preference ordering says B should have > won, there's your reversal. > > And I've mentioned it before, but I suppose I can do so again, since we're > talking about two-method runoffs :-) From time to time I've thought about > the idea of having a runoff using a strategy-resistant method and a method > that provides good results under honesty. This could be useful in a society > where people have become used to strategizing. If they strategize wildly, > then the honest method fails but the resistant method keeps the result from > being too bad; and if they don't, then the honest method's candidate wins > the runoff and all is good. > > In that case MJ could be the strategy resistant method, and ordinary range the method with good honesty results. Range nominates the candidate with the highest grade point average, while MJ nominates the candidate with the highest median grade. Here's another suggestion: Use five slot ballots as in MJ for the strategic ballots. The voters are allowed to choose their approval cutoffs. Candidate A is the approval winner for all of the ballots that set the cutoff somewhere strictly below the middle level. Candidate B is the approval winner for all of the ballots that set the cutoff at C or above. Use the sincere rankings to choose between candidates A and B. A check box allows voters to indicate that their strategic and sincere ratings are the same, thereby cutting their work in half if they feel no need to strategize. On Sun, Apr 21, 2013 at 1:27 AM, Kristofer Munsterhjelm < km_el...@lavabit.com> wrote: > On 04/20/2013 12:09 AM, Forest Simmons wrote: > >> Suppose the two methods were IRV and Approval, and that each voter could >> choose which of the two methods to vote on their strategic ballots, and >> then rank the candidates non-strategically as well for the choice >> between the two method winners. >> >> We would learn something about the popularity of the two methods, which >> one chose the final winner the most often, which one elicited the most >> order reversals, etc. >> >> The same experiment could be done with any two methods. >> > > For that matter, the experiment could be done with ordinary runoff to > check if the voters change their minds between the rounds of the runoff. > > The experiment would go like this: first round, the voters vote using the > two methods in question, and also give a honest preference ordering for a > virtual runoff. Second round, they vote in the actual runoff between the > winner candidates (or some complex tiebreak if the winner is the same for > both methods). Then one compares the preference orderings with the runoff > results. If the runoff is A vs B, A won, but the preference ordering says B > should have won, there's your reversal. > > And I've mentioned it before, but I suppose I can do so again, since we're > talking about two-method runoffs :-) From time to time I've thought about > the idea of having a runoff using a strategy-resistant method and a method > that provides good results under honesty. This could be useful in a society > where people have become used to strategizing. If they strategize wildly, > then the honest method fails but the resistant method keeps the result from > being too bad; and if they don't, then the honest method's candidate wins > the runoff and all is good. > > Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Instead of Top 2
Suppose the two methods were IRV and Approval, and that each voter could choose which of the two methods to vote on their strategic ballots, and then rank the candidates non-strategically as well for the choice between the two method winners. We would learn something about the popularity of the two methods, which one chose the final winner the most often, which one elicited the most order reversals, etc. The same experiment could be done with any two methods. On Fri, Apr 19, 2013 at 12:56 PM, Forest Simmons wrote: > Methods that choose between top 2 Approval, top 2 Plurality, Top 2 > Bucklin, etc. have problems that we are all familiar with, in particular > clones mess them up. > > But what if our method elects the pairwise preference between > the method A winner and the method B winner? If the two winners are the > same, then the common winner is elected. This idea seems to avoid the > problems associated with top2 methods. > > What would you suggest for methods A and B? > > I would suggest MJ type grade ballots. Then some good possibilities for > Method A or B would be MJ itself, XA (chiastic approval), Approval with > various possibilities for approval cutoff level, etc. > > My personal favorite version is to elect the pairwise preferred of the XA > winner and the candidate with the fewest F's. > > Forest > Election-Methods mailing list - see http://electorama.com/em for list info
[EM] Instead of Top 2
Methods that choose between top 2 Approval, top 2 Plurality, Top 2 Bucklin, etc. have problems that we are all familiar with, in particular clones mess them up. But what if our method elects the pairwise preference between the method A winner and the method B winner? If the two winners are the same, then the common winner is elected. This idea seems to avoid the problems associated with top2 methods. What would you suggest for methods A and B? I would suggest MJ type grade ballots. Then some good possibilities for Method A or B would be MJ itself, XA (chiastic approval), Approval with various possibilities for approval cutoff level, etc. My personal favorite version is to elect the pairwise preferred of the XA winner and the candidate with the fewest F's. Forest Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Consensus threshold
