I told why CC is incompatible with FBC. Why the continuing debate?

MIKE OSSIPOFF nkklrp at hotmail.com
Tue, 18 Oct 2005 02:22:17 +0000

I told why Condorcet's Criterion is incompatible with FBC. In case that posting wasn't noticed, I'm re-posting that demonstation here. I'm not saying that it wasn't already demonstrated by Kevin. But I've posted my demonstration too, and here is is again:

Say there are two voters whose preferences are B>A>everyone else.

Say that, with the configuration of all the other voters' votes (which may or may not be sincere), and sincere ballots by those two voters, there is a circular tie.

Since the method is anonymous and neutral, then, whatever method is used for solving circular ties, it's possible that someone other than A or B could win. Someone who is, for those two voters, worse than A. So let's say that that is so.

Say that, with that votes configuration for the other voters, and with sincere voting by those two voters, A beats everyone but B.

Say that results in a cycle in which the completion method in use, the circular tiebreaker, chooses someone other than A and B, someone whom those two voters like less than A.

But B doesn't beat A by very much. He only beats him by 3 votes.

The two voters, by changing their voting, could change that pairwise result. Let's say that that is the only pairwise result that is close enough that they could influence it.

If those two voters ranked A and B equal 1st place, B would only beat A by one vote.

Then, if they reversed their preferences between A and B, and voted A>B>everyone else, then they make A into BeatsAll winner. Because the method is a Condorcet Criterion method, A wins.

They've improved the results for themselves by voting someone over their favorite.

The method therefore fails FBC. And I didn't say exactly what the method is. All that was assumed was that the method is one that meets the Condorcet Criterion.

Therefore the Condorcet Criterion is incompatible with FBC.

Mike Ossipoff

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