Talk:Condorcet criterion

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Smith set with two members[edit source]

"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_termsPsephomancy (talk) 00:21, 6 January 2020 (UTC)

Generalization[edit source]

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)

I can't seem to find a source saying indifference counts in favor of both, so I've changed the phrasing. Kristomun (talk) 14:45, 24 March 2020 (UTC)