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Copeland's method: Difference between revisions
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{{Wikipedia}}
'''Copeland's method''' is a [[Smith criterion|Smith-efficient]]<ref>http://dss.in.tum.de/files/brandt-research/choicesets.pdf "The Copeland set C is given
by [...] i.e., the set of alternatives with maximal Copeland score." "Theorem 1. The Copeland set [...] [is] contained
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}}</ref>
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.
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Example:
{{ballots|
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.
==Criterion compliances==
Copeland's method passes the [[Smith criterion]] because any candidate in the Smith set by definition beats everybody outside of the Smith set, but no candidate outside of it does so. Thus, every candidate in the Smith set must have a greater Copeland score than any candidate outside of it. Furthermore, the Copeland ranking of candidates (the ordering of candidates based on Copeland score) is a [[Smith set ranking]], since every candidate in the nth Smith set defeats every candidate in the (n+1)-th Smith set, and thus a candidate from the former set always has a greater Copeland score than a candidate from the latter. ▼
===Smith===
▲Copeland's method passes the [[Smith criterion]] because any candidate in the Smith set by definition beats everybody outside of the Smith set, but no candidate outside of it does so. For any candidate X in the Smith set and Y outside of it, Y is defeated by at least as many candidates as X, and X defeats at least one candidate that Y doesn't. Thus, every candidate in the Smith set must have a greater Copeland score than any candidate outside of it. Furthermore, the Copeland ranking of candidates (the ordering of candidates based on Copeland score) is a [[Smith set ranking]], since
(Example showing Smith members having only 2 points more than non-Smith members: Suppose there are two candidates, one of whom is the Condorcet winner, and thus the only candidate in the Smith set. The CW has one victory and no defeats for a Copeland score of 1, while the other candidate has no victories and one defeat for a score of -1.)
===Independence of Smith-dominated alternatives===
Copeland's method also passes [[ISDA]]: since every candidate in the Smith set beats everybody outside it, eliminating a candidate outside of the Smith set will subtract one win from the score of every candidate in that Smith set. Thus eliminating a Smith-dominated candidate can never change the relative Copeland scores of candidates in the Smith set, and thus not change the winner either.
===Uncovered set===
Further, Copeland always elects from the [[uncovered set]], and the Copeland ranking is an uncovered set ranking. This is because when one candidate covers another, the former candidate pairwise beats all candidates pairwise beaten by the latter candidate, and also either pairwise beats the latter candidate or beats someone who beats the latter candidate. Because of this, the covering candidate will have a minimum Copeland score of ((number of candidates beaten by latter candidate) + 1) - (number of candidates beating former candidate)), and the covered candidate will have a maximal Copeland score of ((number of candidates beaten by latter candidate) - ((number of candidates beating former candidate) + 1), resulting in the covering candidate having at least 2 more points than the covered candidate. This type of logic can be used to simplify the above Smith set-related proofs too.
==Criterion failures==
== Generalizations==▼
=== Schwartz ===
In the election
{{ballots|
1: A > B3 > B1 > B2 > C
1: B3 > B1 > B2 > C > A
2: C > A > B2 > B1 > B3
}}
the Schwartz winner is C, but Copeland elects A.
=== Independence of clones ===
Copeland's method is vulnerable to crowding. This example is [[w:Independence_of_clones_criterion#Copeland|due to Wikipedia]].
First consider the election
{{ballots|
1: A>B>C
1: B>C>A
2: C>A>B}}
C is the Condorcet winner and thus also the Copeland winner. Now clone B into B1, B2, and B3:
{{ballots|
1: A > B3 > B1 > B2 > C
1: B3 > B1 > B2 > C > A
2: C > A > B2 > B1 > B3
}}
The Copeland winner changes to A.
Since the Copeland winner is unique in both cases, every method that elects from the Copeland set must also fail clone independence.
=== Independence of covered alternatives ===
Unlike the Smith set, the Copeland set is not independent of alternatives not in it. It's possible for an election to contain a candidate X that is not in the Copeland set, but when removed, changes what candidates are in that set. An election set by Forest Simmons<ref>{{cite web|url=http://lists.electorama.com/pipermail/election-methods-electorama.com/2010-July/091997.html|title=independence form covered alternatives is incompatible with monotonicity|website=Election-methods mailing list archives|date=2010-07-16|last=Simmons|first=F.}}</ref> can be used to show that Copeland is not independent of covered alternatives. First, consider the election
{{ballots|
40: D>B>C>A
30: A>B>C>D
30: C>A>D>B
}}
In this election, the Copeland set consists of A and C. The uncovered set is {A, B, C}. Now eliminate D, who is a covered candidate, and the election becomes
{{ballots|
40: B>C>A
30: A>B>C
30: C>A>B
}}
where every candidate is in the Copeland set. Thus eliminating a covered alternative changed the Copeland set.
=== Dominant mutual third burial resistance ===
Copeland fails dominant mutual third burial resistance.<ref>{{cite web|url=http://lists.electorama.com/pipermail/election-methods-electorama.com/2022-June/003987.html|title=Copeland DMTBR (DMTCBR) failure|website=Election-methods mailing list archives|date=2022-06-26|last=Munsterhjelm|first=K.}}</ref>
Zoghi et al have developed a multi-armed bandit variant of Copeland's method. It can be used to determine a winner in a multi-armed bandit setting, even if a Condorcet winner does not necessarily exist.<ref>{{Cite journal|last=Zoghi|first=Masrour|last2=Karnin|first2=Zohar|last3=Whiteson|first3=Shimon|last4=de Rijke|first4=Maarten|date=2015-05-31|title=Copeland Dueling Bandits|url=http://arxiv.org/abs/1506.00312|journal=arXiv:1506.00312 [cs]}}</ref>
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==See also==
# E Stensholt, "[http://www.electoral-reform.org.uk/publications/votingmatters/P2.HTM Nonmonotonicity in AV]"; Electoral Reform Society ''Voting matters'' - Issue 15, June 2002 (online).
#A.H. Copeland, A 'reasonable' social welfare function, Seminar on Mathematics in Social Sciences, University of Michigan, 1951.
# V.R. Merlin, and D.G. Saari, "Copeland Method. II. Manipulation, Monotonicity, and Paradoxes"; Journal of Economic Theory; Vol. 72, No. 1; January, 1997; 148-172.
#D.G. Saari. and V.R. Merlin, 'The Copeland Method. I. Relationships and the Dictionary'; Economic Theory; Vol. 8, No. l; June, 1996; 51-76.
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[[Category:Smith-efficient Condorcet methods]]
[[Category:Condorcet-related concepts]]
[[Category:Ranked voting methods]]
[[Category:Condorcet methods]]
[[Category:Monotonic electoral systems]]
{{fromwikipedia}}
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