# Participation criterion

The participation criterion is a voting system criterion. Voting systems that fail the participation criterion are said to exhibit the no show paradox[1] and allow a particularly unusual strategy of tactical voting: abstaining from an election can help a voter's preferred choice win. The criterion has been defined[2] as follows:

• In a deterministic framework, the participation criterion says that the addition of a ballot, where candidate A is strictly preferred to candidate B, to an existing tally of votes should not change the winner from candidate A to candidate B.
• In a probabilistic framework, the participation criterion says that the addition of a ballot, where each candidate of the set X is strictly preferred to each other candidate, to an existing tally of votes should not reduce the probability that the winner is chosen from the set X.

Plurality voting, approval voting, range voting, and the Borda count all satisfy the participation criterion. All Condorcet methods,[3][4] Bucklin voting,[5] and IRV[6] fail.

The participation criterion for voting systems is one example of a rational participation constraint for social choice mechanisms in general.

## Quorum requirements

The most common failure of the participation criterion is not in the use of particular voting systems, but in simple yes or no measures that place quorum requirements. A public referendum, for example, if it required majority approval and a certain number of voters to participate in order to pass, would fail the participation criterion, as a minority of voters preferring the "no" option could cause the measure to fail by simply not voting rather than voting no. In other words, the addition of a "no" vote may make the measure more likely to pass. A referendum that required a minimum number of yes votes (not counting no votes), by contrast, would pass the participation criterion.

## Incompatibility with the Condorcet criterion

Hervé Moulin showed in 1988 that whenever there are at least 4 candidates and at least 25 voters, no resolute (single-valued) Condorcet consistent voting rule satisfies the participation criterion.[3] However, when there are at most 3 candidates, the minimax method (with some fixed tie-breaking) satisfies both the Condorcet and the participation criterion.[3] Similarly, when there are 4 candidates and at most 11 voters, there is a voting rule that satisfies both criteria,[7] but no such rule exists for 4 candidates and 12 voters.[7] Similar incompatibilities have also been proven for set-valued voting rules.[7][8][9]

Certain conditions that are weaker than the participation criterion are also incompatible with the Condorcet criterion. For example, weak positive involvement requires that adding a ballot in which candidate A is most-preferred does not change the winner away from A; similarly, weak negative involvement requires that adding a ballot in which A is least-preferred does not make A the winner if it was not the winner before. Both conditions are incompatible with the Condorcet criterion if one allows ballots to include ties.[10] Another condition that is weaker than participation is halfway-monotonicity, which requires that a voter cannot be better off by completely reversing his ballot. Again, halfway-monotonicity is incompatible with the Condorcet criterion.[11]

## Examples

### Copeland

Main article: Copeland's method

This example shows that Copeland's method violates the Participation criterion. Assume four candidates A, B, C and D with 13 potential voters and the following preferences:

# of voters Preferences
3 A > B > C > D
1 A > C > D > B
1 A > D > C > B
4 B > A > C > D
4 D > C > B > A

The three voters with preferences A > B > C > D are unconfident whether to participate in the election.

#### Voters not participating

Assume the 3 voters would not show up at the polling place.

The preferences of the remaining 10 voters would be:

# of voters Preferences
1 A > C > D > B
1 A > D > C > B
4 B > A > C > D
4 D > C > B > A

The results would be tabulated as follows:

 X A B C D Y A [X] 8 [Y] 2 [X] 4 [Y] 6 [X] 4 [Y] 6 B [X] 2 [Y] 8 [X] 6 [Y] 4 [X] 6 [Y] 4 C [X] 6 [Y] 4 [X] 4 [Y] 6 [X] 5 [Y] 5 D [X] 6 [Y] 4 [X] 4 [Y] 6 [X] 5 [Y] 5 Pairwise election results (won-tied-lost): 2-0-1 1-0-2 1-1-1 1-1-1

Result: A can defeat two of the three opponents, whereas no other candidate wins against more than one opponent. Thus, A is elected Copeland winner.

