Hi, I thought about this a bit. Consider this election:
49 A 24 B>E 27 C>D>B>E C has 3 wins, and is the only Copeland winner. Woodall's plurality criterion is violated, since there's no way to raise C in rankings including C so that C has even the first preference count that A starts with. Consider this: 49 A>F 24 B 27 C>G>B A has 3 wins, and is the only Copeland winner. Eppley's minimal defense criterion is violated, because there is no way for the C>G>B voters (with the B voters) to at least elect B, without insincerely ranking B above G (for a tie with A) or both C and G (to win). So when I told Rob I couldn't advocate Copeland unless the tie-breaker satisfied minimal defense, I was talking about something impossible. Kevin Venzke ___________________________________________________________________________ Appel audio GRATUIT partout dans le monde avec le nouveau Yahoo! Messenger Téléchargez cette version sur http://fr.messenger.yahoo.com ---- Election-methods mailing list - see http://electorama.com/em for list info