Consistency criterion

Examples

Copeland

This example shows that Copeland's method violates the consistency criterion. Assume five candidates A, B, C, D and E with 27 voters with the following preferences:

Now, the set of all voters is divided into two groups at the bold line. The voters over the line are the first group of voters; the others are the second group of voters.

First group of voters

In the following the Copeland winner for the first group of voters is determined.

The results would be tabulated as follows:

Result: With the votes of the first group of voters, A can defeat three of the four opponents, whereas no other candidate wins against more than two opponents. Thus, A is elected Copeland winner by the first group of voters.

Second group of voters

Now, the Copeland winner for the second group of voters is determined.

The results would be tabulated as follows:

Result: Taking only the votes of the second group in account, again, A can defeat three of the four opponents, whereas no other candidate wins against more than two opponents. Thus, A is elected Copeland winner by the second group of voters.

All voters

Finally, the Copeland winner of the complete set of voters is determined.

The results would be tabulated as follows:

Result: C is the Condorcet winner, thus Copeland chooses C as winner.

Instant-runoff voting

This example shows that Instant-runoff voting violates the consistency criterion. Assume three candidates A, B and C and 23 voters with the following preferences:

Now, the set of all voters is divided into two groups at the bold line. The voters over the line are the first group of voters; the others are the second group of voters.

First group of voters

In the following the instant-runoff winner for the first group of voters is determined.

B has only 2 votes and is eliminated first. Its votes are transferred to A. Now, A has 6 votes and wins against C with 4 votes.

Result: A wins against C, after B has been eliminated.

Second group of voters

Now, the instant-runoff winner for the second group of voters is determined.

C has the fewest votes, a count of 3, and is eliminated. A benefits from that, gathering all the votes from C. Now, with 7 votes A wins against B with 6 votes.

Result: A wins against B, after C has been eliminated.

All voters

Finally, the instant runoff winner of the complete set of voters is determined.

C has the fewest first preferences and so is eliminated first, its votes are split: 4 are transferred to B and 3 to A. Thus, B wins with 12 votes against 11 votes of A.

Result: B wins against A, after C is eliminated.

Conclusion

A is the instant-runoff winner within the first group of voters and also within the second group of voters. However, both groups combined elect B as the instant-runoff winner. Thus, instant-runoff voting fails the consistency criterion.

Kemeny method

This example shows that the Kemeny method violates the consistency criterion. Assume three candidates A, B and C and 38 voters with the following preferences:

Now, the set of all voters is divided into two groups at the bold line. The voters over the line are the first group of voters; the others are the second group of voters.

First group of voters

In the following the Kemeny winner for the first group of voters is determined.

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

The ranking scores of all possible rankings are:

Result: The ranking A > B > C has the highest ranking score. Thus, A wins ahead of B and C.

Second group of voters

Now, the Kemeny winner for the second group of voters is determined.

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

The ranking scores of all possible rankings are:

Result: The ranking A > C > B has the highest ranking score. Hence, A wins ahead of C and B.

All voters

Finally, the Kemeny winner of the complete set of voters is determined.

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

The ranking scores of all possible rankings are:

Result: The ranking B > A > C has the highest ranking score. So, B wins ahead of A and C.

Conclusion

A is the Kemeny winner within the first group of voters and also within the second group of voters. However, both groups combined elect B as the Kemeny winner. Thus, the Kemeny method fails the reinforcement criterion.

Ranking consistency

The Kemeny method satisfies ranking consistency; that is, if the electorate is divided arbitrarily into two parts and separate elections in each part result in the same ranking being selected, an election of the entire electorate also selects that ranking. In fact, it is the only Condorcet method that satisfies ranking consistency.

Informal proof

The Kemeny score of a ranking ${\displaystyle \mathcal{R}}$ is computed by summing up the number of pairwise comparisons on each ballot that match the ranking ${\displaystyle \mathcal{R}}$. Thus, the Kemeny score ${\displaystyle s_V(\mathcal{R})}$ for an electorate ${\displaystyle V}$ can be computed by separating the electorate into disjoint subsets ${\displaystyle V=V_1\cup V_2}$ (with ${\displaystyle V_1\cap V_2=\mathrm{\varnothing }}$), computing the Kemeny scores for these subsets and adding it up:

${\displaystyle \text{(I)}{s}_V(\mathcal{R})=s_{V_1}(\mathcal{R})+s_{V_2}(\mathcal{R})}$.

