Condorcet winner criterion: Difference between revisions
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The '''Condorcet candidate''', '''Pairwise Champion''' (PC), '''beats-all winner''', or '''Condorcet winner''' (CW) of an [[election]] is the candidate who is preferred by more voters than any other candidate. This is determined by observing whether more voters rank or score the Condorcet winner higher than each of the other candidates than the other way around.
The '''Condorcet criterion''' for a [[voting system]] is that it chooses the beats-all winner when one exists. Any method conforming to the Condorcet criterion is known as a [[Condorcet method]].
On a one-dimensional [[political spectrum]], the
Mainly because of [[Condorcet paradox|Condorcet's voting paradox]], a
== A more general wording of Condorcet criterion definition ==
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It has been recently argued that the definition of the verb "beat" should be regarded as external to the Condorcet Criterion...and that "beat should be defined in a way that interprets equal-top ranking consistent with the actual preferences, intent and wishes of the equal-top-ranking voters. When such a definition of "beat" is used in the Condorcet Criterion definition, then the Condorcet Criterion is compatible with [[FBC]], and there are Condorcet methods that pass [[FBC]]. Discussion and arguments on that matter can be found at the [[Symmetrical ICT]] article.
[[Approval voting]], [[Range voting]], [[Borda count]], [[plurality voting]], and [[instant-runoff voting]] do not comply with the Condorcet Criterion. However, any voting method that collects enough information to detect pairwise preferences (i.e. scoring or ranking methods) can be "forced" to comply with the Condorcet criterion by automatically electing the Condorcet winner if one exists (or alternatively, eliminating all candidates not in the Smith Set) before doing anything else.
==Commentary==
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according to the Condorcet criteria, B should win. By contrast, according to the rules of IRV, B is ranked first by the fewest voters and is eliminated, and C wins with the transferred voted from B; in plurality voting A wins with the most first choices.
Range voting does not comply because it allows for the difference between 'rankings' to matter. E.g. 51 people might rate A at 100, and B at 90, while 49 people rate A at 0, and B at 100. Condorcet would consider this 51 people voting A>B, and 49 voting B>A, and A would win. Range voting would see this as A having support of 5100/100 = 51%, and B support of (51*90+49*100)/100 = 94.9%; range voting advocates would probably say that in this case the Condorcet winner is not the socially ideal winner. In general however, it is expected that the Condorcet winner (and Smith Set candidates in general) will almost always be very high-utility when compared to the utilitarian winner.
Sometimes there is no Condorcet winner, but there may be candidate(s) who are preferred by at least as many voters as all other candidates (i.e. as many voters rank or score them higher or equally as each of the other candidates as the other way around), who are known as weak Condorcet winners. While it may thus seem reasonable that a Condorcet method should pass a condition of always electing solely from the set of weak Condorcet winners when no regular Condorcet winner exists and at least one weak Condorcet winner exists, this guaranteeably leads to failures of reversal symmetry and clone immunity. Example (parentheses are used to indicate implied rankings):<blockquote>3 A(>B1=B2=B3)
1 B1>B2>B3(>A)
1 B2>B3>B1(>A)
1 B3>B1>B2(>A)</blockquote>A is the only weak Condorcet winner here, as they tie 3 to 3 when compared to B1, B2, and B3, and each of the latter 3 candidates suffer at least one pairwise defeat, and since there is no regular Condorcet winner (since every candidate has at least one pairwise tie or defeat), by the above-proposed condition A must win. To show a failure of reversal symmetry, suppose the ballots are reversed:<blockquote>3 B1=B2=B3(>A)
1 A>B3>B2>B1
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>
== Multi-winner generalizations ==
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