Re: [Election-Methods] Correction of false statements by Ossipff & Schudy about range voting.
>>"Range voting is a generalisation of approval voting where you can >> give each candidate any score >>between 0 and 1. Optimal strategies never vote anything other than >> 0 or 1, so range voting >>complicates ballots and confuses voters for little or no gain." >> >>Ossipoff: Warren Schude's statement was correct >--CORRECTION: optimal strategies can vote other than 0 and 1, and >voting 0 or 1 can be suboptimal. Seems you were not so assiduous as to actually read the footnotes in Warren Schudy's paper, which in this particular case (footnote number 1) reads: "As long as the population is sufficiently big and uncertain." And in that case we may nicely agree, I guess. I think Michael Ossipoff's analysis: "Suppose that the method is 0-10 RV... Now, suppose that you consider the points that you're awarding one-at-a-time, as if it were a series of 10 Approval elections... We're assuming that it's a public election, so that there are so many voters that your own votes have no significant effect on the probabilities. Your Approval strategy is based on two things: The candidates' utility to you, and the probabilities that you estimate... Your utilities don't change during that series of 10 Approval elections that you vote. The probability estimates don't change either... If you give to a candidate any points at all, you give hir 10 points." is informal, but fully correct. (Which doesn't mean I agree with him in everything else.) And those who refer to linear programming essentially say the same thing. If the field is big enough, a small part of it may be considered as having linear probability-distribution (or whatever) functions, so the optimum lies somewhere on the border. So if you started to go to a direction you have to stay on that course. So, they say if the number of voters goes toward infinity, the probability of a case where Approval-style voting is suboptimal goes toward zero. The counterexamples? most of them have extremally small number of votes. And even which does not so, uses the less-then infinite, non-linear attributes or simply wrong. http://beyondpolitics.org/Range2Utility.htm when shifts from Range(0,1,2) to Approval, calculates like those "odd number" cases would simply vanish. And vanishing some good vote value, the average worsens when Approval becomes the method. But those cases don't vanish. The logical statistic assumption is that they evenly distribute themselves among the neighboring cases. And some of previously irrelevant cases become relevant cases, so the probability of decisive vote rises. This rise exactly compensates for the loss of utility rise. As for http://rangevoting.org/RVstrat5.html it's more reality, but only by using the three-candidate-tie event, attributed with a T probability. If T=0, the classic case happens: giving the in-between B candidate max or min is optimal, or all the same. And if the number of voters goes toward infinity, T goes toward zero. So, please, don't infer "which graduation is best for range voting" type statements from these calculations. We can go back to the consensus (used even in your simulations) that _essentially_ a strategic Range vote is an Approval vote. Which doesn't decide which one is better. Valid arguments exists on both sides. Range voters can choose from more possibilities, but is this choice a pleasant one? I can be a "sucker" or a "cheater", maybe I would be more glad without it. I think they are so close that even their fans can be close and fight side by side. I'm looking for the future when TV-personalities as well as people at the coffee machine dispute about whether Approval or Range is better method. Peter Barath Tavaszig, most minden féláron! ADSL Internet már 1 745 Ft/hó -tól. Keresse ajánlatunkat a http://www.freestart.hu oldalon! Election-Methods mailing list - see http://electorama.com/em for list info
Re: [Election-Methods] When Voters Strategize, Approval Voting Elects Condorcet Winners but Condorcet Methods can Elect Condorcet Losers
On 2007 July 17 Tuesday 20.21, Abd ul-Rahman Lomax wrote: >It's a little-noticed fact that, in Warren Smith's IEVS simulations, >which generate sets of voters with simulated utilities, then apply >various voting methods and strategies, (see rangevoting.org), Range >Voting, when voters "strategize," is actually beaten by Range with a >top-two runoff. It's little-noticed, indeed. Thank you for the information. I talked to a friend who happens to be a member of Mensa Hungary and now he wants to write an article about voting methods for the local Mensa periodical! He has two questions: >From your words I guess you propose a real runoff in a later time, when the results of the first round are already known. Is this real and principal decision, or only practical? Would it be possible either in Range or in Approval to gain this pairwise comparison information from the already cast votes? Maybe by Condorcet-style votes or something? (People don't like to vote twice.) The second question: As far as we understand, Steven Brams and Peter Fishburn can be considered as foundig fathers of proposing Approval vote in modern days (Wikipedia - Voting system says it was in use in Venice in the 13th century!). Who can be honoured as proposer of Approval+2 and Range+2? Do they have the same origin? Peter Barath P.S.: For your interest: Mensa Hungary uses plurality to elect its leaders. Mensa International uses instant runoff for the same aim. Tavaszig, most minden féláron! ADSL Internet már 1 745 Ft/hó -tól. Keresse ajánlatunkat a http://www.freestart.hu oldalon! Election-Methods mailing list - see http://electorama.com/em for list info
Re: [Election-Methods] When Voters Strategize, Approval Voting Elects Condorcet Winners but Condorcet Methods can Elect Condorcet Losers
