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 passesis [[ISDA]];Smith-efficient sinceis as follows: every candidate in the CopelandSmith winnerset ishas alwaysa pairwise victory over every candidate not in the Smith set by definition, and at most has a pairwise defeat against all candidatesbut one candidate other than themselves not in the Smith set must(since, haveif higherthey Copelandhad scoresa thanpairwise defeat against all candidates notother than themselves in the Smith set, andthen sincethey themselves would not be in the Smith set by definition), so all candidates in the Smith set have a pairwiseCopeland victoryscore againstof everyat candidateleast ((number of candidates not in the Smith set, adding or removing) any- ((number of candidates in the Smith set) - 1). Every candidate not in the Smith set willhas onlya resultpairwise indefeat against every candidate in the Smith set havingby thatdefinition, numberand ofcan at most have pairwise victories addedagainst orevery subtractedcandidate fromother theirthan total;themselves sincenot in the originalSmith set, thus their Copeland winnerscore mustcan haveat hadmost abe higher((number Copelandof scorecandidates thannot allin otherthe Smith set) - 1) - (number of candidates in orderSmith toset). winThus, theythe members of the Smith set will stillalways have a higher Copeland score andat thusleast still2 winpoints higher than the candidates not in the Smith set.