The votes are cast as follows:
AC 498 BC 497 CA 3 CB 2
Under condorcet candidate C as the least objectionable wins.
Under what method did you compute C as the winner?
Assuming I entered things correctly, you have a pairwise matrix of:
Option A B C A 0 0 498 B 0 0 497 C 3 2 0
Now, using my web site to compute the winner (http://www.ericgorr.net/condorcet/), Basic (Plain) Condorcet, Beatpath Winner and Ranked Pairs(Deterministic #1 - what Mike Ossipoff was calling it), I show that A & B tie in all three methods.
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