**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.**

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?

- Range voting (mean)
- Range voting (median)
- Borda Count
- Condorcet
- Instant runoff
- Plurality

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):

#voters | Their 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 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 h4184 5 17 15 0 13 14 3 01451112108 9 2 012201011918191 2 01217111163 1 0 1 4 2 316155

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.