Condorcet winner criterion: Difference between revisions
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1 A>B1>B3>B2
1 A>B2>B1>B3</blockquote>Again A is the only weak CW here, with there being no regular CW, so the same condition holds that A must still win. To show a failure of clone immunity, suppose B2 and B3 drop out of the race:<blockquote>3 A(>B1)
3 B1(>A)</blockquote>Now both of A and B1 are weak CWs, because they both pairwise tie each other. In this particular example, since there is nothing that distinguishes either candidate from the other, the neutrality criterion requires that both A and B1 must have an equal probability of winning i.e. both must have a 50% chance. This means that removing clones of B1 increased B1's chances of winning (which were originally at 0%, since A was guaranteed to win earlier i.e. had a 100% chance of winning.) <ref>https://arxiv.org/abs/1804.02973v6 p. 206-207</ref>
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