Talk:Condorcet winner criterion
Smith set with two members
"If the pairwise champion exists, they will be the only candidate in the Smith set; otherwise, the Smith set will have three or more members."
If there are two weak Condorcet winners, wouldn't the Smith Set have exactly two members? BetterVotingAdvocacy (talk) 02:18, 4 January 2020 (UTC)
- It's pretty much impossible to have a Smith set of two members, because it's impossible to have a pairwise cycle with only two members. The Smith set can have three members (A, B, and C) when A beats B, B beats C, and C beats A. But it's not possible to have a Smith set with only two members (A and B), because if A beats B, then B can't beat A. -- RobLa (talk) 04:32, 5 January 2020 (UTC)
- Doesn't "weak" mean the two are tied? — Psephomancy (talk) 06:23, 5 January 2020 (UTC)
- Hmm, I guess you're right. I suppose it's possible to have a three-candidate election where two candidates have a Copeland win-loss-tie score of 1-0-1, and one candidate has a Copeland score of 0-2-0. The Smith set would have the two candidates with the 1-0-1 Copeland score. I vaguely recall thinking about this someone added this, but not objecting because it intuitively seemed correct.
- I admittedly had to look up what a "weak Condorcet winner" meant, which took me back to w:Condorcet_method, which has this definition:
- a candidate who beats or ties with every other candidate in a pairwise matchup. There can be more than one weak Condorcet winner.
- That article doesn't have a citation for that definition, though. What is the best citation for "weak Condorcet method"? -- RobLa (talk) 23:50, 5 January 2020 (UTC)
- Just to give an example, suppose you have 2 voters bullet voting A, 2 bullet voting B, and 1 bullet voting C. A and B pairwise tie, but both pairwise beat all other candidates (C). BetterVotingAdvocacy (talk) 19:26, 27 February 2020 (UTC)
I don't know of a citation, but it's listed here, too: Condorcet_method#Related_terms — Psephomancy (talk) 00:21, 6 January 2020 (UTC)
Generalization
User:Kristomun, your edit
- A method passes the M-seat Condorcet criterion [...] When M=1, the generalization reduces to the ordinary Condorcet criterion as long as the method passes the majority criterion.
has a bit of an issue, since the majority criterion seems to speak of absolute majorities, not majorities of voters with any preference between the candidates. Thus, if 35% of voters prefer A>B, 25% prefer B>A, and 40% have no preference, with A and B being the only candidates, A is a plurality's 1st choice, and the 1st choice of a majority of voters with any preference between A and B, but not an absolute majority. Should we make an edit addressing this? BetterVotingAdvocacy (talk) 03:08, 19 March 2020 (UTC)
Social utility efficiency
Some of Colin Champion's work might be relevant to show that Condorcet can produce high efficiencies, but is still bounded away from maximum efficiency. E.g. "Better than Condorcet?" on EM, 2022-02-01, does this for a contrived mixture model. Kristomun (talk) 12:11, 16 April 2024 (UTC)