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:
Plurality voting, approval voting, range voting, and the Borda count all satisfy the participation criterion.^{[citation needed]} 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.
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.^{[citation needed]} 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.
Hervé Moulin showed in 1988 that whenever there are at least four candidates and at least 25 voters, no resolute (singlevalued) Condorcet consistent voting rule satisfies the participation criterion.^{[3]} However, when there are at most three candidates, the minimax method (with some fixed tiebreaking) satisfies both the Condorcet and the participation criterion.^{[3]} Similarly, when there are four candidates and at most 11 voters, there is a voting rule that satisfies both criteria,^{[7]} but no such rule exists for four candidates and 12 voters.^{[7]} Similar incompatibilities have also been proven for setvalued 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 mostpreferred does not change the winner away from A; similarly, weak negative involvement requires that adding a ballot in which A is leastpreferred 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 their ballot. Again, halfway monotonicity is incompatible with the Condorcet criterion.^{[11]}
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:
Preferences  # of voters 

A > B > C > D  3 
A > C > D > B  1 
A > D > C > B  1 
B > A > C > D  4 
D > C > B > A  4 
The three voters with preferences A > B > C > D are unconfident whether to participate in the election.
Assume the 3 voters would not show up at the polling place.
The preferences of the remaining 10 voters would be:
Preferences  # of voters 

A > C > D > B  1 
A > D > C > B  1 
B > A > C > D  4 
D > C > B > A  4 
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 results for X, wontiedlost 
201  102  111  111 
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.
Now, consider the three unconfident voters decide to participate:
Preferences  # of voters 

A > B > C > D  3 
A > C > D > B  1 
A > D > C > B  1 
B > A > C > D  4 
D > C > B > A  4 
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 results for X, wontiedlost 
201  300  102  003 
Result: B is the Condorcet winner and thus, B is Copeland winner, too.
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.
Main article: Instantrunoff voting 
This example shows that instantrunoff voting violates the participation criterion. Assume three candidates A, B and C and 15 potential voters, two of them (in blue) unconfident whether to vote.
Preferences  # of voters 

A > B > C  2 
A > B > C  3 
B > C > A  4 
C > A > B  6 
If they don't show up at the election the remaining voters would be:
Preferences  # of voters 

A > B > C  3 
B > C > A  4 
C > A > B  6 
The following outcome results:
Candidate  Votes for round  

1st  2nd  
A  3  
B  4  7 
C  6  6 
Result: After A is eliminated first, B gets their votes and wins.
If they participate in the election, the preferences list is:
Preferences  # of voters 

A > B > C  5 
B > C > A  4 
C > A > B  6 
The outcome changes as follows:
Candidate  Votes for round  

1st  2nd  
A  5  5 
B  4  
C  6  10 
Result: Now, B is eliminated first and C gets their votes and wins.
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, instantrunoff voting fails the participation criterion.
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:
Preferences  # of voters 

A > B > C > D  3 
A > C > B > D  3 
A > D > C > B  4 
B > A > D > C  4 
C > B > A > D  2 
D > B > A > C  2 
D > C > B > A  3 
The three voters with preferences A > B > C > D are unconfident whether to participate in the election.
Assume the 3 voters would not show up at the polling place.
The preferences of the remaining 18 voters would be:
Preferences  # of voters 

A > C > B > D  3 
A > D > C > B  4 
B > A > D > C  4 
C > B > A > D  2 
D > B > A > C  2 
D > C > B > A  3 
The Kemeny–Young method arranges the pairwise comparison counts in the following tally table:
Candidate pairs  Number who prefer…  

X  Y  X  Neither  Y 
A  B  7  0  11 
A  C  13  0  5 
A  D  13  0  5 
B  C  6  0  12 
B  D  9  0  9 
C  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 KemenyYoung winner.
Now, consider the 3 unconfident voters decide to participate:
Preferences  # of voters 

A > B > C > D  3 
A > C > B > D  3 
A > D > C > B  4 
B > A > D > C  4 
C > B > A > D  2 
D > B > A > C  2 
D > C > B > A  3 
The Kemeny–Young method arranges the pairwise comparison counts in the following tally table:
Candidate pairs  Number who prefer…  