>>Would it be possible either in Range or in Approval to >>gain this pairwise comparison information from the >>already cast votes? Maybe by Condorcet-style votes >>or something? (People don't like to vote twice.) >Of course they don't. However, if you choose the pairwise winner, >you might as well use a Condorcet method, but you'll lose the >optimization of overall satisfaction. Well, if you remind me to Condorcet, I have to admit having a bit of doubt about this loss. I don't remember seeing the result of this particular simulation. (I mean Condorc-Appr.) I remember a simulation with full strategic voters but without make-a-cutoff-and-if-no-condorcet-winner-then-approval method, and another one which maybe included it but it came with an arbitrarily chosen 50 percent honest voter, which is unrealistic by my guess. (Warren D. Smith gives somewhere an example of distorted - probably unintentionally - poll answers. In a place where Nader was not allowed to vote for, people were asked would they if possible. They gave an unrealisticly high proportion of "yes". So I don't think polls are good for estimating honesty.) My guess is Condorcet is a super method as long as we stay in situations where a Condorcet winner exists. Theorem (Barath's first and last theorem): It's not possible by changing honest votes into dishonest ones to change from a Condorcet-winner to a better Condorcet-winner. (In other words: Condorcet-strategizing can be based on the fact that there is not always Condorcet-winner, so by changing from C-exist to no-C or from no-C to C-exist or from no-C to another no-C situation.) Proof sketch: If such move exists, it means there are ballots dishonestly changed and made Bea win instead of Al. If Bea is better for these voters then on their sincere votes Bea must have been in higher position than Al. If with these honest votes Al beat Bea (like every others a Cond-w must), then Al still must beat Bea with the dishonest ones, because on these ballots Bea's position compared to Al cannot be improved. And if Bea doesn't beat Al, C-winning is out of question. Maybe by mixing Condorcet and Approval - a highly "honest" method - we can get a method which gives more than sincere approval sortings: sincere preference orders! Before Warren D. Smith or anybody else would make a simulation for this - with supposedly "strategizing" voters - could somebody give an example of order-reversal in this Condorcet-cutoff-Approval method? Peter Barath Tavaszig, most minden féláron! ADSL Internet már 1 745 Ft/hó -tól. Keresse ajánlatunkat a http://www.freestart.hu oldalon! Election-Methods mailing list - see http://electorama.com/em for list info
Re: [Election-Methods] response to Schudy re Range vs Approval voting
>*2. So for example, if >49% voted Bush=99, Gore=0, Nader=53(avg), and >49% voted Gore=99, Bush=0, Nader=53(avg), and >2% voted Nader=99, Gore=20, Bush=0 >then Nader would win. > >This structure is a realistic possibility that totally contradicts >the assertion RV >"gives power to the candidate whose supporters >have the most black and white, polarized view of the world." >In this case, Nader is winning despite a severe lack of polarized >Nader supporters. Well, in this example Nader supporters were more "polarized" than either Bush or Gore ones. Anyway, picked up examples don't refute hypotheses about tendencies. >*3. If we also add, say, Badnarik with scores not of 53 like for >Nader, but rather, say, 20, then Badnarik would not win, >but still would get a total range-voting >score in the same ballpark as Bush, Gore, and Nader, thus permitting >him to claim he has a lot of popular support, and thus allowing his >party to try to get money and support for future elections. I wouln't vote that way. If concerned with these smaller scores, I would ask the question: will Badnarik get more or less points than I think he deserves? If think: more, I would give him 0 points to lesser; if think: less, I would give him full points to enhance. >Note, it was an "immediate" bad effect that (above) >Approval caused Nader to lose >when Range vould have caused him to win. Not necessarily. A lot of 53 percenters would approve him. >In our study of the 2004 US election, we were not able >to find any evidence that >Bush voters were either more or less "polarized" and >"strategically exaggerating" >than Gore voters. (Perhaps they were, but if so the >effect was too small for our >statistics to see.) In other words: yes, correlation between degree of strategizing and political stance makes Bayesian regret bigger. But we don't have reason to be afraid of big correlations of that kind. Peter Barath Tavaszig, most minden féláron! ADSL Internet már 1 745 Ft/hó -tól. Keresse ajánlatunkat a http://www.freestart.hu oldalon! Election-Methods mailing list - see http://electorama.com/em for list info
Re: [Election-Methods] response to Schudy re Range vs Approval voting
>In other words: yes, correlation between degree of >strategizing and political stance makes Bayesian regret >bigger. But we don't have reason to be afraid of big >correlations of that kind. Still thinking that, I feel like addig something: If I don't vote and at the end a bad candidate wins, I will kick myself for having not voted. And I will find little comfort in the fact that if all voter groups abstain in the same proportion, the result is the same as if they all voted. If in a Plurality I vote for Calero and he gets almost nothing and Bush wins instead of Gore, I will kick myself for having wasted my vote on an almost chanceless candidate. If in a Range(1,2,3,...,100) I give 100 to Calero, 10 to Gore and nothing to Bush, and Bush wins instead of Gore, I will still kick myself for having wasted nine-tenth of my vote on an almost chanceless candidate. Peter Barath Megújult a Jobline.hu! 2600 állásból most még egyszerűbben válogathatsz és önéletrajzodat akár 1 perc alatt is feltöltheted. http://ad.adverticum.net/b/cl,1,6022,191642,246289/click.prm Election-Methods mailing list - see http://electorama.com/em for list info
Re: [Election-Methods] US court rejects threat to 'vote-trading' websites
http://www.vnunet.com/vnunet/news/2195759/court-rejects-threat-vote >Yes. Very interesting. I think the decision of the Court reasonably >obvious; however, they consider that the issues were sufficiently >unsettled at law that they did not find the Secretary of State of >California, the defendant, to be liable for damages to the A little help, please. My Mensan friend who is to write an article about voting methods, in a Mensa periodical, wants to mention the incident. But we don't have the slightest idea what a "Secretary of State of California" means. "Secretary of State of US" is US term for European "Minister of Foreign Affairs", but we have doubt that a US member state has such thing. Is it like a "Justice Minister", or a "Minister of Home affairs" (police and/or municipalities), or "Senior State Prosecutor", "Head of Supreme Court" or something else? Peter Barath Tavaszig, most minden féláron! ADSL Internet már 1 745 Ft/hó -tól. Keresse ajánlatunkat a http://www.freestart.hu oldalon! Election-Methods mailing list - see http://electorama.com/em for list info
Re: [Election-Methods] US court rejects threat to 'vote-trading' websites
