View source for Multiple districts paradox

{{Short description|Property of electoral systems}}
{{Electoral systems sidebar|expanded=Paradox}}

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''.{{ref}} It is a stronger form of the [[participation criterion]]. Systems that fail the consistency criterion (such as [[Instant-runoff voting]] or [[Condorcet method|Condorcet methods]]) are susceptible to the '''multiple-district paradox''', a [[Pathological (mathematics)|pathological]] behavior where a candidate can win an election without carrying even a single precinct.<ref name=":1" /> Conversely, it can be seen as allowing 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]].<ref name=":1">{{Cite journal |last1=Franceschini |first1=Fiorenzo |last2=Maisano |first2=Domenico A. |date=2022-06-01 |title=Analysing paradoxes in design decisions: the case of "multiple-district" paradox |journal=International Journal on Interactive Design and Manufacturing (IJIDeM) |language=en |volume=16 |issue=2 |pages=677–689 |doi=10.1007/s12008-022-00860-x |issn=1955-2505|doi-access=free }}</ref>

Rules susceptible to the multiple-districts paradox include all [[Majority preferred candidate|majority-rule methods]]<ref name=":0" /> and [[Instant-runoff voting|instant-runoff (or ranked-choice) voting]]. Rules that are not susceptible to it include all [[Positional voting|positional voting rules]] (such as [[first-preference plurality]] and the [[Borda count]]) as well as [[score voting]] and [[approval voting]].

== Variants ==
There are three variants of join-consistency:

# '''Winner-consistency'''—if two districts elect the same winner ''A'', ''A'' also wins in a new district formed by combining the two.
# '''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.
# '''Grading-consistency'''—if two districts assign a candidate the same [[Rated voting|overall grade]], then the combined district assigns that candidate the same grade.

When mentioned without 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]]).<ref>{{Cite book |last1=Balinski |first1=Michel |url=http://dx.doi.org/10.7551/mitpress/9780262015134.001.0001 |title=Majority Judgment |last2=Laraki |first2=Rida |date=2011-01-28 |publisher=The MIT Press |doi=10.7551/mitpress/9780262015134.001.0001 |isbn=978-0-262-01513-4}}</ref><ref name=":0" />

As shown below for [[#Kemeny-Young method|Kemeny-Young]] and [[Highest median voting rules|majority judgment]], these three variants do not always agree with each other (which contrasts with most other voting criteria). Kemeny-Young is the only ranking-consistent Condorcet method, and no Condorcet method can be winner-consistent.<ref name=":0">{{Cite journal |last1=Young |first1=H. P. |last2=Levenglick |first2=A. |date=1978 |title=A Consistent Extension of Condorcet's Election Principle |url=http://pure.iiasa.ac.at/700/1/RR-77-015.pdf |journal=SIAM Journal on Applied Mathematics |volume=35 |issue=2 |pages=285–300 |doi=10.1137/0135023 |jstor=2100667 |issn=0036-1399}}</ref>

{{TOC limit|limit=3}}

== Examples ==<!-- Replace with single unified example for all Condorcet rules -->

=== Copeland ===
{{Main|Copeland's method}}

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:
{| class="wikitable"
|-
! 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
|- style="border-top: 2px black solid"
| 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.