Good Points! On Thu, Apr 11, 2013 at 12:16 AM, Michael Allan wrote: > > The psychological value of this method is that it appeals to our > > natural community spirit which includes a willingness to go along > > with the group consensus when the consensus is strong enough, as > > long as there is no hope for a better consensus, and as long as it > > isn't a candidate that we would rate at zero. > > > > Comments? > > I think there is a general williness to *consider* a consensus, but > not a general willingness to follow it blindly. The popularity of a > candidate is a recommendation to look more closely at that candidate > given the fact of his/her popularity. Here popularity directly serves > only to arouse my curiosity, "Why is this candidate more popular? > What do others know that I don't know?" > > On learning the answer, I decide whether to follow the consensus. > > The proposed method differs in asking me to make the same decision, > but without knowing the reason for the candidate's popularity. It > invites me to act irrationally and enshrines that action as normal > human behaviour. > > As a counter-proposal, consider a broader rationalization of the > electoral design. Rather than overloading a single election with > expectations it cannot fulfil, factor it into two elections: (1) a > continuous, advisory primary to flush out consensus and dissensus, to > give people time to talk things over, and decide what to do; followed > by (2) a decisive election in which they express the decision. This > solves the problem of systematic irrationality by allowing for a real > consensus in the primary, one with reasons behind it, the validity of > which can be discussed and debated before making a decision. > > -- > Michael Allan > > Toronto, +1 416-699-9528 > http://zelea.com/ > > > Forest Simmons said: > > Jobst has suggested that ballots be used to elicit voter's "consensus > > thresholds" for the various candidates. > > > > If your consensus threshold for candidate X is 80 percent, that means > that > > you would be willing to support candidate X if more than 80 percent of > the > > other voters were also willing to support candidate X, but would forbid > > your vote from counting towards the election of X if the total support > for > > X would end up short of 80 percent. > > > > The higher the threshold that you give to X the more reluctant you are to > > join in a consensus, but as long as your threshold t for X is less than > > than 100 percent, a sufficiently large consensus (i.e. larger than t > > percent) would garner your support, as long as it it is the largest > > consensus that qualifies for your support. > > > > A threshold of zero signifies that you are willing to support X no matter > > how small the consensus, as long as no larger consensus qualifies for > your > > support. > > > > I suggest that we use score ballots on a scale of 0 to 100 with the > > convention that the score and the threshold for a candidate are related > by > > s+t=100. > > > > So given the score ballots, here's how the method is counted: > > > > For each candidate X let p(X) be the largest number p between 0 and 100 > > such that p(X) ballots award a score strictly greater than 100-p to > > candidate X. > > > > The candidate X with the largest value of p(X) wins the election. > > > > If there are two or more candidates that share this maximum value of p, > > then choose from the tied set the candidate ranked the highest in the > > following order: > > > > Candidate X precedes candidate Y if X is scored above zero on more > ballots > > than Y. If this doesn't break the tie, then X precedes Y if X is scored > > above one on more ballots than Y. If that still doesn't break the tie, > > then X precedes Y if X is scored above two on more ballots than Y, etc. > > > > In the unlikely event that the tie isn't broken before you get to 100, > > choose the winner from the remaining tied candidates by random ballot. > > > > The psychological value of this method is that it appeals to our natural > > community spirit which includes a willingness to go along with the group > > consensus when the consensus is strong enough, as long as there is no > hope > > for a better consensus, and as long as it isn't a candidate that we would > > rate at zero. > > > > Comments? > > > > Forest > Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Condorcet IRV Hybrid
On Fri, Apr 5, 2013 at 2:43 PM, Kristofer Munsterhjelm wrote:
> On 04/05/2013 09:37 PM, Forest Simmons wrote:
>
> The following observation about Condorcet IRV Hybrids has probably
>> already been made (but I have been gone for a while):
>>
>> These hybrids have no good defense against burying. For example
>>
>> Sincere ballots:
>>
>> 40 A>C
>> 35 B>C
>> 25 C>A
>>
>> If the A faction decides to bury C, there is nothing the C faction can
>> do about it unilaterally. They have to depend on the willingness of the
>> B faction to elevate their compromise over favorite.