>>But we don't have the slightest idea what a "Secretary of State >>of California" means. >Suggestion. Don't know what a term means, look it up on Wikipedia. >The most common, and arguably the most important, function held by >Secretaries of State is to serve as the state's chief elections Thank all of you for the informations. (And I'm really ashamed a little bit for not finding for myself.) Peter Barath Tavaszig, most minden féláron! ADSL Internet már 1 745 Ft/hó -tól. Keresse ajánlatunkat a http://www.freestart.hu oldalon! Election-Methods mailing list - see http://electorama.com/em for list info
Re: [Election-Methods] Improved Approval Runoff
>47: Bush >> Gore > Nader >27: Gore > Nader >> Bush (honest); Gore >> Nader > Bush (strategic) >26: Nader > Gore >> Buch (honest); Nader >> Gore > Bush (strategic) >Bush wins the first rount, but loses for Gore in a runoff (IAR). >With strategic voting, the spoiler effect is possible under simple >approval. But this example was made possible only by the fact that many voters made bad estimates. It's a well accepted fact that in the most cases (no, not in all cases, as Warren D. Smith has low probability counter-examples for that) a strategic Approval vote is made by: - making a honest preference order - putting a cutoff mark between the two candidates with the best chances - if the better one seems to have more chance, put the cutoff immediately under her/him/it - if the worse one seems to have more chance, put the cutoff immediately above her/him/it But rarely is the question asked: what do we mean by the "two candidates with the best chances"? I think we are best to measure this only by the fact of how we expect all other voters to vote. Which means that the concept of "two candidates with the best chances" depends not solely on the candidates themselves but theoretically possibly on the voting method too! In your example with 47 percent firm Bush supporters the voters were very wrong in supposing that he is a harmless candidate. In reality, in this case the strategic votes would be identical to the honest ones. However, I think your example did point to a widely ignored fact: that the ugly dilemma of the Plurality vote: "how will other voters vote?" does exist in Approval, even if it's smaller. So it's plausible to mix Approval. My favourite (at this moment) is a preference ranking with an approval cutoff. For me it's interesting enough that it can be used in two ways: 1. If there is a Condorcet-winner she/he/it wins. If there is not, the Approval winner wins. 2. Calculate the two candidates with the most Approval points and the pairwise winner of them wins. Peter Barath Tavaszig, most minden féláron! ADSL Internet már 1 745 Ft/hó -tól. Keresse ajánlatunkat a http://www.freestart.hu oldalon! Election-Methods mailing list - see http://electorama.com/em for list info
Re: [Election-Methods] Improved Approval Runoff
>By the way, electing from the Condorcet top tier using approval >would be called Smith//Approval or Schwartz//Approval depending on >which top tier is used. I don't typically consider these methods >because they are more complicated than Condorcet//Approval and >can't be adjusted to satisfy FBC. And what about the method (I don't know the name) in which the least approved candidate is eliminated until there is a Condorcet-winner? Does it also fail FBC? Did somebody analyse the strategy incentives then? (And here I don't think of a method in which all ranked candidates are considered as approved, but a whole preference order with a cutoff mark somewhere between.) Peter Barath Tavaszig, most minden féláron! ADSL Internet már 1 745 Ft/hó -tól. Keresse ajánlatunkat a http://www.freestart.hu oldalon! Election-Methods mailing list - see http://electorama.com/em for list info
Re: [Election-Methods] Improved Approval Runoff
>>1. If there is a Condorcet-winner she/he/it wins. If there is >>not, the Approval winner wins. >Sure. That's been proposed many times. However, >it's not a very good method. First of all, it is >blatantly obvious, if you care to look, that the >Condorcet winner is sometimes *not* the best >winner, by far. I guess this is an unjust blame because this thing affect all voting methods. Even in a two-candidate contest where every considerable method becomes Plurality, it's possible that the minority has stronger preference, so the winner is not the social optimum. (The only defense against this is the money voting, the Clarke-tax, which is - I think - treated also a little unjustly. At least the theoretical honor should be given for showing the possibility of strategy-freeness. And who knows, one day it can be proven even practical in some circumstances.) Peter Barath Tavaszig, most minden féláron! ADSL Internet már 1 745 Ft/hó -tól. Keresse ajánlatunkat a http://www.freestart.hu oldalon! Election-Methods mailing list - see http://electorama.com/em for list info
Re: [Election-Methods] Spearman-unbiased apportionment
>* r is the rounding function r(x) = x > m(floor(x), ceil(x)) ? > ceil(x) : floor(x), where m is a generalized mean function Not about this problem, but I am reminded of a conversation many years ago. I don't remember who mentioned, without any practical reference, only as a mathemathical curiosity, that if we are to elect numbers of representatives in differently populated districts, the "true" method is when the number of the representatives is proportional to the square root of the population. As far as I can remember, it was something about the probability of being a pivotal voter. (Which also gets a big role in the theorization of the Clarke-tax.) I'm not sure if I would be able to check it, but somehow my feeling is that it can easily be well grounded. At least if we think of the representatives as clones who always vote the same way. If there are parties and the representatives are party-proportional, this remark seems invalid to me. Peter Barath Tavaszig, most minden féláron! ADSL Internet már 1 745 Ft/hó -tól. Keresse ajánlatunkat a http://www.freestart.hu oldalon! Election-Methods mailing list - see http://electorama.com/em for list info
[Election-Methods] the ladder vote
