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In social choice theory, the best-is-worst paradox occurs when a voting rule declares the same candidate to be both the best and worst possible winner. The worst candidate can be identified by reversing each voter's ballot (to rank candidates from worst-to-best), then applying the voting rule to the reversed ballots find a new "anti-winner".[1][2]
Rules that never exhibit a best-is-worst paradox are said to satisfy the reversal criterion, which states that if every voter's opinions on each candidate are perfectly reversed (i.e. they rank candidates from worst to best), the outcome of the election should be reversed as well, meaning the first- and last- place finishers should switch places.[2] In other words, the results of the election should not depend arbitrarily on whether voters rank candidates from best to worst (and then select the best candidate), or whether we ask them to rank candidates from worst to best (and then select the least-bad candidate).
Methods that satisfy reversal symmetry include the Borda count, ranked pairs, Kemeny–Young, and Schulze. Most rated voting systems, including approval and score voting, satisfy the criterion as well. Best-is-worst paradoxes can occur in ranked-choice runoff voting (RCV) and minimax. A well-known example is the 2022 Alaska special election, where candidate Mary Peltola was both the winner and anti-winner.
Examples
Instant-runoff voting
Consider a preferential system where 11 voters express their preferences as:
- 5 voters prefer A then B then C
- 4 voters prefer B then C then A
- 2 voters prefer C then A then B
With the Borda count A would get 23 points (5×3+4×1+2×2), B would get 24 points, and C would get 19 points, so B would be elected. In instant-runoff, C would be eliminated in the first round and A would be elected in the second round by 7 votes to 4.
Now reversing the preferences:
- 5 voters prefer C then B then A
- 4 voters prefer A then C then B
- 2 voters prefer B then A then C
With the Borda count A would get 21 points (5×1+4×3+2×2), B would get 20 points, and C would get 25 points, so this time C would be elected. In instant-runoff, B would be eliminated in the first round and A would as before be elected in the second round, this time by 6 votes to 5.
Minimax
This example shows that the Minimax method violates the Reversal symmetry criterion. Assume four candidates A, B, C and D with 14 voters with the following preferences:
| # of voters | Preferences |
|---|---|
| 4 | A > B > D > C |
| 4 | B > C > A > D |
| 2 | C > D > A > B |
| 1 | D > A > B > C |
| 1 | D > B > C > A |
| 2 | D > C > A > B |
Since all preferences are strict rankings (no equals are present), all three Minimax methods (winning votes, margins and pairwise opposite) elect the same winners.
Now, the winners are determined for the normal and the reversed order.
Normal order
| # of voters | Preferences |
|---|---|
| 4 | A > B > D > C |
| 4 | B > C > A > D |
| 2 | C > D > A > B |
| 1 | D > A > B > C |
| 1 | D > B > C > A |
| 2 | D > C > A > B |
The results would be tabulated as follows:
| X | |||||
| A | B | C | D | ||
| Y | A | [X] 5 [Y] 9 |
[X] 9 [Y] 5 |
[X] 6 [Y] 8 | |
| B | [X] 9 [Y] 5 |
[X] 4 [Y] 10 |
[X] 6 [Y] 8 | ||
| C | [X] 5 [Y] 9 |
[X] 10 [Y] 4 |
[X] 8 [Y] 6 | ||
| D | [X] 8 [Y] 6 |
[X] 8 [Y] 6 |
[X] 6 [Y] 8 |
||
| Pairwise election results (won-tied-lost): | 2-0-1 | 2-0-1 | 1-0-2 | 1-0-2 | |
| worst pairwise defeat (winning votes): | 9 | 9 | 10 | 8 | |
| worst pairwise defeat (margins): | 4 | 4 | 6 | 2 | |
| worst pairwise opposition: | 9 | 9 | 10 | 8 | |
- [X] indicates voters who preferred the candidate listed in the column caption to the candidate listed in the row caption
- [Y] indicates voters who preferred the candidate listed in the row caption to the candidate listed in the column caption
Result: The candidates A, B, and C form a cycle with clear defeats. D benefits from that since its two losses are relatively close and therefore D's biggest defeat is the closest of all candidates. Thus, D is elected Minimax winner.