Now, consider an election with electorate ${\displaystyle V}$. The premise of reinforcement is to divide the electorate arbitrarily into two parts ${\displaystyle V=V_1\cup V_2}$, and in each part the same ranking ${\displaystyle \mathcal{R}}$ is selected. This means, that the Kemeny score for the ranking ${\displaystyle \mathcal{R}}$ in each electorate is bigger than for every other ranking ${\displaystyle {\mathcal{R}}^{\prime }}$:

${\displaystyle \begin{array}{cc}\text{(II)}\mathrm{\forall }{\mathcal{R}}^{\prime }:& s_{V_1}(\mathcal{R})>s_{V_1}({\mathcal{R}}^{\prime })\\ \text{(III)}\mathrm{\forall }{\mathcal{R}}^{\prime }:& s_{V_2}(\mathcal{R})>s_{V_2}({\mathcal{R}}^{\prime })\end{array}}$

Now, it has to be shown, that the Kemeny score of the ranking ${\displaystyle \mathcal{R}}$ in the entire electorate is bigger than the Kemeny score of every other ranking ${\displaystyle {\mathcal{R}}^{\prime }}$:

${\displaystyle s_V(\mathcal{R})\stackrel{(I)}{=}{s}_{V_1}(\mathcal{R})+s_{V_2}(\mathcal{R})\stackrel{(II)}{>}{s}_{V_1}({\mathcal{R}}^{\prime })+s_{V_2}(\mathcal{R})\stackrel{(III)}{>}{s}_{V_1}({\mathcal{R}}^{\prime })+s_{V_2}({\mathcal{R}}^{\prime })\stackrel{(I)}{=}{s}_V({\mathcal{R}}^{\prime })q.e.d.}$

Thus, the Kemeny method is consistent with respect to complete rankings.

Majority Judgment

This example shows that majority judgment violates reinforcement. Assume two candidates A and B and 10 voters with the following ratings:

Now, the set of all voters is divided into two groups at the bold line. The voters over the line are the first group of voters; the others are the second group of voters.

First group of voters

In the following the majority judgment winner for the first group of voters is determined.

The sorted ratings would be as follows:

Result: With the votes of the first group of voters, A has the median rating of "Excellent" and B has the median rating of "Fair". Thus, A is elected majority judgment winner by the first group of voters.

Second group of voters

Now, the majority judgment winner for the second group of voters is determined.

The sorted ratings would be as follows:

Result: Taking only the votes of the second group in account, A has the median rating of "Fair" and B the median rating of "Poor". Thus, A is elected majority judgment winner by the second group of voters.

All voters

Finally, the majority judgment winner of the complete set of voters is determined.

The sorted ratings would be as follows:

The median ratings for A and B are both "Fair". Since there is a tie, "Fair" ratings are removed from both, until their medians become different. After removing 20% "Fair" ratings from the votes of each, the sorted ratings are now:

Result: Now, the median rating of A is "Poor" and the median rating of B is "Fair". Thus, B is elected majority judgment winner.

Conclusion

A is the majority judgment winner within the first group of voters and also within the second group of voters. However, both groups combined elect B as the Majority Judgment winner. Thus, Majority Judgment fails the consistency criterion.

Ranked Pairs

This example shows that the ranked pairs method violates the consistency criterion. Assume three candidates A, B and C with 39 voters with the following preferences:

Now, the set of all voters is divided into two groups at the bold line. The voters over the line are the first group of voters; the others are the second group of voters.

First group of voters

In the following the ranked pairs winner for the first group of voters is determined.

The results would be tabulated as follows:

The sorted list of victories would be:

Result: B > C and A > B are locked in first (and C > A can't be locked in after that), so the full ranking is A > B > C. Thus, A is elected ranked pairs winner by the first group of voters.

Second group of voters

Now, the ranked pairs winner for the second group of voters is determined.

The results would be tabulated as follows:

The sorted list of victories would be:

Result: Taking only the votes of the second group in account, A > C and C > B are locked in first (and B > A can't be locked in after that), so the full ranking is A > C > B. Thus, A is elected ranked pairs winner by the second group of voters.

All voters

Finally, the ranked pairs winner of the complete set of voters is determined.

The results would be tabulated as follows:

The sorted list of victories would be:

Result: Now, all three pairs (A > C, B > C and B > A) can be locked in without a cycle. The full ranking is B > A > C. Thus, ranked pairs chooses B as winner, which is the Condorcet winner, due to the lack of a cycle.

Conclusion

A is the ranked pairs winner within the first group of voters and also within the second group of voters. However, both groups combined elect B as the ranked pairs winner. Thus, the ranked pairs method fails the consistency criterion.