Sorry if the thing already has a name. Let's suppose there is a vote, where the voters are to chose from a number of numbers. For example, the membership fee of the club, the minimum age of application, the size of the office, anything. Something where we can suppose: If a person prefers number a over number b, and a > b > c then she will prefer b over c (because c is even further from a). Also, if she prefers a over b and a < b < c then she will prefer b over c (because c is even further from a). Scenario one. They vote by everyone giving her first preference then they search the median value. For example: Let the membership fee be: 200$ 7 votes --- (-) 180$ 16 votes --- (-) 150$ 23 votes --- (+) 140$ 9 votes --- (+) 100$ 32 votes --- (+) 70$ 2 votes In this ladder-like scheme I put a sign at the end of each separating line: the sign is - if there are more votes under the line, and is + if there are mor votes above the line. The winner is the 150$ because there is the change, so more than half of the voters wants the fee be 150$ or higher, and more than half wants it to be 150$ or lower. Scenario two. They vote Condorcet, while they keep the above mentioned convention: if somebody sees 180$ as the best option she must prefer 140$ over 70$ etc. The convention says nothing about the preference between, say, 200$ and 70$ in this case. Theorem 1: If in scenario one there is a winner (no ties) then in scenario two there is a Condorcet-winner and is the same as the "ladder" winner in scenario one. Theorem 2: The ladder voting is strategy-free. I don't waste the space with proofs, they seem pretty obvious. The theoretical (well, maybe sometimes practical, yes, we used sometimes to vote about numbers) significance I see mainly in the second theorem. We don't have many strategy- free voting methods, only this, the Clarke-tax and the random methods mentioned usually with the Gibbard-Satterthwaite theorem: (random vote and runoff between two random candidates) for more than 2 choseable possibilities. Peter Barath Tavaszig, most minden féláron! ADSL Internet már 1 745 Ft/hó -tól. Keresse ajánlatunkat a http://www.freestart.hu oldalon! Election-Methods mailing list - see http://electorama.com/em for list info
Re: [Election-Methods] a strategy-free range voting variant?
>4. For each option, determine the probability P(Y) of being a > randomly chosen "benchmark" voter's favourite. These probabilities > build the "benchmark lottery". > >5. Finally, the voting accounts are adjusted like this: > >a) Each deciding voter's account is increased by an amount equal to > the total rating difference between the winner and the benchmark > lottery amoung the *other* deciding voters, minus some fixed fee F, > say 10*N^(1/2). (Note that the resulting adjustment may be positive > or negative.) This is the part I understand the least. Let's imagine the following votes from the deciding voters: 10 million: Nader: 10 Gore: 5 Bush: 0 41 million: Gore: 10 Nader: 5 Bush: 0 49 million: Bush: 10 Gore: 5 Nader: 0 Let's say that the lottery winner was Bush. The real winner is going to be Gore, with 705 million voting money units, while Bush has only 490 million. The total rating difference is 215 million. Do you want to modify each deciding voter's account with that big amount? You can try to diminish this modification by the fixed fee but I guess the modification will still be very high, because you are not able to precisely predict the votes not to mention who the lottery winner is going to be. And I guess if you try to eliminate this huge voting money transfer by some averaging operation, you will bite your other finger by ruining the strategy-freeness. Even if these worries are valid, this random partitioning of the electorate looks a witty idea, worth some other trials. I also like the idea of voting money, but with some reservations; if the value of the voting money is not bound exactly to some real value, then good-bye, strategy-freeness, I guess. Otherways, voting money can be used even with the classical Clarke-tax. Yes, Clarke-tax goes to one direction, but every voter on every day can get one voting money unit. If my voting money does not grows (except by votings), I will use the most amount of it when I'm afraid to die soon - why keep them if I can't use them?. So there seems to be some extra voting power on the part of the deadly ill. Peter Barath Tavaszig, most minden féláron! ADSL Internet már 1 745 Ft/hó -tól. Keresse ajánlatunkat a http://www.freestart.hu oldalon! Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Fixing Range Voting
>Once upon a time, I designed an election method to fix the strategy >problem with Range Voting. > >The strategy problem: >You shouldn't cast a ballot with your honest ratings, you should >maximize them along Approval strategy lines. > >It also fixes the counting problem of how if someone does cast votes >throughout the range, they might have done better in the end by >different values. > >The method I call "Instant Runoff Normalized Ratings" (IRNR): >1. Collect ratings ballots >2. Normalize each ballot so that each has an equal magnitude >3. Sum up normalized ballots >4. If there are more than two choices, drop the one with the > smallest sum. If there are two choices remaining, one is the > winner. 5. Re-normalize from original ballot values but as if > dropped choices weren't there >6. Go to 3 > > >I think it gets very near to a utilitarian ideal solution ( > http://bolson.org/voting/twographs.html ) and encourages people to > vote honestly and uses those honest votes to the best possible > effect. I'm not sure I would vote honestly in such circumstance. Let my "honest" rangings be: 100 percent for my favourite but almost chanceless Robin Hood 20 percent for the frontrunner Cinderella 0 percent for the other frontrunner Ugly Duckling I think I would vote: 100 Robin Hood; 99 Cinderella; 0 Ugly Duckling If I'm really sure that the race decides between Cinderella and Ugly Duckling, why care too much for poor Robin Hood? And what, if I'm not really sure, because that's the situation which multi-candidate voting is really about? If I lower Cinderella's 99 to her honest 20, I make Robin Hood a little bit more hopeful not to drop first. But more hopeful against whom? Cinderella, of course, because I didn't change Robin and Ugly's obvious rangings. So I made more probable a situation in which more than 50 percent is the probability that the worst candidate wins. This is a doubtful advantage. On the other side, there is the effect that by rising Cinderella's points from the honest 20 to 99 I made more probable the similarly unlikely but positively desirable effect of Ugly dropping first instead of her. So, which does have more weigh? The doubtful little hope for Robin Hood, or the clear little hope against Ugly Duckling? I think the latter. Maybe at some point, let's say Cinderella's 5 percent, I like Robin so much more that I chose the first one. In that case I probably would vote 100-1-0 These voting are not the "honest" although by one percent "honer" than the simple Approval voting. But I would be open for persuasion. Peter Barath Tavaszig, most minden féláron! ADSL Internet már 1 745 Ft/hó -tól. Keresse ajánlatunkat a http://www.freestart.hu oldalon! Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] A new simulation
