Puzzle: Find an example election in which Approval Voting, Borda Voting, Plurality Voting, Plurality Voting with Top-Two Runoff, Instant Runoff Voting, and Condorcet methods, all return different unique-winners. Election examples with fewer voters and voter-types are preferred as "more elegant."
Solution by Warren D. Smith by making several improvements on a suboptimal solution by Joe Malkevitch:
| #voters | Their Vote |
|---|---|
| 4 | A>D>E>C>B>F |
| 3 | B>E>D>F>C>A |
| 2 | C>B>E>D>F>A |
| 1 | D>C>F>A>E>B |
| 1 | F>C>E>D>B>A |
11 voters, 5 voter-types, and 6 candidates (A,B,C,D,E, and F) in total.
Plurality: A wins (most top-rank votes with 4).
Plurality+Runoff among top two: B wins over A in the runoff, 6-to-5.
IRV: C wins
(eliminate E, D, F, and B in that order – it does not matter in which way you break the
DF tie – then C beats A in the final round by 7 to 4).
IRV repeatedly zaps the candidate with fewest
top-rank votes, then the remaining one wins.
Borda: D wins.
(D's Borda score is 16+9+4+5+2=36 versus E with 12+12+6+1+3=34 and
with lower scores for A, B, C, and F.)
Condorcet: E wins. (Since E pairwise-beats each other candidate, e.g. beating
A 6:5, B 6:5, C 7:4, D 6:5, and F 9:2.)
Approval Voting:
If all the red candidates are approved, then F wins.
And with 0-99 Range Voting, it would in fact be possible to make any of the 6 candidates win, depending on how the voters chose the scores compatibly with the orderings above.
Optimality:
Obviously, 6 candidates is minimum possible in order to get 6 different winners.
We now shall prove that 11 voters also is minimum possible, and so is 5 voter types.
First, in order to have an untied plurality winner A, he must
have more votes (here 4) than any other. Then the plur+runoff winner
B, in order also to be untied, must have more votes (here 3) than any other besides A.
Finally, the IRV winner C must, after benefitting from any vote transfers,
have more top-rank votes than B, i.e. must have at least 4 votes.
Hence there must be at least 4+3+4=11 voters if A has 4 votes, merely in order
to get distinct and untied Plurality, Plur+runoff, and IRV winners.
(Actually, 4+2+3 might also have been possible, except that it isn't, see below.)
On the other hand, if A had only 3 or 2 or 1 votes, then 4+3+4=11 would become 3+2+3=8
or 2+1+2=5 or 1+0+1=2 voters.
But C in order to be the IRV winner
cannot be eliminated, so C must have more top-rank votes than somebody at every round,
but less than B at the first round.
That means the top-rank votes for C,D,E,F must be distributed like 2,1,1,0 in the 4+3+4=11 case
(as they are), or like 1,0,0,0 in the 3+2+3=8 or 4+2+3 or
other cases (which will not work since these cannot sum to at least 3 or 2 as they must).
Hence 11 is the minimum possible number of voters.
Finally, the above argument has also shown as a side effect that
5 is the minimum possible number of voter types. Note that this optimality proof
did not even have to mention Borda, Condorcet, or Approval.
Q.E.D.