# [EM] Smith Sets with >3 members

Alex Small asmall at physics.ucsb.edu
Tue Feb 26 12:36:23 PST 2002

```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

```