Participation criterion

The participation criterion is a voting system criterion. It is also known as the "no show paradox". It has been defined^[1] 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.^{[citation needed]} All Condorcet methods,^[2]^[3] Bucklin voting,^[4] and IRV^[5] fail.

Voting systems that fail the participation criterion allow a particularly unusual strategy of not voting to, in some circumstances, help a voter's preferred choice win.

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.^{[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.

Examples

Copeland

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:

Pairwise election results
		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:

Pairwise election results
		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

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:

Votes in round/ 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:

Votes in round/ 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

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
All possible pairs of choice names		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 + 7 + 13 + 9 + 12); against e.g. 65 (= 11 + 9 + 6 + 13 + 13 + 13) 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
All possible pairs of choice names		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 (= 11 + 12 + 9 + 16 + 16 + 13); against e.g. 76 (= 16 + 16 + 10 + 13 + 9 + 12) 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

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

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:

Pairwise election results
		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:

Pairwise election results
		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

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 would 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:

Pairwise election results
		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:

Pairwise election results
		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 giving 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

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

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External links

A Strong No Show paradox is a common flaw in Condorcet voting correspondences by Joaquin Perez.

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v t e Voting systems
Part of the Politics and Election series
Single-winner voting system	Approval voting Borda count Bucklin voting Contingent vote Coombs' method Copeland's method Exhaustive ballot First-past-the-post voting Instant-runoff voting Kemeny–Young method Simple majoritarianism Minimax Condorcet Nanson's method Plurality Positional voting system Range voting Ranked pairs Schulze method Two-round system
Proportional representation	Mixed-member Party-list Single transferable vote Schulze STV CPO-STV Highest averages method Sainte-Laguë D'Hondt Largest remainder method Alternative vote Plus Closed list Open list Overhang seat Underhang seat
Semi-proportional representation	Parallel voting Single non-transferable vote Cumulative voting Limited voting Proportional approval voting Sequential proportional approval voting Satisfaction approval voting
Usage	Table of voting systems by country
Voting system criterion	Condorcet criterion Condorcet loser criterion Consistency criterion Independence of clones Independence of irrelevant alternatives Independence of Smith-dominated alternatives Later-no-harm criterion Majority criterion Majority loser criterion Monotonicity criterion Mutual majority criterion Pareto efficiency Participation criterion Plurality criterion Resolvability criterion Reversal symmetry Smith criterion
Voting system quotas	Droop quota Hagenbach-Bischoff quota Hare quota Imperiali quota
Other	Ballot Vote counting system Election threshold First-preference votes Spoilt vote Sortition
Portal — Project

Participation criterion

Contents

Quorum requirements

Examples

Copeland

Voters not participating

Voters participating

Conclusion

Instant-runoff voting

Voters not participating

Voters participating

Conclusion

Kemeny–Young method

Voters not participating

Voters participating

Conclusion

Majority Judgment

Voters not participating

Voters participating

Conclusion

Minimax

Voters not participating

Voters participating

Conclusion

Ranked pairs

Voters not participating

Voters participating

Conclusion

Schulze method

Voters not participating

Voters participating

Conclusion

See also

References

Further reading

External links

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