==== First group of voters ====
In the following the Copeland winner for the first group of voters is determined.
{| class="wikitable"
|-
! 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:
{| class="wikitable" style="text-align:center"
|+ Pairwise preferences
! {{diagonal split header
|2=X{{NoteTag|indicates voters who preferred the candidate listed in the column caption to the candidate listed in the row caption}}
|1=Y{{NoteTag|indicates voters who preferred the candidate listed in the row caption to the candidate listed in the column caption}}
}}
! style="background-color:#c0c0ff" | A
! style="background-color:#c0c0ff" | B
! style="background-color:#c0c0ff" | C
! style="background-color:#c0c0ff" | D
! style="background-color:#c0c0ff" | E
|-
! style="background-color:#ffc0c0" | A
|
| {{diagonal split header|5|9|#e0e0ff}}
| {{diagonal split header|8|6|#ffe0e0}}
| {{diagonal split header|11|3|#ffe0e0}}
| {{diagonal split header|8|6|#ffe0e0}}
|-
! style="background-color:#ffc0c0" | B
| {{diagonal split header|9|5|#ffe0e0}}
|
| {{diagonal split header|6|8|#e0e0ff}}
| {{diagonal split header|6|8|#e0e0ff}}
| {{diagonal split header|9|5|#ffe0e0}}
|-
! style="background-color:#ffc0c0" | C
| {{diagonal split header|6|8|#e0e0ff}}
| {{diagonal split header|8|6|#ffe0e0}}
|
| {{diagonal split header|9|5|#ffe0e0}}
| {{diagonal split header|6|8|#e0e0ff}}
|-
! style="background-color:#ffc0c0" | D
| {{diagonal split header|3|11|#e0e0ff}}
| {{diagonal split header|8|6|#ffe0e0}}
| {{diagonal split header|5|9|#e0e0ff}}
|
| {{diagonal split header|11|3|#ffe0e0}}
|-
| style="background-color:#ffc0c0" | E
| {{diagonal split header|6|8|#e0e0ff}}
| {{diagonal split header|5|9|#e0e0ff}}
| {{diagonal split header|8|6|#ffe0e0}}
| {{diagonal split header|3|11|#e0e0ff}}
|
|-
! style="background-color:#c0c0ff" | Pairwise election results<br/>(won-tied-lost)
| 3–0–1
| 2–0–2
| 2–0–2
| 2–0–2
| 1–0–3
|}
{{NoteFoot}}

'''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.
{| class="wikitable"
|-
! 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:
{| class=wikitable
|+ Pairwise election results
! {{diagonal split header|2=X|1=Y}}
! style="background-color:#c0c0ff" | A
! style="background-color:#c0c0ff" | B
! style="background-color:#c0c0ff" | C
! style="background-color:#c0c0ff" | D
! style="background-color:#c0c0ff" | E
|-
! style="background-color:#ffc0c0" | A
|
| {{diagonal split header|7|6|#ffe0e0}}
| {{diagonal split header|4|9|#e0e0ff}}
| {{diagonal split header|10|3|#ffe0e0}}
| {{diagonal split header|7|6|#ffe0e0}}
|-
! style="background-color:#ffc0c0" | B
| {{diagonal split header|6|7|#e0e0ff}}
|
| {{diagonal split header|7|6|#ffe0e0}}
| {{diagonal split header|9|4|#ffe0e0}}
| {{diagonal split header|6|7|#e0e0ff}}
|-
! style="background-color:#ffc0c0" | C
| {{diagonal split header|9|4|#ffe0e0}}
| {{diagonal split header|6|7|#e0e0ff}}
|
| {{diagonal split header|6|7|#e0e0ff}}
| {{diagonal split header|9|4|#ffe0e0}}
|-
! style="background-color:#ffc0c0" | D
| {{diagonal split header|3|10|#e0e0ff}}
| {{diagonal split header|4|9|#e0e0ff}}
| {{diagonal split header|7|6|#ffe0e0}}
|
| {{diagonal split header|10|3|#ffe0e0}}
|-
! style="background-color:#ffc0c0" | E
| {{diagonal split header|6|7|#e0e0ff}}
| {{diagonal split header|7|6|#ffe0e0}}
| {{diagonal split header|4|9|#e0e0ff}}
| {{diagonal split header|3|10|#e0e0ff}}
|
|-
! style="background-color:#c0c0ff" | Pairwise election results<br/>(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.

==== All voters ====
Finally, the Copeland winner of the complete set of voters is determined.
{| class="wikitable"
|-
! 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:
{| class=wikitable
|+ Pairwise election results
! {{diagonal split header|2=X|1=Y}}
! style="background-color:#c0c0ff" | A
! style="background-color:#c0c0ff" | B
! style="background-color:#c0c0ff" | C
! style="background-color:#c0c0ff" | D
! style="background-color:#c0c0ff" | E
|-
! style="background-color:#ffc0c0" | A
|
| {{diagonal split header|12|15|#e0e0ff}}
| {{diagonal split header|12|15|#e0e0ff}}
| {{diagonal split header|21|6|#ffe0e0}}
| {{diagonal split header|15|12|#ffe0e0}}
|-
! style="background-color:#ffc0c0" | B
| {{diagonal split header|15|12|#ffe0e0}}
|
| {{diagonal split header|13|14|#e0e0ff}}
| {{diagonal split header|15|12|#ffe0e0}}
| {{diagonal split header|15|12|#ffe0e0}}
|-
! style="background-color:#ffc0c0" | C
| {{diagonal split header|15|12|#ffe0e0}}
| {{diagonal split header|14|13|#ffe0e0}}
|
| {{diagonal split header|15|12|#ffe0e0}}
| {{diagonal split header|15|12|#ffe0e0}}
|-
! style="background-color:#ffc0c0" | D
| {{diagonal split header|6|21|#e0e0ff}}
| {{diagonal split header|12|15|#e0e0ff}}
| {{diagonal split header|12|15|#e0e0ff}}
|
| {{diagonal split header|21|6|#ffe0e0}}
|-
! style="background-color:#ffc0c0" | E
| {{diagonal split header|12|15|#e0e0ff}}
| {{diagonal split header|12|15|#e0e0ff}}
| {{diagonal split header|12|15|#e0e0ff}}
| {{diagonal split header|6|21|#e0e0ff}}
|
|-
! style="background-color:#c0c0ff" | Pairwise election results<br/>(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.