X  Y  X  Neither  Y 
A  B  10  0  11 
A  C  16  0  5 
A  D  16  0  5 
B  C  9  0  12 
B  D  12  0  9 
C  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 KemenyYoung winner.
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, KemenyYoung fails the participation criterion.
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  
Excellent  Good  2 
Fair  Poor  2 
Poor  Good  1 
The two voters rating A "Excellent" are unconfident whether to participate in the election.
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  
Fair  Poor  2 
Poor  Good  1 
The sorted ratings would be as follows:
Candidate 
 
A 
 
B 
 

Result: A has the median rating of "Fair" and B has the median rating of "Poor". Thus, A is elected majority judgment winner.
Now, consider the 2 unconfident voters decide to participate:
Candidates  # of voters  

A  B  
Excellent  Good  2 
Fair  Poor  2 
Poor  Good  1 
The sorted ratings would be as follows:
Candidate 
 
A 
 
B 
 

Result: A has the median rating of "Fair" and B has the median rating of "Good". Thus, B is the majority judgment winner.
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.
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:
Preferences  # of voters 

A > B > C > D  2 
A > B > D > C  2 
B > D > C > A  6 
C > A > B > D  5 
D > A > B > C  1 
D > C > A > B  2 
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 (in blue) with preferences A > B > C > D are unconfident whether to participate in the election.
Assume the two voters would not show up at the polling place.
The preferences of the remaining 16 voters would be:
Preferences  # of voters 

A > B > D > C  2 
B > D > C > A  6 
C > A > B > D  5 
D > A > B > C  1 
D > C > A > B  2 
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 results for X, wontiedlost 
102  201  102  201  
Worst opposing votes  13  10  11  13  
Worst margin  10  4  6  10  
Worst opposition  13  10  11  13 
Result: B has the closest biggest defeat. Thus, B is elected minimax winner.
Now, consider the two unconfident voters decide to participate:
Preferences  # of voters 

A > B > C > D  2 
A > B > D > C  2 
B > D > C > A  6 
C > A > B > D  5 
D > A > B > C  1 
D > C > A > B  2 
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 results for X, wontiedlost 
111  201  102  111  
Worst opposing votes  13  12  11  15  
Worst margin  8  6  4  8  
Worst opposition  13  12  11  15 
Result: C has the closest biggest defeat. Thus, C is elected minimax winner.
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.
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:
Preferences  # of voters 

A > B > C > D  4 
A > D > B > C  8 
B > C > A > D  7 
C > D > B > A  7 
The four voters with preferences A > B > C > D are unconfident whether to participate in the election.
Assume the 4 voters do not show up at the polling place.
The preferences of the remaining 22 voters would be:
Preferences  # of voters 

A > D > B > C  8 
B > C > A > D  7 
C > D > B > A  7 
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 results for X, wontiedlost 
102  201  201  102 
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.
Now, consider the 4 unconfident voters decide to participate:
Preferences  # of voters 

A > B > C > D  4 
A > D > B > C  8 
B > C > A > D  7 
C > D > B > A  7 
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 results for X, wontiedlost 
102  201  201  102 
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.
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.
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:
Preferences  # of voters 

A > B > C > D  2 
B > A > D > C  7 
B > C > A > D  1 
B > D > C > A  2 
C > A > D > B  7 
D > B > A > C  2 
D > C > A > B  4 
The two voters with preferences A > B > C > D are unconfident whether to participate in the election.
Assume the two voters would not show up at the polling place.
The preferences of the remaining 23 voters would be:
Preferences  # of voters 

B > A > D > C  7 
B > C > A > D  1 
B > D > C > A  2 
C > A > D > B  7 
D > B > A > C  2 
D > C > A > B  4 
The pairwise preferences would be tabulated as follows:
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).
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.
Now, consider the 2 unconfident voters decide to participate:
Preferences  # of voters 

A > B > C > D  2 
B > A > D > C  7 
B > C > A > D  1 
B > D > C > A  2 
C > A > D > B  7 
D > B > A > C  2 
D > C > A > B  4 
The pairwise preferences would be tabulated as follows:
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.
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.
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.