>Condorcet Ranked Pairs obviates the need for tie breakers anyway. > >(I realize now that perhaps I wasn't too clear when I said I thought any >Condorcet method was better than IRV, but that you needed a way to break >out of Condorcet "ties" (cycles, really); Ranked Pairs is a Condorcet >sub-method which does that. Apoligies for any confusion I caused.) My understanding was that there is a possibility of a three-way tie, where a given three entrants would each be ranked higher than one of the other two and lower than the other, resulting in an M.C. Escheresque circular paradox, thus requiring some kind of run-off vote. So I am mistaken and the ranked pairs method can handle this?