Copeland's method: Difference between revisions

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).

 

The reasoning for why Copeland's method is Smith-efficient is as follows: every candidate in the Smith set has a pairwise victory over every candidate not in the Smith set by definition, and at most has a pairwise defeat against all but one candidate other than themselves not in the Smith set (since, if they had a pairwise defeat against all candidates other than themselves in the Smith set, then they themselves would not be in the Smith set by definition), so all candidates in the Smith set have a Copeland score of at least ((number of candidates not in the Smith set) - ((number of candidates in the Smith set) - 1). Every candidate not in the Smith set has a pairwise defeat against every candidate in the Smith set by definition, and can at most have pairwise victories against every candidate other than themselves not in the Smith set, thus their Copeland score can at most be ((number of candidates not in the Smith set) - 1) - (number of candidates in Smith set). Thus, the members of the Smith set will always have a Copeland score at least 2 points higher than the candidates not in the Smith set.