Rob LeGrand wrote (11 Feb 2010):
<snip>
35:A
32:B>C
33:C,
by which I mean
35:A>B=C
32:B>C>A
33:C>A=B.
In this example, C is the Condorcet winner even though C does not have a
majority over B. I can see how this example could be seen as an
embarrassment to the Condorcet criterion, in that a good method might not
choose C as the winner.
<end quoted message>
Rob,
Well I can't. Electing A would be a violation of the Minmal Defense criterion,
and electing B would violate Woodall's Plurality criterion and Condorcet Loser.
What "good method" do you have in mind that might not elect C?
And what's good about it?
Chris Benham
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