Multiple districts paradox

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A voting system satisfies join-consistency (also called the reinforcement criterion) if combining two sets of votes, both electing A over B, always results in a combined electorate that ranks A over B.^[1] It is a stronger form of the participation criterion. Systems that fail the consistency criterion (such as ranked-choice voting or Condorcet methods) are susceptible to the multiple-district paradox, which allows for a particularly egregious kind of gerrymander: it is possible to draw boundaries in such a way that a candidate who wins the overall election fails to carry even a single electoral district.^[1]

There are three variants of join-consistency:

1. Winner-consistency: if two districts elect the same winner A, A also wins in the combined district.
2. Ranking-consistency: if two districts rank a set of candidates exactly the same way, then the combined district returns the same ranking of all candidates.
3. Grading-consistency: if two different districts assign a candidate the same overall grade to a candidate, the overall grade for the candidate must still be the same.

A voting system is winner-consistent if and only if it is a point-summing method; in other words, it must be a positional voting system or score voting (including approval voting).^[2][3]

As shown below under Kemeny-Young, whether a system passes reinforcement can depend on whether the election selects a single winner or a full ranking of the candidates (sometimes referred to as ranking consistency): in some methods, two electorates with the same winner but different rankings may, when added together, lead to a different winner. Kemeny-Young is the only ranking-consistent Condorcet method, and no Condorcet method can be winner-consistent.^[3]

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:

Preferences	Voters
A > D > B > E > C	3
A > D > E > C > B	2
B > A > C > D > E	3
C > D > B > E > A	3
E > C > B > A > D	3
A > D > C > E > B	3
A > D > E > B > C	1
B > D > C > E > A	3
C > A > B > D > E	3
E > B > C > A > D	3

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.

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

Preferences	Voters
A > D > B > E > C	3
A > D > E > C > B	2
B > A > C > D > E	3
C > D > B > E > A	3
E > C > B > A > D	3

The results would be tabulated as follows:

Pairwise preferences

X[note 1] Y[note 2]	A	B	C	D	E
A		9 5	6 8	3 11	6 8
B	5 9		8 6	8 6	5 9
C	8 6	6 8		5 9	8 6
D	11 3	6 8	9 5		3 11
E	8 6	9 5	6 8	11 3
Pairwise election results (won-tied-lost)	3–0–1	2–0–2	2–0–2	2–0–2	1–0–3

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.

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

Preferences	Voters
A > D > C > E > B	3
A > D > E > B > C	1
B > D > C > E > A	3
C > A > B > D > E	3
E > B > C > A > D	3

The results would be tabulated as follows:

Pairwise election results

X Y	A	B	C	D	E
A		6 7	9 4	3 10	6 7
B	7 6		6 7	4 9	7 6
C	4 9	7 6		7 6	4 9
D	10 3	9 4	6 7		3 10
E	7 6	6 7	9 4	10 3
Pairwise election results (won-tied-lost)	3–0–1	2–0–2	2–0–2	2–0–2	1–0–3

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.

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

Preferences	Voters
A > D > B > E > C	3
A > D > C > E > B	3
A > D > E > B > C	1
A > D > E > C > B	2
B > A > C > D > E	3
B > D > C > E > A	3
C > A > B > D > E	3
C > D > B > E > A	3
E > B > C > A > D	3
E > C > B > A > D	3

The results would be tabulated as follows:

Pairwise election results

X Y	A	B	C	D	E
A		15 12	15 12	6 21	12 15
B	12 15		14 13	12 15	12 15
C	12 15	13 14		12 15	12 15
D	21 6	15 12	15 12		6 21
E	15 12	15 12	15 12	21 6
Pairwise election results (won-tied-lost)	2–0–2	3–0–1	4–0–0	1–0–3	0–0–4

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

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:

Preferences	Voters
A > B > C	4
B > A > C	2
C > B > A	4
A > B > C	4
B > A > C	6
C > A > B	3

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.

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

Preferences	Voters
A > B > C	4
B > A > C	2
C > B > A	4

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.

Candidate	Votes in round
	1st	2nd
A	4	6
B	2
C	4	4

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

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

Preferences	Voters
A > B > C	4
B > A > C	6
C > A > B	3

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.

Candidate	Votes in round
	1st	2nd
A	4	7
B	6	6
C	3

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

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

Preferences	Voters
A > B > C	8
B > A > C	8
C > A > B	3
C > B > A	4

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.

Candidate	Votes in round
	1st	2nd
A	8	11
B	8	12
C	7

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

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.

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

Group	Preferences	Voters
1st	A > B > C	7
	B > C > A	6
	C > A > B	3
2nd	A > C > B	8
	B > A > C	7
	C > B > A	7

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.

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

Preferences	Voters
A > B > C	7
B > C > A	6
C > A > B	3

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

Pairs of choices		Voters who prefer
X	Y	X over Y	Neither	Y over X
A	B	10	0	6
A	C	7	0	9
B	C	13	0	3

The ranking scores of all possible rankings are:

Preferences	1 vs 2	1 vs 3	2 vs 3	Total
A > B > C	10	7	13	30
A > C > B	7	10	3	20
B > A > C	6	13	7	26
B > C > A	13	6	9	28
C > A > B	9	3	10	22
C > B > A	3	9	6	18

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

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

Preferences	Voters
A > C > B	8
B > A > C	7
C > B > A	7

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

Pairs of choices		Voters who prefer
X	Y	X over Y	Neither	Y over X
A	B	8	0	14
A	C	15	0	7
B	C	7	0	15

The ranking scores of all possible rankings are:

Preferences	1 vs 2	1 vs 3	2 vs 3	Total
A > B > C	8	15	7	30
A > C > B	15	8	15	38
B > A > C	14	7	15	36
B > C > A	7	14	7	28
C > A > B	7	15	8	30
C > B > A	15	7	14	36

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

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

Preferences	Voters
A > B > C	7
A > C > B	8
B > A > C	7
B > C > A	6
C > A > B	3
C > B > A	7

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

Pairs of choices		Voters who prefer
X	Y	X over Y	Neither	Y over X
A	B	18	0	20
A	C	22	0	16
B	C	20	0	18

The ranking scores of all possible rankings are:

Preferences	1 vs 2	1 vs 3	2 vs 3	Total
A > B > C	18	22	20	60
A > C > B	22	18	18	58
B > A > C	20	20	22	62
B > C > A	20	20	16	56
C > A > B	16	18	18	52
C > B > A	18	16	20	54

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

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

The Kemeny-Young 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.

The Kemeny-Young 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-Young 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 ${\displaystyle V_{1}\cap V_{2}=\emptyset }$), computing the Kemeny-Young scores for these subsets and adding it up:

${\displaystyle {\text{(I)}}\quad s_{V}({\mathcal {R}})=s_{V_{1}}({\mathcal {R}})+s_{V_{2}}({\mathcal {R}})}$.

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-Young score for the ranking ${\displaystyle {\mathcal {R}}}$ in each electorate is bigger than for every other ranking ${\displaystyle {\mathcal {R}}'}$:

${\displaystyle {\begin{aligned}{\text{(II)}}\quad \forall {\mathcal {R}}':{}&s_{V_{1}}({\mathcal {R}})>s_{V_{1}}({\mathcal {R}}')\\{\text{(III)}}\quad \forall {\mathcal {R}}':{}&s_{V_{2}}({\mathcal {R}})>s_{V_{2}}({\mathcal {R}}')\end{aligned}}}$

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

${\displaystyle s_{V}({\mathcal {R}})\ {\stackrel {(I)}{=}}\ s_{V_{1}}({\mathcal {R}})+s_{V_{2}}({\mathcal {R}})\ {\stackrel {(II)}{>}}\ s_{V_{1}}({\mathcal {R}}')+s_{V_{2}}({\mathcal {R}})\ {\stackrel {(III)}{>}}\ s_{V_{1}}({\mathcal {R}}')+s_{V_{2}}({\mathcal {R}}')\ {\stackrel {(I)}{=}}\ s_{V}({\mathcal {R}}')\quad q.e.d.}$

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

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

Candidate		Voters
A	B
Excellent	Fair	3
Poor	Fair	2
Fair	Poor	3
Poor	Fair	2

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.

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

Candidates		Voters
A	B
Excellent	Fair	3
Poor	Fair	2

The sorted ratings would be as follows:

Candidate	↓ Median point
A
B
	Excellent   Good   Fair   Poor

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.

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

Candidates		Voters
A	B
Fair	Poor	3
Poor	Fair	2

The sorted ratings would be as follows:

Candidate	↓ Median point
A
B
	Excellent   Good   Fair   Poor

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.

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

Candidates		Voters
A	B
Excellent	Fair	3
Fair	Poor	3
Poor	Fair	4

The sorted ratings would be as follows:

Candidate	↓ Median point
A
B
	Excellent   Good   Fair   Poor

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:

Candidate	↓ Median point
A
B

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

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.

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:

Preferences	Voters
A > B > C	7
B > C > A	6
C > A > B	3
A > C > B	9
B > A > C	8
C > B > A	6

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.

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

Preferences	Voters
A > B > C	7
B > C > A	6
C > A > B	3

The results would be tabulated as follows:

Pairwise election results

X Y	A	B	C
A		6 10	9 7
B	10 6		3 13
C	7 9	13 3
Pairwise election results (won-tied-lost)	1–0–1	1–0–1	1–0–1

The sorted list of victories would be:

Pair	Winner
B (13) vs C (3)	B 13
A (10) vs B (6)	A 10
A (7) vs C (9)	C 9

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.

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

Preferences	Voters
A > C > B	9
B > A > C	8
C > B > A	6

The results would be tabulated as follows:

Pairwise election results

X Y	A	B	C
A		14 9	6 17
B	9 14		15 8
C	17 6	8 15
Pairwise election results (won-tied-lost)	1–0–1	1–0–1	1–0–1

The sorted list of victories would be:

Pair	Winner
A (17) vs C (6)	A 17
B (8) vs C (15)	C 15
A (9) vs B (14)	B 14

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.

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

Preferences	Voters
A > B > C	7
A > C > B	9
B > A > C	8
B > C > A	6
C > A > B	3
C > B > A	6

The results would be tabulated as follows:

Pairwise election results

X Y	A	B	C
A		20 19	15 24
B	19 20		18 21
C	24 15	21 18
Pairwise election results (won-tied-lost)	1–0–1	2–0–0	0–0–2

The sorted list of victories would be:

Pair	Winner
A (25) vs C (15)	A 24
B (21) vs C (18)	B 21
A (19) vs B (20)	B 20

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.

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.

1. ^ John H Smith, "Aggregation of preferences with variable electorate", Econometrica, Vol. 41 (1973), pp. 1027–1041.
2. ^ D. R. Woodall, "Properties of preferential election rules", Voting matters, Issue 3 (December 1994), pp. 8–15.
3. ^ H. P. Young, "Social Choice Scoring Functions", SIAM Journal on Applied Mathematics Vol. 28, No. 4 (1975), pp. 824–838.