Independence of Irrelevant Alternatives and the Condorcet Loser
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https://electowiki.org/wiki/Condorcet_loser_criterion says:
Given a three-candidate Condorcet cycle, it's always possible to eliminate a candidate who didn't win so that the winner changes. Thus the Condorcet loser criterion is incompatible with independence of irrelevant alternatives.
Is this correct?
Consider two systems. System A is straight Score and System B is to eliminate the Condorcet loser (beaten-by-all-pairwise candidate) if there is one repeatedly and then apply Score to the rest.
If there are only three candidates in the election and they form a Condorcet cycle, there is no Condorcet loser so systems A and B produce the same outcome. How does this show a dependence on irrelevant alternatives? Isn't plain Score famously compliant with IAA?
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@jack-waugh You missed an important part of the quote.
In an election with only two candidates, the Condorcet loser criterion implies the majority criterion. Given a three-candidate Condorcet cycle, it's always possible to eliminate a candidate who didn't win so that the winner changes. Thus the Condorcet loser criterion is incompatible with independence of irrelevant alternatives.
Say you've got an A>B>C>A cycle. Without loss of generality, A wins in some method that passes the Condorcet loser criterion.
Now remove B. C beats A head-to-head so must now win (A is the Condorcet loser with only A and C). Given that removing B has changed the winner from A to C, the method must fail IIA.