Answer to Puzzle #2: election with 6 different winners in 6 different voting systems

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:

#votersTheir 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.


Carl Bialik, the "Numbers Guy" at the Wall Street Journal, asked me in Feb. 2009:

Can you think of an example of a set of ballots with six candidates in which each candidate wins under exactly one of the following systems?

The same example-election we gave above also works to solve Bialik's problem if we choose the range-voting scores as follows (this is for 0-99 range voting; it might also be possible to accomplish Bialik's goal using 0-9, i.e "single digit," range voting, but I have not looked into that):

#votersTheir Vote
4 A99>D4>E3>C2>B1>F0
3 B99>E67>D66>F65>C1>A0
1 C99>B98>E67>D66>F65>A0
1 C99>B98>E53>D52>F51>A0
1 D99>C98>F65>A2>E1>B0
1 F99>C4>E3>D2>B1>A0

In this election, B wins using mean-based range voting, with a mean score of 45.27 (beating F with 43.18 and D with 39.45) but F wins using median-based range voting (F's median score is 65, beating D with 52). Note the candidates range-scored above 50 are the approved ones.

Bialik's 8-winner solution (Feb. 2009)

Bialik modified our original solution in a different way [using 0-20 range voting and/or rank-order voting systems] to make it have eight different winners under the 8 voting systems, counting range-mean and range-median as the two other kinds of winners. It's below. Same six winners for the six systems described at top, plus g wins on range-mean, h on range-median.

He noted: Use same approvals as in the old example, and make g and h not approved on the first two ballot sets (so each disapproved by seven, so they don't win approval voting). That is, in the below table of scores, the underlined votes are "approved."

#voters  a       b       c       d       e       f       g       h
4        18      4       5       17      15      0       13      14
3        0       14      5       11      12      10      8       9
2        0       12      20      10      11      9       18      19
1        2       0       12      17      1       11      16      3
1        0       1       4       2       3       16      15      5

As rank-order ballots (omitting the numerical scores) this is:

4: a>d>e>h>g>c>b>f
3: b>e>d>f>h>g>c>a
2: c>h>g>b>e>d>f>a
1: d>g>c>f>h>a>e>b
1: f>g>h>c>e>d>b>a

Verification of winners:
Plurality: a has four first-place votes, more than any rival.
Top-2-runoff: b (The top 2 in the first round are a and b; then b is preferred by three of the four voters who ranked other candidates first, so beats a, 6-5.)
IRV: c (after e,g,h, then d and f, then b, then a eliminated)
Borda: d (with Borda score 52, followed by e in second place with Borda score 48)
Condorcet: e is a beats-all winner (beats a, b, and d by 6:5 majorities each; beats others by larger majorities)
Approval: f (with 7 approvals; d is next with 6)
Range-mean: g with ColumnSum=70 beating d(57) and h(50).
Range-median: h with median=14 beating g(13), e(12), d(11).

Bialik's election then was printed in his Wall Street Journal column (page A9, on 6 Feb 2009) except unfortunately with some misprints.

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