Copeland's method: Difference between revisions
→Criteria
No edit summary |
|||
Line 14:
| page=317-328
| url=https://www.researchgate.net/publication/220007301_Ramon_Llull_From_Ars_Electionis_to_Social_Choice_Theory
}}</ref>
(Some alternative versions of Copeland don't count pairwise defeats, and others give each candidate in a pairwise tie half a point each.)
Proponents argue that this method is more understandable to the general populace, which is generally familiar with the sporting equivalent. In many team sports, the teams with the greatest number of victories in regular season matchups make it to the playoffs.
Line 21 ⟶ 23:
== Criteria ==
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, they can't be pairwise defeated by themselves, and 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) -
Copeland's method also passes [[ISDA]]; since the Copeland winner is always in the Smith set, all candidates in the Smith set must have higher Copeland scores than all candidates not in the Smith set, and since by definition candidates in the Smith set have a pairwise victory against every candidate not in the Smith set, adding or removing any number of candidates not in the Smith set will only result in every candidate in the Smith set having that number of pairwise victories added or subtracted from their total; since the original Copeland winner must have had a higher Copeland score than all other Smith set candidates in order to win, they will still have a higher Copeland score and thus still win.
| |||