Participation criterion

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The participation criterion, also called vote or population monotonicity, is a voting system criterion that says that a candidate should never lose an election because they have "too much support." It says that adding voters who support A over B should not cause A to lose the election to B. [1]

Contents

Voting systems that fail the participation criterion exhibit the no-show paradox, and allow voters to do better by not showing up to vote. [2] Such voting systems arguably violate the principle of one man, one vote, as participation failures are cases where "one man has negative-one votes."

Positional and score voting methods satisfy the participation criterion, while all Condorcet methods [3] [4] and highest median methods [5] can fail in some situations.

Common failure modes

The most common cause of participation failures is caused by the use of runoff systems (including instant-runoff voting).

Incompatibility with the Condorcet criterion

With 3 candidates, Minimax Condorcet and its variants (including Ranked Pairs and Schulze's method) satisfy the participation criterion. [3] However, with more than 3 candidates, every Condorcet system is guaranteed to fail the participation criterion. [3] [4] Similar incompatibilities have also been proven for set-valued voting rules. [4] [6] [7]

Certain conditions 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 one of the voter's most-preferred candidates does not change the winner away from A. Similarly, weak negative involvement requires that adding a ballot in which A is one of the voter's 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. [8] If the ballots must express A as their sole favorite, then there exist Condorcet methods that pass.

Another condition that is weaker than participation is half-way monotonicity, which requires that a voter cannot be better off by completely reversing their ballot.[ clarification needed ] Again, half-way monotonicity is incompatible with the Condorcet criterion. [9]

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:

Preferences# of voters
A > B > C > D3
A > C > D > B1
A > D > C > B1
B > A > C > D4
D > C > B > A4

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:

Preferences# of voters
A > C > D > B1
A > D > C > B1
B > A > C > D4
D > C > B > A4

The results would be tabulated as follows:

Pairwise election results
X
ABCD
YA[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,
won-tied-lost
2-0-11-0-21-1-11-1-1

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 three unconfident voters decide to participate:

Preferences# of voters
A > B > C > D3
A > C > D > B1
A > D > C > B1
B > A > C > D4
D > C > B > A4

The results would be tabulated as follows:

Pairwise election results
X
ABCD
YA[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,
won-tied-lost
2-0-13-0-01-0-20-0-3

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 (in blue) unconfident whether to vote.

Preferences# of voters
A > B > C2
A > B > C3
B > C > A4
C > A > B6

Voters not participating

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

Preferences# of voters
A > B > C3
B > C > A4
C > A > B6

The following outcome results:

CandidateVotes for round
1st2nd
A3
B47
C66

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

Voters participating

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

Preferences# of voters
A > B > C5
B > C > A4
C > A > B6

The outcome changes as follows:

CandidateVotes for round
1st2nd
A55
B4
C610

Result: Now, B is eliminated first and C gets their 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:

Preferences# of voters
A > B > C > D3
A > C > B > D3
A > D > C > B4
B > A > D > C4
C > B > A > D2
D > B > A > C2
D > C > B > A3

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:

Preferences# of voters
A > C > B > D3
A > D > C > B4
B > A > D > C4
C > B > A > D2
D > B > A > C2
D > C > B > A3

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

Candidate pairsNumber who prefer…
XYXNeitherY
AB7011
AC1305
AD1305
BC6012
BD909
CD5013

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:

Preferences# of voters
A > B > C > D3
A > C > B > D3
A > D > C > B4
B > A > D > C4
C > B > A > D2
D > B > A > C2
D > C > B > A3

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

Candidate pairsNumber who prefer…
XYXNeitherY
AB10011
AC1605
AD1605
BC9012
BD1209
CD8013

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

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
AB
ExcellentGood2
FairPoor2
PoorGood1

The two voters rating A "Excellent" are unsure 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
AB
FairPoor2
PoorGood1

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 voters decide to participate:

Candidates# of
voters
AB
ExcellentGood2
FairPoor2
PoorGood1

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:

Preferences# of voters
A > B > C > D2
A > B > D > C2
B > D > C > A6
C > A > B > D5
D > A > B > C1
D > C > A > B2

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.

Voters not participating

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 > C2
B > D > C > A6
C > A > B > D5
D > A > B > C1
D > C > A > B2

The results would be tabulated as follows:

Pairwise election results
X
ABCD
YA[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,
won-tied-lost
1-0-22-0-11-0-22-0-1
Worst opposing votes13101113
Worst margin104610
Worst opposition13101113
  • [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:

Preferences# of voters
A > B > C > D2
A > B > D > C2
B > D > C > A6
C > A > B > D5
D > A > B > C1
D > C > A > B2

The results would be tabulated as follows:

Pairwise election results
X
ABCD
YA[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,
won-tied-lost
1-1-12-0-11-0-21-1-1
Worst opposing votes13121115
Worst margin8648
Worst opposition13121115

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:

Preferences# of voters
A > B > C > D4
A > D > B > C8
B > C > A > D7
C > D > B > A7

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:

Preferences# of voters
A > D > B > C8
B > C > A > D7
C > D > B > A7

The results would be tabulated as follows:

Pairwise election results
X
ABCD
YA[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,
won-tied-lost
1-0-22-0-12-0-11-0-2

The sorted list of victories would be:

PairWinner
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:

Preferences# of voters
A > B > C > D4
A > D > B > C8
B > C > A > D7
C > D > B > A7

The results would be tabulated as follows:

Pairwise election results
X
ABCD
YA[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,
won-tied-lost
1-0-22-0-12-0-11-0-2

The sorted list of victories would be:

PairWinner
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

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 > D2
B > A > D > C7
B > C > A > D1
B > D > C > A2
C > A > D > B7
D > B > A > C2
D > C > A > B4

The two voters 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 23 voters would be:

Preferences# of voters
B > A > D > C7
B > C > A > D1
B > D > C > A2
C > A > D > B7
D > B > A > C2
D > C > A > B4

The pairwise preferences would be tabulated as follows:

Matrix of pairwise preferences
d[·, A]d[·, B]d[·, C]d[·, D]
d[A, ·]11915
d[B, ·]121210
d[C, ·]14118
d[D, ·]81315

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, ·]131515
p[B, ·]121212
p[C, ·]141314
p[D, ·]141315

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:

Preferences# of voters
A > B > C > D2
B > A > D > C7
B > C > A > D1
B > D > C > A2
C > A > D > B7
D > B > A > C2
D > C > A > B4

The pairwise preferences would be tabulated as follows:

Matrix of pairwise preferences
d[·, A]d[·, B]d[·, C]d[·, D]
d[A, ·]131117
d[B, ·]121412
d[C, ·]141110
d[D, ·]81315

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, ·]131517
p[B, ·]141414
p[C, ·]141314
p[D, ·]141315

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.

See also

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Further reading