> 1. Random Ballot performs quite well: On average, the Random Ballot > lottery achieved 93.7% the total utility of the option with the > largest total utility, and 98.4% the Gini welfare value of the > option with the largest Gini welfare value. > In only 5% of all situations, this measure of relative > performance was less than 88.3% resp. 94.5%! I don't feel so happy about it. We should care not only for how a method influences the result but also for how a method influences what the options will be. In a real world, if we used Random Ballot, on many ballot paper's first option would be: "I become dictator" so we would end up soon having a dictator. Peter Barath Tavaszig, most minden féláron! ADSL Internet már 1 745 Ft/hó -tól. Keresse ajánlatunkat a http://www.freestart.hu oldalon! Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Auctions
>> The simple, humble Clarke tax method does seem a bituh unfair to >> me. There are various ways to remedy this problem like having it >> be based on the log of your income or making it based on how much >> income you have left as a means of judging how much you have >> contributed. There is a near endless number of ways to tweak this >> in order to amount to something useful. Do you think that if we vote simply with money, the rich have more power than the poor? Yes, it is true. And exactly that is the situation right now, too. > The big issue is that according to economic theory, nobody would > vote. If someone is in a polling booth, they have already ignored > economic theory and thus there is no way they are going to care > about a tiny probability of paying. I feel here some mixing up economicness and selfishness. Yes, when I go to vote, I have so little probability to save myself from being killed by Nazis that even this big stake maybe doesn't make my voting economic. But I also save many others while hoping that they also save me. In other words, I can be altruistic, so consider for some extent other peoples' interests as my interest. And I can even express my interests (including altrusitic ones) by money units, and bear fully economically rational decisions about paying probabilities and other things. (Altruism is not a very frequent behaviour among humans. But: if a person rationally can expect others to return her altruism, the behaviour becomes much more likely. And I go to vote with the strong belief that others do it too.) Peter Barath Tavaszig, most minden féláron! ADSL Internet már 1 745 Ft/hó -tól. Keresse ajánlatunkat a http://www.freestart.hu oldalon! Election-Methods mailing list - see http://electorama.com/em for list info
[EM] Square Vote: an alternative to the Clarke-tax
No mistake: I'm a big fan of the Clarke-tax. And I don't think the fact that in the Clarke-tax real payment is rare, is a really big problem. But here is a proposal to solve this problem: I think the attribute of the Clarke tax that you pay only if you become a pivotal voter, can be substituted with every voter on the winning side having to pay, but only a small amount. The following are for only the situation with two options. Why work out more if nobody is interested? Besides, I don't yet have any idea how to do for more than two. We ask the voter to declare on the ballot, which is her option, and how much more money that option is worth for her, compared to the other one. In other words she have to price the difference, exactly as in the Clarke-tax. The option with the bigger sum of declared money wins and the winners pay for the treasury. But how much? If they have to pay the whole amount, voting will be uneconomic for them. What if we divide the amount with a constant? That wouldn't be good, this is why: We can make an estimation amout the probability distribution. Let D be the (money for option A) - (money for option B) difference. Let d be the probability of D being between $0 and $1. I call it the probability density, although I guess it has already a name but I don't know it. In a short area the density can be considered to be constant. So the probability of D being somewhere between $0 and $a is d*a If we make the winning voter to pay her declared money divided by a fixed number, her decision will be uncertain, because if for example she doubles the declared, the probability of making her option win is also doubles, and the money she pays in case of winning also doubles. No countable optimum. So let's make the declared -> payed function non-linear. It seems good if the function starts from the origo horizontally and its steepness grows. With very low declarations, the payment is almost 0, so it seems rational to declare more than 0. But in high numbers, a small increase of the declaration leads to very big increase of payment. An optimum must exist between. The simplest such function is the square ( x*x or x^2 ) function. Let the method be: - The voting authority makes an estimation for the d density and makes it public. - Every voter declares which option she wants and the value for that. - The option with the bigger sum wins. - If a voter is on the winning side, and she declared value a, she must pay to the treasury d*a^2 This will usually be a very small payment, because d is a very small number - the probability of a $0-$1 margin. What is the rational decision for the declared a money? Let P be the probability that her choice wins even without her. Let V be the value of winning for her. There are three cases: - Her side loses. Probability: 1-P-d*a She does not win the V value and doesn't pay anything. - Her side wins and would have won even without her. Probability: P She didn't win the V value (because her vote was unnecessary for that). And she have to pay d*a^2 so her gain is -d*a^2 - She becomes a pivotal voter. Probability: d*a Now she really gained the V value, and also have to pay so her gain is V-d*a^2 So her expected gain, the sum of gains multiplied with their respective probabilities is d*a*(V - d*a^2) - P*d*a^2 If we consider it a function of a and search for the maximum gain, we derivate (or how it called in English) it and search the a at which the derivated function has 0 value. It will be an equasion of a second degree, and the solution: a = ( SQRT( P^2 + 3*d*V ) - P ) / (3*d) There is a possible approximation that if X>>y then SQRT(X^2+y)-X approximately equal to y/(2*x) so here the optimal a approximately equal to ((3*d*V)/(2*P))/(3*d) that is V/(2*P) In a relatively close election P can be considered around 1/2 so the optimal declared a is closely equal to the real V value of the choice. I think we can not leave out the approximation that P is about 1/2. It would be somehow impolite if the election authority predicted which choice were more likely to win. But we can adjust the approximation in the optimum calculation if we use a more precisely chosen payment function, an infinite sum. instead of the d*a^2 the winner payment would be: (1/(4*d)) * ((2*d*a)^2 - (2*d*a)^3 + (2*d*a)^4 - (2*d*a)^5 + ) Peter Barath Tavaszig, most minden féláron! ADSL Internet már 1 745 Ft/hó -tól. Keresse ajánlatunkat a http://www.freestart.hu oldalon! Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Square Vote