=== Instant-runoff voting ===
{{Main|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:
{| class="wikitable"
|-
! Preferences !! Voters
|-
| A > B > C   || 4
|-
| B > A > C   || 2
|-
| C > B > A   || 4
|- style="border-top: 2px black solid"
| 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.

==== First group of voters ====
In the following the instant-runoff winner for the first group of voters is determined.
{| class="wikitable"
|-
! 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.
{| class="wikitable"
|-
! rowspan=2 | Candidate
! colspan=2 | Votes in round
|-
! 1st !! 2nd
|-
| A || bgcolor=#ddffbb | 4 || bgcolor=#bbffbb | '''6'''
|-
| B || bgcolor=#ffbbbb | ''2''
|-
| C || bgcolor=#ddffbb | 4 || bgcolor=#ffbbbb | ''4''
|}

'''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.
{| class="wikitable"
|-
! 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.
{| class="wikitable"
|-
! rowspan=2 | Candidate
! colspan=2 | Votes in round
|-
! 1st !! 2nd
|-
| A || bgcolor=#ddffbb|4 || bgcolor=#bbffbb|'''7'''
|-
| B || bgcolor=#ddffbb|6 || bgcolor=#ffbbbb|''6''
|-
| C || bgcolor=#ffbbbb|''3''
|}

'''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.

{| class="wikitable"
|-
! 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.
{| class="wikitable"
|-
! rowspan=2 | Candidate
! colspan=2 | Votes in round
|-
! 1st !! 2nd
|-
| A || bgcolor=#ddffbb|8 || bgcolor=#ffbbbb|''11''
|-
| B || bgcolor=#ddffbb|8 || bgcolor=#bbffbb|'''12'''
|-
| C || bgcolor=#ffbbbb|''7''
|}

'''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-Young method ===
{{Main|Kemeny-Young method}}

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:
{| class="wikitable"
|-
! Group
! Preferences || Voters
|-
| rowspan=3 | 1st
| A > B > C   || 7
|-
| B > C > A   || 6
|-
| C > A > B   || 3
|- style="border-top:2px black solid"
| rowspan=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.

==== First group of voters ====
In the following the Kemeny-Young winner for the first group of voters is determined.
{| class="wikitable"
|-
! 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:

{| class="wikitable"
|-
! colspan=2 | Pairs of choices
! colspan=3 | 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:
{| class="wikitable"
|-
! Preferences !! 1 vs 2 !! 1 vs 3 !! 2 vs 3 !! Total
|-
| A > B > C   || 10       ||  7       || 13       || {{yes|'''30'''}}
|-
| A > C > B   ||  7       || 10       ||  3       || {{no|20}}
|-
| B > A > C   ||  6       || 13       ||  7       || {{no|26}}
|-
| B > C > A   || 13       ||  6       ||  9       || {{no|28}}
|-
| C > A > B   ||  9       ||  3       || 10       || {{no|22}}
|-
| C > B > A   ||  3       ||  9       ||  6       || {{no|18}}
|}

'''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-Young winner for the second group of voters is determined.
{| class="wikitable"
|-
! 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:

{| class="wikitable"
|-
! colspan=2 | Pairs of choices
! colspan=3 | 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:
{| class="wikitable"
|-
! Preferences !! 1 vs 2 !! 1 vs 3 !! 2 vs 3 !! Total
|-
| A > B > C   ||  8       || 15       ||  7       || {{no|30}}
|-
| A > C > B   || 15       ||  8       || 15       || {{yes|'''38'''}}
|-
| B > A > C   || 14       ||  7       || 15       || {{no|36}}
|-
| B > C > A   ||  7       || 14       ||  7       || {{no|28}}
|-
| C > A > B   ||  7       || 15       ||  8       || {{no|30}}
|-
| C > B > A   || 15       ||  7       || 14       || {{no|36}}
|}

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

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

{| class="wikitable"
|-
! 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:

{| class="wikitable"
|-
! colspan=2 | Pairs of choices
! colspan=3 | 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:
{| class="wikitable"
|-
! Preferences !! 1 vs 2 !! 1 vs 3 !! 2 vs 3 !! Total
|-
| A > B > C   || 18       || 22       || 20       || {{no|60}}
|-
| A > C > B   || 22       || 18       || 18       || {{no|58}}
|-
| B > A > C   || 20       || 20       || 22       || {{yes|'''62'''}}
|-
| B > C > A   || 20       || 20       || 16       || {{no|56}}
|-
| C > A > B   || 16       || 18       || 18       || {{no|52}}
|-
| C > B > A   || 18       || 16       || 20       || {{no|54}}
|}

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

==== Conclusion ====
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.

==== Ranking consistency ====
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.

===== Informal proof =====
The Kemeny-Young score of a ranking <math>\mathcal{R}</math> is computed by summing up the number of pairwise comparisons on each ballot that match the ranking <math>\mathcal{R}</math>. Thus, the Kemeny-Young score <math>s_V(\mathcal{R})</math> for an electorate <math>V</math> can be computed by separating the electorate into disjoint subsets <math>V = V_1 \cup V_2</math> (with <math>V_1 \cap V_2 = \emptyset</math>), computing the Kemeny-Young scores for these subsets and adding it up:
:<math>\text{(I)} \quad s_V(\mathcal{R}) = s_{V_1}(\mathcal{R}) + s_{V_2}(\mathcal{R})</math>.

Now, consider an election with electorate <math>V</math>. The premise of reinforcement is to divide the electorate arbitrarily into two parts <math>V = V_1 \cup V_2</math>, and in each part the same ranking <math>\mathcal{R}</math> is selected. This means, that the Kemeny-Young score for the ranking <math>\mathcal{R}</math> in each electorate is bigger than for every other ranking <math>\mathcal{R}'</math>:
:<math>\begin{align}
   \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{align}</math>

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

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

=== Majority Judgment ===
{{Main|Majority Judgment}}

This example shows that majority judgment violates reinforcement. Assume two candidates A and B and 10 voters with the following ratings:
{| class="wikitable"
|-
! colspan=2 | Candidate
! rowspan=2 | Voters
|-
! A !! B
|-
| bgcolor="green"     | Excellent || bgcolor="yellow"    | Fair || 3
|-
| bgcolor="orangered" | Poor      || bgcolor="yellow"    | Fair || 2
|- style="border-top: 2px black solid"
| bgcolor="yellow"    | Fair      || bgcolor="orangered" | Poor || 3
|-
| bgcolor="orangered" | Poor      || bgcolor="yellow"    | 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.

==== First group of voters ====
In the following the majority judgment winner for the first group of voters is determined.
{| class="wikitable"
|-
! colspan=2 | Candidates
! rowspan=2 | Voters
|-
! A !! B
|-
| bgcolor="green"     | Excellent || bgcolor="yellow" | Fair || 3
|-
| bgcolor="orangered" | Poor      || bgcolor="yellow" | Fair || 2
|}

The sorted ratings would be as follows:
{|
|-
| align=right | Candidate&nbsp;&nbsp;&nbsp;
|
{| cellpadding=0 width=500 border=0 cellspacing=0
|-
| width=49% | &nbsp;
| width=2% textalign=center | ↓
| width=49% | Median point
|}
|-
| align=right | A
|
{| cellpadding=0 width=500 border=0 cellspacing=0
|-
| bgcolor=green width=60% | &nbsp;
| bgcolor=orangered width=40% |
|}
|-
| align=right | B
|
{| cellpadding=0 width=500 border=0 cellspacing=0
|-
| bgcolor=yellow width=100% | &nbsp;
|}
|-
| &nbsp;
| &nbsp;
|-
| &nbsp;
|
{{legend inline|green|Excellent}}
{{legend inline|yellowgreen|Good}}
{{legend inline|yellow|Fair}}
{{legend inline|orangered|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.

