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Reversal symmetry

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Electoral system criterion
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Thereversal symmetry criterion is avoting system criterion which says that if every voter's opinions on each of the candidates is perfectly reversed (i.e. they rank candidates in order from worst to best), the outcome of the election should be reversed as well, i.e. the first- and last-place finishers should switch places.[1] 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).

Another, equivalent way to motivate the criterion is to say that a voting system should never elect the worst candidate, according to the method itself (as doing so suggests the method is, in some sense, self-contradictory). The worst candidate can be identified by reversing all ballots (to rank candidates from worst-to-best) and then running the algorithm to find a single worst candidate.[2]

Situations where the same candidate is elected when all ballots are reversed are sometimes calledbest-is-worst paradoxes, and can occur ininstant-runoff voting andminimax. Methods that satisfy reversal symmetry include theBorda count,ranked pairs,Kemeny–Young, andSchulze. Mostrated voting systems, includingapproval andscore voting, satisfy the criterion as well.

Examples

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Instant-runoff voting

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Main article: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

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Main article:Minimax Condorcet

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 votersPreferences
4A > B > D > C
4B > C > A > D
2C > D > A > B
1D > A > B > C
1D > B > C > A
2D > 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

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# of votersPreferences
4A > B > D > C
4B > C > A > D
2C > D > A > B
1D > A > B > C
1D > B > C > A
2D > C > A > B

The results would be tabulated as follows:

Pairwise election results
X
ABCD
YA[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):2-0-12-0-11-0-21-0-2
worst pairwise defeat (winning votes):99108
worst pairwise defeat (margins):4462
worst pairwise opposition:99108
  • [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

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# of votersPreferences
4C > D > B > A
4D > A > C > B
2B > A > D > C
1C > B > A > D
1A > C > B > D
2B > A > C > D

The results would be tabulated as follows:

Pairwise election results
X
ABCD
YA[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):1-0-21-0-22-0-12-0-1
worst pairwise defeat (winning votes):91098
worst pairwise defeat (margins):4642
worst pairwise opposition:91098

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, andD is elected Minimax winner.

Conclusion

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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

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Main article:Plurality voting system

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 votersPreferences
1A > B > C
1C > B > A
1B > A > C
1C > 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 votersFavorite
1A
1B
2C

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

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Main article: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

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Given a 3-candidate election between candidatesA,B, andC:

CandidateBallots
A55222
B00333
C11204

The results are tabulated below:

CandidateTotal ScorePreferred vs
ABC
A16--23
B93--2
C813--

Result: In the election, candidatesA andB have the highest scores, and advance to the runoff round.B wins being preferred overA 3 votes to 2.

Reversed score

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Reversing the ballots by subtracting each score from 5 (the maximum score in STAR) gives the following:

CandidateBallots
A00333
B55222
C44351

The results are tabulated below:

CandidateTotal ScorePreferred vs
ABC
A9--31
B162--3
C1732--

Result: In the reversed ballots,B andC have the highest total score, andB wins being preferred toC 3 votes to 2.

References

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  1. ^Saari, Donald G. (2012-12-06).Geometry of Voting. Springer Science & Business Media.ISBN 978-3-642-48644-9.
  2. ^Schulze, Markus (2024-03-03),The Schulze Method of Voting,arXiv:1804.02973, retrieved2024-07-27
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