>>
>
> That's strange, because one of the points of James Green-Armytage in his
> voting strategy paper was that the Condorcet-IRV hybrids were significantly
> less prone to burying than ordinary Condorcet methods. Quoting,
>
> "All Condorcet-efficient methods are vulnerable to burying, but this
> vulnerability seems to be substantially less frequent in the Condorcet-Hare
> hybrids than in most other Condorcet methods. The reason for this is that
> voters who prefer q to w will already have ranked q ahead of w, so that
> further burying w will not affect w's plurality score unless q has already
> been eliminated."
>
> ("Four Condorcet-Hare Hybrid Methods for Single-Winner Elections",
> http://www.econ.ucsb.edu/~**armytage/hybrids.pdf<http://www.econ.ucsb.edu/~armytage/hybrids.pdf>,
> p. 8)
>
> Or are we talking about different things? Perhaps C/IRV methods are less
> vulnerable to burying in the first place, but when they are, it's harder to
> employ defensive strategy to correct the burial?
>
> -
>
> Also, I seem to recall that Uncovered,X is generally more susceptible to
> burial than is X for various types of X, unless X is already rather
> susceptible to burial. It might be interesting to run a JGA type analysis
> on your "eliminate until covering" method, and compare to the Smith-IRV
> methods.
>
> The method I proposed is not of the type "Uncovered X." The first
candidate to cover the remaining candidates may well be a covered candidate
herself.
Election-Methods mailing list - see http://electorama.com/em for list info
[EM] Condorcet IRV Hybrid
Here's a Condorcet IRV Hybrid similar to Benham's version: Use IRV to eliminate candidates until one of the remaining candidates covers the other remaining candidates. The rationale for this proposal is that a candidate who merely beats (pairwise) the other remaining candidates may not be considered strong enough for an exemption from elimination, but such a candidate has a stronger case if she also beats every candidate that is beaten by any of the remaining candidates. What made me think of this version is that our DMC (Democratic Majority Choice) proposal to Toby Nixon eliminated low approval candidates until there was a beats all candidate among the remaining. In my opinion the DMC is much improved if you keep eliminating low approval candidates until one of the remaining candidates covers the others: you tend to get higher up the approval chain. In this modified version of DMC if there is no Condorcet candidate, then most of the time the highest approval Smith candidate will win. In particular, if (as in a cycle of three) no member of the Smith set covers any other member of the Smith set, then the highest Approval Smith candidate wins. The following observation about Condorcet IRV Hybrids has probably already been made (but I have been gone for a while): These hybrids have no good defense against burying. For example Sincere ballots: 40 A>C 35 B>C 25 C>A If the A faction decides to bury C, there is nothing the C faction can do about it unilaterally. They have to depend on the willingness of the B faction to elevate their compromise over favorite. Forest Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Election-Methods Digest, Vol 106, Issue 2
Kris, Optimal MJ strategy is still approval strategy. You can instruct the voters to make absolute choices, but you cannot enforce it. Their willingness to abide by the instructions is purely psychological. The same psychology will work, only better for Consensus Threshold Approval. Forest On Thu, Apr 4, 2013 at 2:02 AM, Kristofer Munsterhjelm wrote: > On 04/04/2013 02:40 AM, Forest Simmons wrote: > >> >> >> On Wed, Apr 3, 2013 at 12:07 AM, Kristofer Munsterhjelm >> mailto:km_el...@lavabit.com>> wrote: >> >> > Perhaps there's some value in making methods that appeal to the >> right sentiment, even if one has to trade off "objective" qualities >> (like BR, strategy resistance or criterion compliance) to get there. >> The trouble is that we can't quantify this, nor how much of >> sentiment-appeal makes up for deficiencies elsewhere, at least not >> without performing costly experiments. >> >> >> If I am not mistaken, both methods (Chiastic and this one) are >> strategically equivalent to Approval from a game theoretic point of >> view. But psychologically they are quite different. I think that this >> new version is much less likely to elicit approval style responses (at >> the extremes) than ordinary Range voting for example, or even the median >> method with J in the title (I can't think of it