On 2008 December 04 Thursday 23.46, Peter Barath wrote: > - The voting authority makes an estimation for the d density > and makes it public. > - Every voter declares which option she wants and the value for > that. > - The option with the bigger sum wins. > - If a voter is on the winning side, and she declared value a, > she must pay to the treasury d*a^2 By thinking a little more I think now: This method can be transformulated: everybody declares, how much she is willing to pay if her side wins, and when counting the votes the voting authority divides every vote with d (the predeclared estimated probability density) and take the square root of the result. Of course, the constant divider does not influence the result, so it can be left out of the play. So the method is: - Every voter declares how much she wants to pay. - The voting authority sums the square roots. - The bigger sum wins. (There would be voters asking: If an issue is twice as important for me, why to pay four times more? I would answer: twice for the twice more importance, and twice for the twice more probability to influence the result.) This way, the burden of estimating the probability density (the probability of the margin being between $0 and $+1) moves from the authority to the voter. (But it was there already: only with the option that she simply believe the "official" one.) The payments will be simply the declared ones. But who is to pay? Now, I don't thik this is a very simple question, but here is what I think: There are three possibilities: - Everybody pay. - The winners pay. - The winners pay, the losers get their declared money. Let's see in the first case, wher everybody pays, what the rational payment (which in this case can be paid when the actual voting takes place) is. Let d be that probability density which were real if the voters declared the square of their V values, V^2 Actually, they will declare a different (much smaller) sum. Let c be the constant averaging this effect so that the voters usually pay c^2 * V^2, so their counted sum is c*V But this way the real probability density will change to d/c Look at a single voter now, what her rational decision for the payment would be? Let z be her personal constant so she pays z^2 * V^2 counts z*V What is her expected value if she pays z^2 * V^2 instead of nothing? - z^2 * V^2 for she paid that amount, plus the bet: the probability multiplied with the value, that is the probability density multiplied with the counted sum multiplied with the value. density: d/c counted: z*V Value:V So her expected value is: (d/c)*z*V*V - z^2*V^2 The maximum is where z = d/(2*c) The equilibrium (is this the proper concept here?) is where the usual constant equals the usual contant, that is where z = c so the stable c is when c = d/(2*c) that is c^2 = d/2 So the rational payment for a voter with V value is (d/2)*V^2 So the counted rational sum is the square root of this, so is proportional to the value of the question, so this voting is sincere. If only the winners pay, the rational payment is (d/2)*V^2/P where P the probability of her side to win. So this is not sincere, it motivates the less likely winning side to "overstate" their value. If the losers get their declared money, the result is even worse. The likely losers will grossly overstate their values in hope of getting that money. Generally, this is my reaction to Abd: if losers get real compensations, I don't know how to defend against false claims. So my Square Vote proposal gets this simple: - Every voter pays whatever she wants to the treasury. - The square roots are counted. - Biggest sum wins. Of course, this way we can vote for more than 2 options, but I don't have enough math to decide whether sincerity remains. Peter Barath Tavaszig, most minden féláron! ADSL Internet már 1 745 Ft/hó -tól. Keresse ajánlatunkat a http://www.freestart.hu oldalon! Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Square Vote
>So my Square Vote proposal gets this simple: > >- Every voter pays whatever she wants to the treasury. >- The square roots are counted. >- Biggest sum wins. Sorry, I failed again to invent something new. As someone kindly informed me, this "...proposal has a resemblance to Hylland, A., Zeckhauser, R., August 1979. A mechanism for selecting public goods when preferences must be elicited. Tech. Rep. KSG Discussion Paper 70D, Harvard University. which is mentioned, among other places, at http://mpra.ub.uni-muenchen.de/3964/ Peter Barath Tavaszig, most minden féláron! ADSL Internet már 1 745 Ft/hó -tól. Keresse ajánlatunkat a http://www.freestart.hu oldalon! Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] IRV and Brown vs. Smallwood
