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
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
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
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
Therefore the Condorcet Criterion is incompatible with FBC.
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