at the moment). >> >> > I found a quite broad reduction for ratings-type methods. I posted it when > I did, but I'll repeat it. > > Say you have a rated ballot set, and candidate C's set of scores is > represented by the vector s_C (first element in the vector is the first > voter's rating of the candidate, and so on), then: > > - if each candidate gets a meta-score, call it m_C, from some function > f(s_C), > - the candidate with the highest meta-score wins, > - and f(s_C) is monotone in the sense that increasing a rating in s_C > never makes f(s_C) evaluate to a lower value than before, and decreasing a > rating in s_C never makes f(s_C) evaluate to a higher value than before, > > - then Approval strategy is optimal. > > The reason is that if a voter likes a candidate X, he can never be worse > off by not giving X a higher score; and if he dislikes X, he can never be > worse off by not giving X a lower score. Thus the scores migrate to the > Approval extremes. > > And unless I'm missing something, both the chiastic method and your method > fulfill the properties above. > > There's a caveat: the "optimality" might be of a form where it never hurts > you to go to an extreme, but it doesn't hurt not to either. To eliminate > that kind of equilibrium, one would have to replace the monotonicity > property with something stronger. > > - > > As for the median method, you're probably thinking of Majority Judgement. > As long as the voters act in the way B&L say they should do, and judge > candidates to absolute grades rather than comparing the candidates to each > other, it avoids Approval reduction. B&L use evidence from France to argue > that enough voters judge to absolute grades that it effectively works this > way. Maybe your method would also do so, but then it would have to be > phrased in terms of grades rather than numbers. > > I think that Range encourages rating at the extremes, though it doesn't > seem to always do so. Web sites lend evidence to both sides: IMDB used (and > uses) Range for the movie ratings, while Youtube switched from ratings to > Approval-style up-and-down voting. IMDB does some filtering on the votes, > though, so perhaps that's what is keeping it from reducing to approval. In > any case, it shouldn't be hard to make a method that's more resistant than > bare Range in that respect. > > Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Election-Methods Digest, Vol 106, Issue 2
For purposes of clarification, I would like to show how Approval, Bucklin, Range, Chiastic Approval, and Consensus Threshold Approval manifest themselves relative to each other visually. I assume versions of these methods that make use of range style ballots on a scale of zero to 100. These methods also have in common that once the ballots are counted each candidate ends up with a score of some kind, and the candidate with the largest score is elected. So let’s concentrate on how each of these methods would assign a score to the same fixed candidate. All of these methods can be explained in terms of the graph of the function F given by p=F(r) is the percentage of the ballots that rate our candidate strictly greater than r. Each point (r, p) of this graph will lie somewhere in the 100 by 100 square with corners at (0,0), (0,100), (100, 0) and (100, 100). Furthermore, the graph will descend from left to right in steps whose widths are whole numbers. The left endpoint of each step will be included but the right end point will not be included. Color this graph blue. Now join the steps with vertical segments. The interior points of the vertical segments are colored red, while the top end point of each red segment will be colored red, and the bottom point will be colored blue. Now the union of the red and blue separates the lower left corner from the upper right corner of the square. Therefore the diagonal from (0, 0) to (100, 100) must cross the colored graph in either a red or blue point. Since the red and blue are non-increasing while the diagonal is strictly increasing, there can be only one point of intersection. The common value of the coordinates of this intersection point is the Chiastic Approval score. Now calculate the area of the region of the square that lies to the lower left of the red/blue diagonal. This area is the average rating of our candidate. So the candidate whose lower left area is greatest is the Range winner. Now bisect our square horizontally with a straight line segment from (0, 50) to (100, 50). The first coordinate (r) of the point of intersection (r, 50) of this line