Copeland's method: Difference between revisions
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This method often leads to ties in cases when there are multiple members of the [[Smith set]]; specifically, there must be at least five or more candidates in the Smith set in order for Copeland to not produce a tie for winner unless there are [[pairwise counting#Terminology|pairwise ties]]. Critics argue that it also puts too much emphasis on the quantity of pairwise victories rather than the magnitude of those victories (or conversely, of the defeats).
Example:
25 A>B>C
40 B>C>A
35 C>A>B
A beats B beats C beats A, so there is a [[Condorcet cycle]] between all candidates. Each candidate has one [[pairwise beat|pairwise victory]] and one defeat, so their Copeland scores are all 0, thus there is a tie. This example demonstrates why Copeland is almost never used for actual elections; it can guarantee someone in the [[Smith set]] will win, but says much less about who.
== Criteria ==
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