# 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)

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)