with the red determines the basic Bucklin score. Ties are broken by various methods. Now bisect the square with a vertical segment from (50, 0) to (50, 100). Assuming an approval cutoff of fifty, the second coordinate (p) of the intersection (50, p) of this segment with the blue is the approval score. Now consider the diagonal from the upper left corner (0, 100) to the lower right corner (100, 0). If this diagonal does not intersect the blue, then the candidate’s Consensus Threshold Approval score is zero. Otherwise it is the second coordinate of the highest (and therefore leftmost) blue point of intersection. In summary, we have bisected the 100 by 100 square vertically, horizontally, and diagonally. The diagonal with positive slope leads us to the chiastic approval winner. The other diagonal leads us to the consensus threshold approval winner. The horizontal bisector leads us to the Bucklin winner. The vertical bisector leads us to the Approval winner. The area cut off by the colored graph determines the Range winner. On Wed, Apr 3, 2013 at 6:18 PM, Jameson Quinn wrote: > > > > 2013/4/3 Forest Simmons > >> >> >> On Wed, Apr 3, 2013 at 12:07 AM, Kristofer Munsterhjelm < >> km_el...@lavabit.com> wrote: >> >>> On 04/03/2013 12:01 AM, Forest Simmons wrote: >>> >>>> Jobst has suggested that ballots be used to elicit voter's "consensus >>>> thresholds" for the various candidates. >>>> >>>> If your consensus threshold for candidate X is 80 percent, that means >>>> that you would be willing to support candidate X if more than 80 percent >>>> of the other voters were also willing to support candidate X, but would >>>> forbid your vote from counting towards the election of X if the total >>>> support for X would end up short of 80 percent. >>>> >>>> The higher the threshold that you give to X the more reluctant you are >>>> to join in a consensus, but as long as your threshold t for X is less >>>> than than 100 percent, a sufficiently large consensus (i.e. larger than >>>> t percent) would garner your support, as long as it it is the largest >>>> consensus that qualifies for your support. >>>> >>>> A threshold of zero signifies that you are willing to support X no >>>> matter how small the consensus, as long as no larger consensus qualifies >>>> for your support. >>>> >>>> I suggest that we use score ballots on a scale of 0 to 100 with the >>>> convention that the score and the threshold for a candidate are related >>>>
Re: [EM] Election-Methods Digest, Vol 106, Issue 2
On Wed, Apr 3, 2013 at 12:07 AM, Kristofer Munsterhjelm < km_el...@lavabit.com> wrote: > On 04/03/2013 12:01 AM, Forest Simmons wrote: > >> Jobst has suggested that ballots be used to elicit voter's "consensus >> thresholds" for the various candidates. >> >> If your consensus threshold for candidate X is 80 percent, that means >> that you would be willing to support candidate X if more than 80 percent >> of the other voters were also willing to support candidate X, but would >> forbid your vote from counting towards the election of X if the total >> support for X would end up short of 80 percent. >> >> The higher the threshold that you give to X the more reluctant you are >> to join in a consensus, but as long as your threshold t for X is less >> than than 100 percent, a sufficiently large consensus (i.e. larger than >> t percent) would garner your support, as long as it it is the largest >> consensus that qualifies for your support. >> >> A threshold of zero signifies that you are willing to support X no >> matter how small the consensus, as long as no larger consensus qualifies >> for your support. >> >> I suggest that we use score ballots on a scale of 0 to 100 with the >> convention that the score and the threshold for a candidate are related >> by s+t=100. >> >> So given the score ballots, here's how the method is counted: >> >> For each candidate X let p(X) be the largest number p between 0 and 100 >> such that p(X) ballots award a score strictly greater than 100-p to >> candidate X. >> >> The candidate X with the largest value of p(X) wins the election. >> > > I think a similar method has been suggested before. I don't remember what > it was called, but it had a very distinct name. > > It went: for each candidate x, let f(x) be the highest number so that at > least f(x)% rate the candidate above f(x). > > I *think* it went like that, at least. Sorry that I don't remember the > details! Good memory, that was Andy Jennings' Chiastic method. Graphically these two methods are based on different diagonals of the same