#### Voters participating

Now, consider the 3 unconfident voters decide to participate:

# of voters Preferences
3 A > B > C > D
1 A > C > D > B
1 A > D > C > B
4 B > A > C > D
4 D > C > B > A

The results would be tabulated as follows:

 X A B C D Y A [X] 8 [Y] 5 [X] 4 [Y] 9 [X] 4 [Y] 9 B [X] 5 [Y] 8 [X] 6 [Y] 7 [X] 6 [Y] 7 C [X] 9 [Y] 4 [X] 7 [Y] 6 [X] 5 [Y] 8 D [X] 9 [Y] 4 [X] 7 [Y] 6 [X] 8 [Y] 5 Pairwise election results (won-tied-lost): 2-0-1 3-0-0 1-0-2 0-0-3

Result: B is the Condorcet winner and thus, B is Copeland winner, too.

#### Conclusion

By participating in the election the three voters supporting A would change A from winner to loser. Their first preferences were not sufficient to change the one pairwise defeat A suffers without their support. But, their second preferences for B turned both defeats B would have suffered into wins and made B Condorcet winner and thus, overcoming A.

Hence, Copeland fails the Participation criterion.

### Instant-runoff voting

Main article: Instant-runoff voting

This example shows that Instant-runoff voting violates the Participation criterion. Assume three candidates A, B and C and 15 potential voters, two of them (marked bold in the table) unconfident whether to vote.

# of voters Preferences
2 A > B > C
3 A > B > C
4 B > C > A
6 C > A > B

#### Voters not participating

If they don't show up at the election the remaining voters would be:

# of voters Preferences
3 A > B > C
4 B > C > A
6 C > A > B

The following outcome results:

Candidate
1st 2nd
A 3
B 4 7
C 6 6

Result: After A is eliminated first, B gets his votes and wins.

#### Voters participating

If they participate in the election, the preferences list is:

# of voters Preferences
5 A > B > C
4 B > C > A
6 C > A > B

The outcome changes as follows:

Candidate
1st 2nd
A 5 5
B 4
C 6 10

Result: Now, B is eliminated first and C gets his votes and wins.

#### Conclusion

The additional votes for A were not sufficient for winning, but for descending to the second round, thereby eliminating the second preference of the voters. Thus, due to participating in the election, the voters changed the winner from their second preference to their strictly least preference.

Thus, Instant-runoff voting fails the Participation criterion.

### Kemeny–Young method

Main article: Kemeny–Young method

This example shows that the Kemeny–Young method violates the Participation criterion. Assume four candidates A, B, C, D with 21 voters and the following preferences:

# of voters Preferences
3 A > B > C > D
3 A > C > B > D
4 A > D > C > B
4 B > A > D > C
2 C > B > A > D
2 D > B > A > C
3 D > C > B > A

The three voters with preferences A > B > C > D are unconfident whether to participate in the election.

#### Voters not participating

Assume the 3 voters would not show up at the polling place.

The preferences of the remaining 18 voters would be:

# of voters Preferences
3 A > C > B > D
4 A > D > C > B
4 B > A > D > C
2 C > B > A > D
2 D > B > A > C
3 D > C > B > A

The Kemeny–Young method arranges the pairwise comparison counts in the following tally table:

All possible pairs
of choice names
Number of votes with indicated preference
Prefer X over Y Equal preference Prefer Y over X
X = A Y = B 7 0 11
X = A Y = C 13 0 5
X = A Y = D 13 0 5
X = B Y = C 6 0 12
X = B Y = D 9 0 9
X = C Y = D 5 0 13

Result: The ranking A > D > C > B has the highest ranking score of 67 (= 13 + 13 + 13 + 12 + 9 + 7); against e.g. 65 (= 13 + 13 + 13 + 11 + 9 + 6) of B > A > D > C. Thus, A is Kemeny-Young winner.

#### Voters participating

Now, consider the 3 unconfident voters decide to participate:

# of voters Preferences
3 A > B > C > D
3 A > C > B > D
4 A > D > C > B
4 B > A > D > C
2 C > B > A > D
2 D > B > A > C
3 D > C > B > A

The Kemeny–Young method arranges the pairwise comparison counts in the following tally table:

All possible pairs
of choice names
Number of votes with indicated preference
Prefer X over Y Equal preference Prefer Y over X
X = A Y = B 10 0 11
X = A Y = C 16 0 5
X = A Y = D 16 0 5
X = B Y = C 9 0 12
X = B Y = D 12 0 9
X = C Y = D 8 0 13

Result: The ranking B > A > D > C has the highest ranking score of 77 (= 16 + 16 + 13 + 12 + 11 + 9); against e.g. 76 (= 16 + 16 + 13 + 12 + 10 + 9) of A > D > C > B. Thus, B is Kemeny-Young winner.