> Yet me give you an example > Vote for mayor: > We have three candidate running for mayor > Vote for one: > [ ] Smith > [ ] Jones > [ ] Johnson > > If Smith has the least number of votes and is eliminated then who > would you vote for. > [ ] Jones > [ ] Johnson > > If Jones has the least number of votes and is eliminated then who > would you vote for. > [ ] Smith > [ ] Johnson > > If Johnson has the least number of votes and is eliminated then who > would you vote for. > [ ] Smith > [ ] Jones > > This is clearly consistent with one-person one-vote. > Every voter votes once and all voters are treated equal. > You could say that each voter is voting twice once in the first part > and a contingency vote in the second part depending on who is > eliminated. This is conceptually the same as IRV method, with one > exception. IRV assumes voters are rational and that the votes are > independent. > In this example a voter could vote for Jones on the first part and > if Smith is eliminated then vote for Johnson in the second part. > It is hard to understand the rationality of this kind of vote. For punctuality's sake: I think this kind of vote would be rare, but sometimes rational. Suppose that by my estimation about the electorate is about 400: Smith, Jones, Johnson 300: Jones, Smith, Johnson 600: Johnson, Jones, Smith and I happen to belong to the latest, estimatedly 600 strong group. I am afraid to vote first for my favourite Johnson, because Jones is likely to drop first, and then the hated Smith is almost sure to win. So I make a compromise and vote for Jones, sacrificing my favourite Johnson. Suppose that my estimation proves to be wrong - it can happen even to a rational person - and Smith drops first. Then I would happily vote for Johnson. The thing can happen even without wrong estimation. If we Johnsonists are smart enough then 150 of us vote for Jones and 450 vote for Johnson, so we _make_ Smith drop first. Of course, we don't have much hope to win in the second round but no reason for not to try. Still I don't think IRV violates the one person one vote principle. This principle is not to apply in a what-would- happen-if-using-not-this-but-another-method style. The question is what a vote is. If a vote is a paper which can be filled, then of course IRV is one person one vote. If a vote is a checkmark on a paper then of course IRV is one person one vote. The principle doesn't say a person have one vote in her entire life! Of course, a paper can contain one first vote, one second vote, one third vote and so on, as long as these possibilities are for everyone. The voter has the right not to use all thiese votes - in the US, voting is not compulsory (yes, I know there are countries where it is). Peter Barath PUMA termékek akár 70%-kal olcsóbban CSAK MOST, rendkívüli akció. Katt ide >> http://www.brands.hu/origo Hetente a legnagyobb márkák akciói webáruházunkban! Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Multiwinner Election Algorithm
> Set X and set Y are adjacent if it is possible to create one group by > changing a single candidate in the other. ...in other words, all the members > are identical but one. > Set X is a local maximum if the utility of every adjacent set is less than > Set X´s utility. > > The utility function is rather simple. > > > for each voter, the utility is ln(1+score_sum/max) > > with score_sum being the score they gave each candidate individually > and max being the maximum rating allowable for a single candidate. > I, however, lack the expertise to prove whether it is possible for multiple > local maxima to occur. I was wondering if anyone could. Let's try it. Let's take the simplest situation where there are non-adjacent sets. The minimum for this is 4 candidates (ABCD), 2 seats. For the simplicity, we suppose the rating can be between 0 to 1. Our voters will all vote approval-style, that is all 0 and 1 ratings. And each of them will vote for exactly two candidates. For example, AB and CD are non-adjacent sets. Let's try it with 2 voters, one votes AB, one CD. The situation is quite simmetrical, so we just compare sets AB and BC for the example. The utility for the AB set is ln(1+2) from the AB voter and zero from the CD voter. The utility for the BC set is ln(1+1) from the AB voter and another ln(1+1) from the CD voter. That is, the utility for the AB set is ln(3) and the utility for the BC set is 2*ln(2) = ln(4) so this is not a counter-example, the non-adjacent AB and CD are not local maxima; the principle of "many dwarfs are stronger than a few giants" worked. So let's try it ortogonally. Now let's have all kind of voters except AB and CD. So we have four voters, each of them votes for one candidate from the AB group and one from the CD group. So we have voters AC,AD,BC,BD. Now, for example the AB group means partial utility for all four voters, while for example BC means more utility for the BC voter, but nothing for the AD voter. The AB set means ln(1+1) for all four voters, so its full utility is 4*ln(1+1) = ln(16) The BC set means ln(1+2) for the BC voter, ln(1+1) for the AC voter, also ln(1+1) for the BD voter and zero for the AD voter. All together ln(3) + 2*ln(2) = ln(3) + ln(4) = ln(12) Similarly for the others, so the utilities for the sets: AB ln(16) AC ln(12) AD ln(12) BC ln(12) BD ln(12) CD ln(16) The non-adjacent AB and CD are local maxima. Peter Barath Elindult a Genertel lakásbiztosítása!Számítsa ki díját, kösse meg lakásbiztosítását online, néhány perc alatt.1 év lakás assistance szolgáltatást most ingyen adunk!http://ad.adverticum.net/b/cl,1,6022,340633,421168/click.prm Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] SEC quickly maximizes total utility in spatial model
Jobst Heitzig wrote: > Dear folks, > > earlier this year Forest and I submitted an article to Social Choice and > Welfare (http://www.fair-chair.de/some_chance_for_consensus.pdf) > describing a very simple democratic method to achieve consensus: I looked at it, and have to admit that my math knowledge is not enough to follow it fully in reasonable time. >> Simple Efficient Consensus (SEC): >> = >> >> 1. Each voter casts two plurality-style ballots: >> A "consensus ballot" which she puts into the "consensus urn", >> and a "favourite ballot" put into the "favourites urn". >> >> 2. If all ballots in the "consensus urn" have the same option ticked, >> that option wins. >> >> 3. Otherwise, a ballot drawn at random from the "favourites urn" >> decides. > > > This method (called the "basic method" in our paper) solves the problem > of how to... > >> make sure option C is elected in the following situation: >> >> a% having true utilities A(100) > C(alpha) > B(0), >> b% having true utilities B(100) > C(beta) > A(0). >> >> with a+b=100 and a*alpha + b*beta > max(a,b)*100. >> (The latter condition means C has the largest total utility.) Still, I have the very strong feeling that that claim is not part of your above mentioned paper and also it is not true. Counter-example: a = 40 b = 60 alpha = 10 beta = 99 the condition is true: max(a,b)*100 = 60*100 = 6000 a*alpha + b*beta = 40*10 + 60*99 = 400 + 5940 = 6340 So C does have the largest total utility. Can be sure option C is elected? As far as I remember, the paper doesn't say anything about the decision-making mechanisms in such situations. It always assumes that enough participants prefer this or that above the lottery. But here in your post you didn't say "above the lottery", you said "has the largest total". And I think in such situation many "A" voters including myself would prefer the lottery with 40% chance to the 100 value option over the sure 10 value. So C wouldn't be elected. > Since then I looked somewhat into spatial models of preferences and > found that also in traditional spatial models, our method has the nice > property of leading to a very quick maximization of total utility (the > most popular utilitarian measure of social welfare): > > Assume the following very common spatial model of preferences: Each > voter and each option has a certain position in an n-dimensional issue > space, and the utility a voter assigns to an option is the negative > squared distance between their respective positions. Also assume that > voters can nominate additional options for any "in-between" position (to > be mathematically precise, any position in the convex hull of the > positions of the original options). > > Traditional theory shows that, given a set of voters and options with > their positions, total utility is maximized by the option closest to the > mean voter position, but many traditional voting methods fail or > struggle to make sure this option is picked. > > With our method SEC, however, total utility will be maximized very > quickly: If the "optimal" option X located at the mean voter position is > already nominated, every voter will have an incentive to tick X on her > "consensus ballot" since she will prefer X to the otherwise realized > fall-back lottery that picks the favourite of a randomly drawn voter. If > X is not already nominated, every voter will have an incentive to > nominate X for the same reason. This makes sure X is elected and thus > total utility is maximized. Still I can't comprehend the full mathemathic background, but look at this example: An economic community with a common wealth decides about their future: Option "Dismiss": dismiss the community by sharing equally the wealth, and everyone does what she wants with it. Option "Salary": work as a cooperative, still common wealth, but members get different payment by their work. Option "Equality": work as a classic kibbutz, equal living conditions, no money. The utility for the 40 "Dismissists": Dismiss(100) Salary(10) Equality(0) For the 20 "Salarists": Dismiss(10) Salary(100) Equality(30) For the 40 "Equalists": Dismiss(0) Salary(80) Equality(100) For me it looks here the Salarists are the median voters, and also the "Salary" option has the largest total. And again, it looks that a typical Dismissist will go for the 40% lottery instead of accepting the low-value compromise. All these don't make the proposals necessarily look bad in
[EM] (no subject)
Admitting that I didn'f fully follow the topic: I think my selfish incentives are enough to make me vote. Maybe I have also altruistic incentives but they are surplus. Also, my selfish incentives in great part have ethical and community nature, but still selfish. How can a selfish motive have ethical nature? Simple. If I don't steal an exotic fruit from the supermarket, my motives have ethical nature, but some of them selfish: I don't want to be punished. So, what are my costs about the voting? Five minute walk to the place, five minute vote, and five minute walk back. (And consider that I like to walk, sometimes I do it just for recreation.) Knowing about politics I don't count as a cost. Even if I didn't have the right to vote, I would know about politics to make decisions about my life, to not look dumb when conversating, and from simple curiosity, which means something like hoping to utilize knowledge maybe somewhere, some time, in some field (but it's not as much a decision, as an instinct, evolution-made - not all of those work well, for example, our instinctous carbohydrate craving can make us less healthy, but curiosity is still okayy). So let's make the cost a dollar (this is not a very high GDP per capita country - Hungary). Let's see the plus side. In my country a million voter minus already would somehow endanger democracy. Let's suppose it's a 0.1 probability of a fascist dictatorship which kills me or makes my life as miserable as death with a 0.1 probability. So roughly my voting makes my life 0.0001 safer. Question is wheter my life is worth a hundred million dollars. I'm not sure. It's also important to note that in such magnitudes utility can not be considered as a linear function of money. (For Bill Gates, a million dollar plus doesn't mean nearly as much as for me would.) As a matter of fact, my voting or not voting makes more than one vote plus or minus because other people tend in this respect more to follow than to counterfollow my example. Since voting is considered as an ethical act, I vote to make my reputation better. I could maybe lie in this respect, but lying also has high costs. My family members know when I'm coming and going, I can be caught if something interesting happens in my voting place and I don't know about. This can be considered like this: for some extent I also protect other people from a fascist dictatorship, and they also protect me. So we have an agreement to vote. But my keeping of the agreement is not fully altruistic because others know what I do. Also, if something interesting happens in my voting place, it's good for me, I can talk about it, I can get some attention in the company, which is such a hard thing to do. Even if nothing particular happens, voting is a little bit fun. Maybe I forgot something, I don't know. Just one more note: Since the previus elections Hungarian Parlament created a law to make cheating less probable. By this, in every constituency, only one voting place is able to get votes from people who don't vote in their own dwelling place. In these places, some people waited for six hours in line to vote. I don't know what I would have done in this situation. Probably would have wait for my turn. Why? Would have this been able to explain by purely selfish motives? Maybe. Becaus if I do my thing in such circumstances, it makes me look even better before others. The probability of something interesting happens grows. And not only something to talk about: it can be an interesting experience. What people say in this situation? How the authorities react? Taking part is sometimes valueable because media often lie about events. If you are there, you have chance to know. Peter Barath http://ad.adverticum.net/b/cl,1,73468,1603402,1600294/click.prm"; target="_blank">Autót vásárol? Balesetmentesen vezet? Genertel kötelező szenzációs kedvezménnyel! Election-Methods mailing list - see http://electorama.com/em for list info