rectangle. > > > If there are two or more candidates that share this maximum value of p, >> then choose from the tied set the candidate ranked the highest in the >> following order: >> >> Candidate X precedes candidate Y if X is scored above zero on more >> ballots than Y. If this doesn't break the tie, then X precedes Y if X >> is scored above one on more ballots than Y. If that still doesn't break >> the tie, then X precedes Y if X is scored above two on more ballots than >> Y, etc. >> >> In the unlikely event that the tie isn't broken before you get to 100, >> choose the winner from the remaining tied candidates by random ballot. >> > > I imagine Random Pair would also work. > > > The psychological value of this method is that it appeals to our natural >> community spirit which includes a willingness to go along with the group >> consensus when the consensus is strong enough, as long as there is no >> hope for a better consensus, and as long as it isn't a candidate that we >> would rate at zero. >> > > That's an interesting point. I don't think that factor has been considered > much in mechanism design in general. Condorcet, say, is usually advocated > on the basis that it provides good results and resists enough strategy, and > then one adds the reasoning "it looks like a tournament, so should be > familiar" afterwards. > > Perhaps there's some value in making methods that appeal to the right > sentiment, even if one has to trade off "objective" qualities (like BR, > strategy resistance or criterion compliance) to get there. The trouble is > that we can't quantify this, nor how much of sentiment-appeal makes up for > deficiencies elsewhere, at least not without performing costly experiments. > If I am not mistaken, both methods (Chiastic and this one) are strategically equivalent to Approval from a game theoretic point of view. But psychologically they are quite different. I think that this new version is much less likely to elicit approval style responses (at the extremes) than ordinary Range voting for example, or even the median method with J in the title (I can't think of it at the moment). Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Election-Methods Digest, Vol 106, Issue 2
Jobst has suggested that ballots be used to elicit voter's "consensus thresholds" for the various candidates. If your consensus threshold for candidate X is 80 percent, that means that you would be willing to support candidate X if more than 80 percent of the other voters were also willing to support candidate X, but would forbid your vote from counting towards the election of X if the total support for X would end up short of 80 percent. The higher the threshold that you give to X the more reluctant you are to join in a consensus, but as long as your threshold t for X is less than than 100 percent, a sufficiently large consensus (i.e. larger than t percent) would garner your support, as long as it it is the largest consensus that qualifies for your support. A threshold of zero signifies that you are willing to support X no matter how small the consensus, as long as no larger consensus qualifies for your support. I suggest that we use score ballots on a scale of 0 to 100 with the convention that the score and the threshold for a candidate are related by s+t=100. So given the score ballots, here's how the method is counted: For each candidate X let p(X) be the largest number p between 0 and 100 such that p(X) ballots award a score strictly greater than 100-p to candidate X. The candidate X with the largest value of p(X) wins the election. If there are two or more candidates that share this maximum value of p, then choose from the tied set the candidate ranked the highest in the following order: Candidate X precedes candidate Y if X is scored above zero on more ballots than Y. If this doesn't break the tie, then X precedes Y if X is scored above one on more ballots than Y. If that still doesn't break the tie, then X precedes Y if X is scored above two on more ballots than Y, etc. In the unlikely event that the tie isn't broken before you get to 100, choose the winner from the remaining tied candidates by random ballot. The psychological value of this method is that it appeals to our natural community spirit which includes a willingness to go along with the group consensus when the consensus is strong enough, as long as there is no hope for a better consensus, and as long as it isn't a candidate that we would rate at zero. Comments? Forest Election-Methods mailing list - see http://electorama.com/em for list info