#### Conclusion

By participating in the election the three voters supporting A would change A from winner to loser. Their ballots support 3 of the 6 pairwise comparisons of the ranking A > D > C >B, but four pairwise comparisons of the ranking B > A > D > C, enough to overcome the first one.

Thus, Kemeny-Young fails the Participation criterion.

### Majority Judgment

Main article: Majority Judgment

This example shows that Majority Judgment violates the Participation criterion. Assume two candidates A and B with 5 potential voters and the following ratings:

Candidates/
# of voters
A B
2 Excellent Good
2 Fair Poor
1 Poor Good

The two voters rating A "Excellent" are unconfident whether to participate in the election.

#### Voters not participating

Assume the 2 voters would not show up at the polling place.

The ratings of the remaining 3 voters would be:

Candidates/
# of voters
A B
2 Fair Poor
1 Poor Good

The sorted ratings would be as follows:

Candidate
 ↓ Median point
A
B

 Excellent Good Fair Poor

Result: A has the median rating of "Fair" and B has the median rating of "Poor". Thus, A is elected Majority Judgment winner.

#### Voters participating

Now, consider the 2 unconfident voters decide to participate:

Candidates/
# of voters
A B
2 Excellent Good
2 Fair Poor
1 Poor Good

The sorted ratings would be as follows:

Candidate
 ↓ Median point
A
B

 Excellent Good Fair Poor

Result: A has the median rating of "Fair" and B has the median rating of "Good". Thus, B is the Majority Judgment winner.

#### Conclusion

By participating in the election the two voters preferring A would change A from winner to loser. Their "Excellent" rating for A was not sufficient to change A's median rating since no other voter rated A higher than "Fair". But, their "Good" rating for B turned B's median rating to "Good", since another voter agreed with this rating.

Thus, Majority Judgment fails the Participation criterion.

### Minimax

Main article: Minimax Condorcet

This example shows that the Minimax method violates the Participation criterion. Assume four candidates A, B, C, D with 18 potential voters and the following preferences:

# of voters Preferences
2 A > B > C > D
2 A > B > D > C
6 B > D > C > A
5 C > A > B > D
1 D > A > B > C
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.

The two voters (marked bold) with preferences A > B > C > D are unconfident whether to participate in the election.

#### Voters not participating

Assume the two voters would not show up at the polling place.

The preferences of the remaining 16 voters would be:

# of voters Preferences
2 A > B > D > C
6 B > D > C > A
5 C > A > B > D
1 D > A > B > C
2 D > C > A > B

The results would be tabulated as follows:

 X A B C D Y A [X] 6 [Y] 10 [X] 13 [Y] 3 [X] 9 [Y] 7 B [X] 10 [Y] 6 [X] 7 [Y] 9 [X] 3 [Y] 13 C [X] 3 [Y] 13 [X] 9 [Y] 7 [X] 11 [Y] 5 D [X] 7 [Y] 9 [X] 13 [Y] 3 [X] 5 [Y] 11 Pairwise election results (won-tied-lost): 1-0-2 2-0-1 1-0-2 2-0-1 worst pairwise defeat (winning votes): 13 10 11 13 worst pairwise defeat (margins): 10 4 6 10 worst pairwise opposition: 13 10 11 13
• [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: B has the closest biggest defeat. Thus, B is elected Minimax winner.

#### Voters participating

Now, consider the two unconfident voters decide to participate:

# of voters Preferences
2 A > B > C > D
2 A > B > D > C
6 B > D > C > A
5 C > A > B > D
1 D > A > B > C
2 D > C > A > B

The results would be tabulated as follows:

 X A B C D Y A [X] 6 [Y] 12 [X] 13 [Y] 5 [X] 9 [Y] 9 B [X] 12 [Y] 6 [X] 7 [Y] 11 [X] 3 [Y] 15 C [X] 5 [Y] 13 [X] 11 [Y] 7 [X] 11 [Y] 7 D [X] 9 [Y] 9 [X] 15 [Y] 3 [X] 7 [Y] 11 Pairwise election results (won-tied-lost): 1-1-1 2-0-1 1-0-2 1-1-1 worst pairwise defeat (winning votes): 13 12 11 15 worst pairwise defeat (margins): 8 6 4 8 worst pairwise opposition: 13 12 11 15

Result: C has the closest biggest defeat. Thus, C is elected Minimax winner.

