Mr. Eppley wrote: Isn't it possible for the votes against an unbeaten candidate in a pair-tie to be worse than the Condorcet score of a beaten candidate? A B C A 50= 2L B 50= 35 C 1 45L ---- ---- ---- ? ? 2 <--- Maximum votes against. Count the ties? A (and only A) is unbeaten and wins by our definition. But A's 50 in the tie is larger than C's 2 in C's only loss. Is A really less beaten than C? I hope I'm not muddying the waters. The odds against a pair-tie in a large election are negligible. ----- Demorep1--I thought that a candidate must get more votes than each other candidate to be the Condorcet winner (i.e. a tie is not a win). Thus, is A the winner ? Does any candidate in the example have majority approval on a yes/no vote ? Are there overlapping votes in the 6 pairings- 1 CA 2 AC 35 BC 45 CB 50 AB 50 BA (such as an ACB vote) ? With a low number of voters, as in many clubs, ties are a distinct possibility.