Copeland's method: Difference between revisions

=== Schwartz ===

In this 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 alternative 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.