Forest, In reference to your new Condorcet method suggestion (pasted at the bottom), which elects an uncovered candidate and if there is none one-at-time disqualifies the Range loser until a remaining candidate X covers all the other remaining candidates and then elects X, you wrote:
"Indeed, the three slot case does appear to satisfy the FBC..". No. Here is my example, based on that Kevin Venke proof you didn't like. Say sincere is 3: B>A 3: A=C 3: B=C 2: A>C 2: B>A 2: C>B 1: C Range (0,1,2) scores: C19, B17, A12. C>B 8-5, B>A 10-5, A>C 7-6. C wins. Now we focus on the 3 B>A preferrers. Suppose (believing the method meets the FBC) they vote B=A. 3: B=A (sincere is B>A) 3: A=C 3: B=C 2: A>C 2: B>A 2: C>B 1: C Range (0,1,2) scores: C19, B17, A15. C>B 8-5, B>A 7-5, A>C 7-6. C still wins. Now suppose they instead rate their sincere favourite Middle: 3: A>B (sincere is B>A) 3: A=C 3: B=C 2: A>C 2: B>A 2: C>B 1: C Range (0,1,2) scores: C19, A15, B12. A>B 8-7, A>C 7-6, C>B 8-5 Now those 3 voters get a result they prefer, the election of their compromise candidate A. Since it is clear they couldn't have got a result for themselves as good or better by voting B>A or B>C or B this is a failure of the FBC. Chris Benham ________________________________ From: "[email protected]" <[email protected]> Sent: Wednesday, 23 November 2011 9:01 AM Subject: Re: An ABE solution voters to avoid the middle slot. Then the method reduces to Approval, which does satisfy the FBC." The FBC doesn't stipulate that all the voters use "optimal strategy", so that isn't relavent. http://wiki.electorama.com/wiki/FBC http://nodesiege.tripod.com/elections/#critfbc Chris Benham Forest Simmons wrote (17 Nov 2011): Here’s my current favorite deterministic proposal: Ballots are Range Style, say three slot for simplicity. When the ballots are collected, the pairwise win/loss/tie relations are determined among the candidates. The covering relations are also determined. Candidate X covers candidate Y if X beats Y as well as every candidate that Y beats. In other words row X of the win/loss/tie matrix dominates row Y. Then starting with the candidates with the lowest Range scores, they are disqualified one by one until one of the remaining candidates X covers any other candidates that might remain. Elect X. You are right that although the method is defined for any number of slots, I suggested three slots as most practical. So my example of two slots was only to disprove the statement the assertion that the method cannot be FBC compliant, since it is obviously compliant in that case. Furthermore something must be wrong with the quoted proof (of the incompatibility of the FBC and the CC) because the winner of the two slot case can be found entirely on the basis of the pairwise matrix. The other escape hatch is to say that two slots are not enough to satisfy anything but the voted ballots version of the Condorcet Criterion. But this applies equally well to the three slot case. Either way the cited "therorem" is not good enough to rule out compliance with the FBC by this new method. Indeed, the three slot case does appear to satisfy the FBC as well. It is an open question. I did not assert that it does. But I did say that "IF" it is strategically equivalent to Approval (as Range is, for example) then for "practical purposes" it satisfies the FBC. Perhaps not the letter of the law, but the spirit of the law. Indeed, in a non-stratetgical environment nobody worries about the FBC, i.e. only strategic voters will betray their favorite. If optimal strategy is approval strategy, and approval strategy requires you to top rate your favorite, then why would you do otherwise? Forest ----- Original Message ----- From: Chris Benham Forest, "When the range ballots have only two slots, the method is simply Approval, which does satisfy the FBC." When you introduced the method you suggested that 3-slot ballots be used "for simplicity". I thought you might be open to say 4-6 slots, but a complicated algorithm on 2-slot ballots that is equivalent to Approval ?? "Now consider the case of range ballots with three slots: and suppose that optimal strategy requires the
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