#### Conclusion

By participating in the election the two voters changed the winner from B to C whilst strictly preferring B to C. Their preferences of B over C and D does not advance B's Minimax value since B's biggest defeat was against A. Also, their preferences of A and B over C does not degrade C's Minimax value since C's biggest defeat was against D. Therefore, only the comparison "A > B" degrade B's value and the comparison "C > D" advanced C's value. This results in C overcoming B.

Thus, the Minimax method fails the Participation criterion.

### Ranked pairs

Main article: Ranked pairs

This example shows that the Ranked pairs method violates the Participation criterion. Assume four candidates A, B, C and D with 26 potential voters and the following preferences:

# of voters Preferences
4 A > B > C > D
8 A > D > B > C
7 B > C > A > D
7 C > D > B > A

The four voters with preferences A > B > C > D are unconfident whether to participate in the election.

#### Voters not participating

Assume the 4 voters do not show up at the polling place.

The preferences of the remaining 22 voters would be:

# of voters Preferences
8 A > D > B > C
7 B > C > A > D
7 C > D > B > A

The results would be tabulated as follows:

 X A B C D Y A [X] 14 [Y] 8 [X] 14 [Y] 8 [X] 7 [Y] 15 B [X] 8 [Y] 14 [X] 7 [Y] 15 [X] 15 [Y] 7 C [X] 8 [Y] 14 [X] 15 [Y] 7 [X] 8 [Y] 14 D [X] 15 [Y] 7 [X] 7 [Y] 15 [X] 14 [Y] 8 Pairwise election results (won-tied-lost): 1-0-2 2-0-1 2-0-1 1-0-2

The sorted list of victories would be:

Pair Winner
A (15) vs. D (7) A 15
B (15) vs. C (7) B 15
B (7) vs. D (15) D 15
A (8) vs. B (14) B 14
A (8) vs. C (14) C 14
C (14) vs. D (8) C 14

Result: A > D, B > C and D > B are locked in (and the other three can't be locked in after that), so the full ranking is A > D > B > C. Thus, A is elected Ranked pairs winner.

#### Voters participating

Now, consider the 4 unconfident voters decide to participate:

# of voters Preferences
4 A > B > C > D
8 A > D > B > C
7 B > C > A > D
7 C > D > B > A

The results would be tabulated as follows:

 X A B C D Y A [X] 14 [Y] 12 [X] 14 [Y] 12 [X] 7 [Y] 19 B [X] 12 [Y] 14 [X] 7 [Y] 19 [X] 15 [Y] 11 C [X] 12 [Y] 14 [X] 19 [Y] 7 [X] 8 [Y] 18 D [X] 19 [Y] 7 [X] 11 [Y] 15 [X] 18 [Y] 8 Pairwise election results (won-tied-lost): 1-0-2 2-0-1 2-0-1 1-0-2

The sorted list of victories would be:

Pair Winner
A (19) vs. D (7) A 19
B (19) vs. C (7) B 19
C (18) vs. D (8) C 18
B (11) vs. D (15) D 15
A (12) vs. B (14) B 14
A (12) vs. C (14) C 14

Result: A > D, B > C and C > D are locked in first. Now, D > B can't be locked in since it would create a cycle B > C > D > B. Finally, B > A and C > A are locked in. Hence, the full ranking is B > C > A > D. Thus, B is elected Ranked pairs winner.

#### Conclusion

By participating in the election the four voters supporting A would change A from winner to loser. The clear victory of D > B was essential for A's win in the first place. The additional votes diminished that victory and at the same time gave a boost to the victory of C > D, turning D > B into the weakest link of the cycle B > C > D > B. Since A had no other victories but the one over D and B had no other losses but the one over D, the elimination of D > B made it impossible for A to win.

Thus, the Ranked pairs method fails the Participation criterion.

