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)