Reversed order
| # of voters | Preferences |
|---|---|
| 4 | C > D > B > A |
| 4 | D > A > C > B |
| 2 | B > A > D > C |
| 1 | C > B > A > D |
| 1 | A > C > B > D |
| 2 | B > A > C > D |
The results would be tabulated as follows:
| X | |||||
| A | B | C | D | ||
| Y | A | [X] 9 [Y] 5 |
[X] 5 [Y] 9 |
[X] 8 [Y] 6 | |
| B | [X] 5 [Y] 9 |
[X] 10 [Y] 4 |
[X] 8 [Y] 6 | ||
| C | [X] 9 [Y] 5 |
[X] 4 [Y] 10 |
[X] 6 [Y] 8 | ||
| D | [X] 6 [Y] 8 |
[X] 6 [Y] 8 |
[X] 8 [Y] 6 |
||
| Pairwise election results (won-tied-lost): | 1-0-2 | 1-0-2 | 2-0-1 | 2-0-1 | |
| worst pairwise defeat (winning votes): | 9 | 10 | 9 | 8 | |
| worst pairwise defeat (margins): | 4 | 6 | 4 | 2 | |
| worst pairwise opposition: | 9 | 10 | 9 | 8 | |
Result: The candidates A, B, and C still form a cycle with clear defeats. Therefore D's biggest defeat is the closest of all candidates, and D is elected Minimax winner.
Conclusion
D is the Minimax winner using the normal preference order and also using the ballots with reversed preference orders. Thus, Minimax fails the Reversal symmetry criterion.
Plurality voting
This example shows that plurality voting violates the Reversal symmetry criterion. Assume three candidates A, B and C and 4 voters with the following preferences:
| # of voters | Preferences |
|---|---|
| 1 | A > B > C |
| 1 | C > B > A |
| 1 | B > A > C |
| 1 | C > A > B |
Note that reversing all the ballots, leads to the same set of ballots, since the reversed preference order of the first voter resembles the preference order of the second, and similarly with the third and fourth.
In the following the Plurality winner is determined. Plurality ballots only contain the single favorite:
| # of voters | Favorite |
|---|---|
| 1 | A |
| 1 | B |
| 2 | C |
Result: The candidates A and B receive 1 vote each, candidate C receives a plurality of 2 votes (50%). Thus, C is elected Plurality winner.
C is the Plurality winner using the normal ballots and also using the reversed ballot. Thus, Plurality fails the Reversal symmetry criterion.
Note, that every voting system that satisfies the Reversal symmetry criterion, would have to lead to a tie in this example (as in every example in which the set of reversed ballots is the same as the set of normal ballots).
STAR voting
This example shows STAR violates the reversal symmetry criterion. In a score ballot, reversed score is calculated as the maximum possible score minus the normal score.
Normal score
Given a 3-candidate election between candidates A, B, and C:
| Candidate | Ballots | ||||
|---|---|---|---|---|---|
| A | 5 | 5 | 2 | 2 | 2 |
| B | 0 | 0 | 3 | 3 | 3 |
| C | 1 | 1 | 2 | 0 | 4 |
The results are tabulated below:
| Candidate | Total Score | Preferred vs | ||
|---|---|---|---|---|
| A | B | C | ||
| A | 16 | -- | 2 | 3 |
| B | 9 | 3 | -- | 2 |
| C | 8 | 1 | 3 | -- |
Result: In the election, candidates A and B have the highest scores, and advance to the runoff round. B wins being preferred over A 3 votes to 2.
Reversed score
Reversing the ballots by subtracting each score from 5 (the maximum score in STAR) gives the following:
| Candidate | Ballots | ||||
|---|---|---|---|---|---|
| A | 0 | 0 | 3 | 3 | 3 |
| B | 5 | 5 | 2 | 2 | 2 |
| C | 4 | 4 | 3 | 5 | 1 |
The results are tabulated below:
| Candidate | Total Score | Preferred vs | ||
|---|---|---|---|---|
| A | B | C | ||
| A | 9 | -- | 3 | 1 |
| B | 16 | 2 | -- | 3 |
| C | 17 | 3 | 2 | -- |
Result: In the reversed ballots, B and C have the highest total score, and B wins being preferred to C 3 votes to 2.
References
- ^ Schulze, Markus (2024-03-03), The Schulze Method of Voting, arXiv:1804.02973
- ^ a b Saari, Donald G. (2012-12-06). Geometry of Voting. Springer Science & Business Media. ISBN 978-3-642-48644-9.
