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 placeIncompatibility with the Condorcet 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: 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: The results would be tabulated as follows: 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: The results would be tabulated as follows: 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.Voters not participating
If they don't show up at the election the remaining voters would be: The following outcome results: Result: After A is eliminated first, B gets their votes and wins.Voters participating
If they participate in the election, the preferences list is: The outcome changes as follows: 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: 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: The Kemeny–Young method arranges the pairwise comparison counts in the following tally table: 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: The Kemeny–Young method arranges the pairwise comparison counts in the following tally table: 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: 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: The sorted ratings would be as follows: 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: The sorted ratings would be as follows: 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: 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: The results would be tabulated as follows: * indicates voters who preferred the candidate listed in the column caption to the candidate listed in the row caption * 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: The results would be tabulated as follows: 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: 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: The results would be tabulated as follows: The sorted list of victories would be: 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: The results would be tabulated as follows: The sorted list of victories would be: 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: 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: The pairwise preferences would be tabulated as follows: 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). 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: The pairwise preferences would be tabulated as follows: Now, the strongest paths have to be identified, e.g. the path C > A > D is stronger than the direct path C > D. 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
*References
Further reading
* Woodall, Douglas R,