There's been a lot of argument on the list over how to resolve the lack of a Condorcet winner. With >3 members in the Smith set I see some easy resolutions that aren't feasible for 3 members.
With 4 members, 2 of them will have 2 victories apiece and 2 of them will have 1 victory apiece (only counting victories against other members of the Smith set). It would seem reasonable to limit our attention to the 2 with the most victories, and then select whichever of those defeats the other. With 5 members, 3 structures are possible: 1) All 5 have 2 victories apiece 2) One has 3 victories, 3 have 2 victories, 1 has 1 victory. 3) 2 have 3 victories, 2 have 1 victory, 1 has 2 victories. For case 2 it seems reasonable to elect the person with 3 victories. For case 3 it seems reasonable to pick whichever of the top 2 can defeat the other. For case 1 the problem is similar to when you have 3 candidates in the Smith set. I won't enumerate all of the possibilities with six members, but by drawing diagrams I've seen that you can have a case where 3 members win equal numbers of victories, and more victories than any other candidate. Among those 3 there can be a "Condorcet Winner," or not, depending on the results. I realize that elections with more than 3 candidates in the Smith set are rather unlikely, but the case with 4 is not totally out of the question, and it seems to present an easy resolution. Does anybody see a problem with that method of resolution for 4 members in the Smith set? I haven't thought about it in depth. Alex Small