Kevin,
You wrote:
I was quite wrong about MMPO being unable to elect the Condorcet Loser unless
all candidates have a majority-strength loss. As an example,
48 A
2 B=A
2 B=C
48 C
MMPO elects B decisively. So MMPO fails Plurality even worse than I thought.
I don't know how I forgot this; two-slot MMPO was the first method I advocated
on this list, and I knew it had this problem then.
A>B 48-2, C>B 48-2, A=C 50-50.
I note that all the candidates are in the CDTT (because none of them
have a majority-strength defeat), and therefore your example also
applies to CDTT,MMPO.
Also we can add one A ballot, making A the CW (and FPP winner), and get
the same result with both methods! (C drops out of the CDTT, but that
doesn't change the winner.)
Even if we also subtract a C ballot (restoring the total to 100
ballots), then the result for both methods is an AB tie, and if that is
resolved by Random Ballot then B will have a
4% chance of winning.
Chris Benham
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