### Schulze method

Main article: Schulze method

This example shows that the Schulze method violates the Participation criterion. Assume four candidates A, B, C and D with 25 potential voters and the following preferences:

# of voters Preferences
2 A > B > C > D
7 B > A > D > C
1 B > C > A > D
2 B > D > C > A
7 C > A > D > B
2 D > B > A > C
4 D > C > A > B

The two voters with preferences A > B > C > D are unconfident whether to participate in the election.

#### Voters not participating

Assume the 2 voters would not show up at the polling place.

The preferences of the remaining 23 voters would be:

# of voters Preferences
7 B > A > D > C
1 B > C > A > D
2 B > D > C > A
7 C > A > D > B
2 D > B > A > C
4 D > C > A > B

The pairwise preferences would be tabulated as follows:

Matrix of pairwise preferences
d[*,A] d[*,B] d[*,C] d[*,D]
d[A,*] 11 9 15
d[B,*] 12 12 10
d[C,*] 14 11 8
d[D,*] 8 13 15

Now, the strongest paths have to be identified, e.g. the path A > D > B is stronger than the direct path A > B (which is nullified, since it is a loss for A).

Strengths of the strongest paths
p[*,A] p[*,B] p[*,C] p[*,D]
p[A,*] 13 15 15
p[B,*] 12 12 12
p[C,*] 14 13 14
p[D,*] 14 13 15

Result: The full ranking is A > D > C > B. Thus, A is elected Schulze winner.

#### Voters participating

Now, consider the 2 unconfident voters decide to participate:

# of voters Preferences
2 A > B > C > D
7 B > A > D > C
1 B > C > A > D
2 B > D > C > A
7 C > A > D > B
2 D > B > A > C
4 D > C > A > B

The pairwise preferences would be tabulated as follows:

Matrix of pairwise preferences
d[*,A] d[*,B] d[*,C] d[*,D]
d[A,*] 13 11 17
d[B,*] 12 14 12
d[C,*] 14 11 10
d[D,*] 8 13 15

Now, the strongest paths have to be identified, e.g. the path C > A > D is stronger than the direct path C > D.

Strengths of the strongest paths
p[*,A] p[*,B] p[*,C] p[*,D]
p[A,*] 13 15 17
p[B,*] 14 14 14
p[C,*] 14 13 14
p[D,*] 14 13 15

Result: The full ranking is B > A > D > C. Thus, B is elected Schulze winner.

#### Conclusion

By participating in the election the two voters supporting A changed the winner from A to B. In fact, the voters can turn the defeat in direct pairwise comparison of A against B into a victory. But in this example, the relation between A and B does not depend on the direct comparison, since the paths A > D > B and B > C > A are stronger. The additional voters diminish D > B, the weakest link of the A > D > B path, while giving a boost to B > C, the weakest link of the path B > C > A.

Thus, the Schulze method fails the Participation criterion.

## References

1. Fishburn, Peter C.; Brams, Steven J. (1983-01-01). "Paradoxes of Preferential Voting". Mathematics Magazine. 56 (4): 207–214. doi:10.2307/2689808. JSTOR 2689808.
2. Douglas Woodall (December 1994). "Properties of Preferential Election Rules, Voting matters - Issue 3, December 1994".
3. Moulin, Hervé (1988-06-01). "Condorcet's principle implies the no show paradox". Journal of Economic Theory. 45 (1): 53–64. doi:10.1016/0022-0531(88)90253-0.
4. "Participation failure" is forced in Condorcet methods with at least 4 candidates". Retrieved 2014-12-24.
5. Markus Schulze (1998-06-12). "Regretted Turnout. Insincere = ranking.". Retrieved 2011-05-14.
6. Warren D. Smith. "Lecture "Mathematics and Democracy"". Retrieved 2011-05-12.
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9. Jimeno, José L.; Pérez, Joaquín; García, Estefanía (2009-01-09). "An extension of the Moulin No Show Paradox for voting correspondences". Social Choice and Welfare. 33 (3): 343–359. doi:10.1007/s00355-008-0360-6. ISSN 0176-1714.
10. Duddy, Conal (2013-11-29). "Condorcet's principle and the strong no-show paradoxes". Theory and Decision. 77 (2): 275–285. doi:10.1007/s11238-013-9401-4. ISSN 0040-5833.
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