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Talk:Condorcet winner criterion: Difference between revisions
→Smith set with two members: can't have a Smith set of only two members
(Created page with ""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...") |
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== 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? [[User:BetterVotingAdvocacy|BetterVotingAdvocacy]] ([[User talk: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. -- [[User:RobLa|RobLa]] ([[User talk:RobLa|talk]]) 04:32, 5 January 2020 (UTC)
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