==== Second group of voters ====
Now, the majority judgment winner for the second group of voters is determined.
{| class="wikitable"
|-
! colspan=2 | Candidates
! rowspan=2 | Voters
|-
! A !! B
|-
| bgcolor="yellow"    | Fair || bgcolor="orangered" | Poor || 3
|-
| bgcolor="orangered" | Poor || bgcolor="yellow"    | Fair || 2
|}

The sorted ratings would be as follows:
{|
|-
| align=right | Candidate&nbsp;&nbsp;&nbsp;
|
{| cellpadding=0 width=500 border=0 cellspacing=0
|-
| width=49% | &nbsp;
| width=2% textalign=center | ↓
| width=49% | Median point
|}
|-
| align=right | A
|
{| cellpadding=0 width=500 border=0 cellspacing=0
|-
| bgcolor=yellow width=60% | &nbsp;
| bgcolor=orangered width=40% |
|}
|-
| align=right | B
|
{| cellpadding=0 width=500 border=0 cellspacing=0
|-
| bgcolor=yellow width=40% | &nbsp;
| bgcolor=orangered width=60% |
|}
|-
| &nbsp;
| &nbsp;
|-
| &nbsp;
|
{{legend inline|green|Excellent}}
{{legend inline|yellowgreen|Good}}
{{legend inline|yellow|Fair}}
{{legend inline|orangered|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.

==== All voters ====
Finally, the majority judgment winner of the complete set of voters is determined.
{| class="wikitable"
|-
! colspan=2 | Candidates
! rowspan=2 | Voters
|-
! A !! B
|-
| bgcolor="green"     | Excellent || bgcolor="yellow"    | Fair || 3
|-
| bgcolor="yellow"    | Fair      || bgcolor="orangered" | Poor || 3
|-
| bgcolor="orangered" | Poor      || bgcolor="yellow"    | Fair || 4
|}

The sorted ratings would be as follows:
{|
|-
| align=right | Candidate&nbsp;&nbsp;&nbsp;
|
{| cellpadding=0 width=500 border=0 cellspacing=0
|-
| width=49% | &nbsp;
| width=2% textalign=center | ↓
| width=49% | Median point
|}
|-
| align=right | A
|
{| cellpadding=0 width=500 border=0 cellspacing=0
|-
| bgcolor=green width=30% | &nbsp;
| bgcolor=yellow width=30% | &nbsp;
| bgcolor=orangered width=40% |
|}
|-
| align=right | B
|
{| cellpadding=0 width=500 border=0 cellspacing=0
|-
| bgcolor=yellow width=70% | &nbsp;
| bgcolor=orangered width=30% |
|}
|-
| &nbsp;
| &nbsp;
|-
| &nbsp;
|
{{legend inline|green|Excellent}}
{{legend inline|yellowgreen|Good}}
{{legend inline|yellow|Fair}}
{{legend inline|orangered|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:
{|
|-
| align=right | Candidate&nbsp;&nbsp;&nbsp;
|
{| cellpadding=0 width=500 border=0 cellspacing=0
|-
| width=49% | &nbsp;
| width=2% textalign=center | ↓
| width=49% | Median point
|}
|-
| align=right | A
|
{| cellpadding=0 width=500 border=0 cellspacing=0
|-
| bgcolor=gray width=10% | &nbsp;
| bgcolor=green width=30% | &nbsp;
| bgcolor=yellow width=10% | &nbsp;
| bgcolor=orangered width=40% |

| bgcolor=gray width=10% |
|}
|-
| align=right | B
|
{| cellpadding=0 width=500 border=0 cellspacing=0
|-
| bgcolor=gray width=10% |
| bgcolor=yellow width=50% | &nbsp;
| bgcolor=orangered width=30% |

| bgcolor=gray width=10% |
|}
|}

'''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:
{| class="wikitable"
|-
! Preferences !! Voters
|-
| A > B > C   || 7
|-
| B > C > A   || 6
|-
| C > A > B   || 3
|- style="border-top: 2px black solid"
| 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.

