Talk:Condorcet winner criterion: Difference between revisions
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(→Smith set with two members: can't have a Smith set of only two members) |
Psephomancy (talk | contribs) (respond to User:RobLa) |
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: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) |
: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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:: Doesn't "weak" mean the two are tied? — [[User:Psephomancy|Psephomancy]] ([[User talk:Psephomancy|talk]]) 06:23, 5 January 2020 (UTC) |
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