Abd wrote: > 34 A > 33 B>C > 33 C>B. > > The Condorcet winner is A, because in the two pairwise > elections involving A, A wins > > A>B, 34:33 > A>C, 34:33.
Assuming that by the above votes you mean 34:A>B=C 33:B>C>A 33:C>B>A, A is not the Condorcet winner and is in fact the Condorcet loser, losing both A:B and A:C by 34:66. Perhaps you had in mind an example like 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. -- Rob LeGrand [email protected] Citizens for Approval Voting http://www.approvalvoting.org/ ---- Election-Methods mailing list - see http://electorama.com/em for list info