==== First group of voters ====
In the following the ranked pairs winner for the first group of voters is determined.
{| class="wikitable"
|-
! Preferences !! Voters
|-
| A > B > C   || 7
|-
| B > C > A   || 6
|-
| C > A > B   || 3
|}

The results would be tabulated as follows:
{| class=wikitable
|+ Pairwise election results
! {{diagonal split header|2=X|1=Y}}
! style="background-color:#c0c0ff" | A
! style="background-color:#c0c0ff" | B
! style="background-color:#c0c0ff" | C
|-
! style="background-color:#ffc0c0" | A
|
| {{diagonal split header|10|6|#ffe0e0}}
| {{diagonal split header|7|9|#e0e0ff}}
|-
! style="background-color:#ffc0c0" | B
| {{diagonal split header|6|10|#e0e0ff}}
|
| {{diagonal split header|13|3|#ffe0e0}}
|-
! style="background-color:#ffc0c0" | C
| {{diagonal split header|9|7|#ffe0e0}}
| {{diagonal split header|3|13|#e0e0ff}}
|
|-
! style="background-color:#c0c0ff" | Pairwise election results<br/>(won-tied-lost)
| 1–0–1
| 1–0–1
| 1–0–1
|}

The sorted list of victories would be:
{| class="wikitable"
! 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.

==== Second group of voters ====
Now, the ranked pairs winner for the second group of voters is determined.
{| class="wikitable"
! Preferences !! Voters
|-
| A > C > B   || 9
|-
| B > A > C   || 8
|-
| C > B > A   || 6
|}

The results would be tabulated as follows:
{| class=wikitable
|+ Pairwise election results
|-
! {{diagonal split header|2=X|1=Y}}
! style="background-color:#c0c0ff" | A
! style="background-color:#c0c0ff" | B
! style="background-color:#c0c0ff" | C
|-
! style="background-color:#ffc0c0" | A
|
| {{diagonal split header|9|14|#e0e0ff}}
| {{diagonal split header|17|6|#ffe0e0}}
|-
! style="background-color:#ffc0c0" | B
| {{diagonal split header|14|9|#ffe0e0}}
|
| {{diagonal split header|8|15|#e0e0ff}}
|-
! style="background-color:#ffc0c0" | C
| {{diagonal split header|6|17|#e0e0ff}}
| {{diagonal split header|15|8|#ffe0e0}}
|
|-
! style="background-color:#c0c0ff" | Pairwise election results<br/>(won-tied-lost)
| 1–0–1
| 1–0–1
| 1–0–1
|}

The sorted list of victories would be:
{| class="wikitable"
! 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.

==== All voters ====
Finally, the ranked pairs winner of the complete set of voters is determined.
{| class="wikitable"
|-
! 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:
{| class=wikitable
|+ Pairwise election results
|-
! {{diagonal split header|2=X|1=Y}}
! style="background-color:#c0c0ff" | A
! style="background-color:#c0c0ff" | B
! style="background-color:#c0c0ff" | C
|-
! style="background-color:#ffc0c0" | A
|
| {{diagonal split header|19|20|#e0e0ff}}
| {{diagonal split header|24|15|#ffe0e0}}
|-
! style="background-color:#ffc0c0" | B
| {{diagonal split header|20|19|#ffe0e0}}
|
| {{diagonal split header|21|18|#ffe0e0}}
|-
! style="background-color:#ffc0c0" | C
| {{diagonal split header|15|24|#e0e0ff}}
| {{diagonal split header|18|21|#e0e0ff}}
|
|-
! style="background-color:#c0c0ff" | Pairwise election results<br/>(won-tied-lost)
| 1–0–1
| 2–0–0
| 0–0–2
|}

The sorted list of victories would be:
{| class="wikitable"
! 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.

==== 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.

==References==
{{Reflist}}
#{{note|Smith}}[[John H Smith (mathematician)|John H Smith]], "Aggregation of preferences with variable electorate", ''Econometrica'', Vol. 41 (1973), pp.&nbsp;1027&ndash;1041.
#{{note|Woodall}}[[Douglas R. Woodall|D. R. Woodall]], "[http://www.votingmatters.org.uk/ISSUE3/P5.HTM Properties of preferential election rules]", ''Voting matters'', Issue 3 (December 1994), pp.&nbsp;8&ndash;15.
#{{note|Young}}[[Peyton Young|H. P. Young]], "Social Choice Scoring Functions", ''SIAM Journal on Applied Mathematics'' Vol. 28, No. 4 (1975), pp.&nbsp;824&ndash;838.

{{voting systems}}

[[Category:Electoral system criteria]]