Section 7.8: Pairwise Comparison Method
- Page ID
- 219586
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Keep in mind that there is no fixed number of voting methods; in principle, infinitely many voting methods exist, although only a small number of representative methods are commonly studied. In the previous sections, we looked at three of those four commonly studied major voting methods. So, now let's introduce our last major voting method called the pairwise comparison method:
The pairwise comparison method is a voting procedure in which voters rank all candidates in order of preference, and every candidate is compared head‑to‑head with each of the others. In each comparison, the candidate preferred by a majority of voters receives 1 point, while a tie awards \(\frac{1}{2}\) to each candidate. Any loss, the candidate receives 0 points. After all pairwise comparisons are completed, the candidate with the greatest total number of points is declared the winner. The winner is referred to as the Condorcet winner.
Note: The Condorcet winner should not be confused with the Condorcet criterion. The Condorcet criterion is a fairness standard used to evaluate voting methods, not a voting method itself. However, both the pairwise comparison method and the Condorcet criterion rely on the same head‑to‑head matchup framework, which is why these concepts are often confused.
Consider the following preference table with Candidates A, B, & C:
| Number of Voters | 5 | 4 | 2 |
|---|---|---|---|
| First Choice | B | C | C |
| Second Choice | A | A | B |
| Third Choice | C | B | A |
Use the pairwise comparison method to determine a winner..
✅ Solution:
Use a table for every possible match-up to record your results.
\(\begin{array}{r|l} \text{A}&\text{B} \\ \hline &5 \\ 4& \\ &2 \\ \hline \large \textbf{4} & \large \textbf{7} \\ \end{array} \) \(\begin{array}{r|l} \text{A}&\text{C} \\ \hline 5& \\ &4 \\ &2 \\ \hline \large \textbf{5} & \large \textbf{6} \\ \end{array} \) \(\begin{array}{r|l} \text{B}&\text{C} \\ \hline 5& \\ &4 \\ &2 \\ \hline \large \textbf{5} & \large \textbf{6} \\ \end{array} \)
B wins A C wins A C wins B
Here are the head-to-head results:
Candidate A: 0 wins = 0 points
Candidate B: 1 win = 1 point
Candidate C: 2 wins = 2 points
Hence, Candidate C has the most points and wins the election by the pairwise comparison method.
Consider the following preference table with Candidates A, B, & C:
| Number of Voters | 5 | 4 | 3 | 2 |
|---|---|---|---|---|
| First Choice | B | C | C | A |
| Second Choice | A | B | A | B |
| Third Choice | C | A | B | C |
Use the pairwise comparison method to determine a winner.
✅ Solution:
Use a table for every possible match-up to record your results.
\(\begin{array}{r|l} \text{A}&\text{B} \\ \hline &5 \\ &4 \\ 3& \\ 2& \\ \hline \large \textbf{9} & \large \textbf{5} \\ \end{array} \) \(\begin{array}{r|l} \text{A}&\text{C} \\ \hline 5& \\ &4 \\ &3 \\ 2& \\ \hline \large \textbf{7} & \large \textbf{7} \\ \end{array} \) \(\begin{array}{r|l} \text{B}&\text{C} \\ \hline 5& \\ &4 \\ &3 \\ 2& \\ \hline \large \textbf{7} & \large \textbf{7} \\ \end{array} \)
A wins B A ties C B ties C
Here are the head-to-head results:
Candidate A: 1 wins & 1 tie = 1 + \(\frac{1}{2}\) = 1.5 points
Candidate B: 0 wins & 1 tie = \(\frac{1}{2}\) point = 0.5 points
Candidate C: 0 wins & 2 ties = \(\frac{1}{2} + \frac{1}{2}\) = 1 point
Hence, Candidate A has the most points and wins the election by the pairwise comparison method.
A group of people is choosing where to eat lunch. The four fast‑food options are: McDonald’s, KFC, Subway, or Taco Bell. Each voter ranks the restaurants from most preferred to least preferred.
| Number of Voters | 6 | 5 | 4 | 3 |
|---|---|---|---|---|
| First Choice | McDonald's | Subway | Taco Bell | KFC |
| Second Choice | Subway | Taco Bell | McDonald's | McDonald's |
| Third Choice | Taco Bell | McDonald's | KFC | Subway |
| Fourth Choice | KFC | KFC | Subway | Taco Bell |
Use the pairwise comparison method to determine a winner.
✅ Solution:
Use a table for every possible match-up to record your results.
\(\begin{array}{c|c} \text{McDonald's}&\text{Subway} \\ \hline 6& \\ &5 \\ 4& \\ 3& \\ \hline \large \textbf{13} & \large \textbf{5} \\ \end{array} \) \(\begin{array}{c|c} \text{McDonald's}&\text{Taco Bell} \\ \hline 6& \\ &5 \\ &4 \\ 3& \\ \hline \large \textbf{9} & \large \textbf{9} \\ \end{array} \) \(\begin{array}{c|c} \text{McDonald's}&\text{KFC} \\ \hline 6& \\ 5& \\ 4& \\ &3 \\ \hline \large \textbf{15} & \large \textbf{3} \\ \end{array} \) \(\begin{array}{c|c} \text{Subway}&\text{Taco Bell} \\ \hline 6& \\ 5& \\ &4 \\ 3& \\ \hline \large \textbf{14} & \large \textbf{4} \\ \end{array} \) \(\begin{array}{c|c} \text{Subway}&\text{KFC} \\ \hline 6& \\ 5& \\ &4 \\ &3 \\ \hline \large \textbf{11} & \large \textbf{7} \\ \end{array} \) \(\begin{array}{c|c} \text{Taco Bell}&\text{KFC} \\ \hline 6& \\ 5& \\ 4& \\ &3 \\ \hline \large \textbf{15} & \large \textbf{3} \\ \end{array} \)
McDonald's wins Subway McDonald's ties Taco Bell McDonald's wins KFC Subway wins Taco Bell Subway wins KFC Taco Bell wins KFC
Here are the head-to-head results:
- McDonald's: 2 wins & 1 tie = 2 + \(\frac{1}{2}\) = 2.5 points
- Subway: 2 wins = 2 points
- Taco Bell: 1 wins & 1 ties = 1 + \(\frac{1}{2}\) = 1.5 points
- KFC: 0 wins = 0 points
Hence, McDonald's has the most points and wins the election by the pairwise comparison method.
Consider the following preference table with Candidates A, B, & C:
| Number of Voters | 8 | 7 | 6 |
|---|---|---|---|
| First Choice | A | B | C |
| Second Choice | B | C | A |
| Third Choice | C | A | B |
Use the pairwise comparison method to determine a winner..
✅ Solution:
Use a table for every possible match-up to record your results.
\(\begin{array}{r|l} \text{A}&\text{B} \\ \hline 8& \\ &7 \\ 6& \\ \hline \large \textbf{14} & \large \textbf{7} \\ \end{array} \) \(\begin{array}{r|l} \text{A}&\text{C} \\ \hline 8& \\ &7 \\ &6 \\ \hline \large \textbf{8} & \large \textbf{13} \\ \end{array} \) \(\begin{array}{r|l} \text{B}&\text{C} \\ \hline 8& \\ 7& \\ &6 \\ \hline \large \textbf{15} & \large \textbf{6} \\ \end{array} \)
A wins B C wins A B wins C
Here are the head-to-head records:
A: 1 win = 1 point
B: 1 win = 1 point
C: 1 win = 1 point
Hence, there is a 3-way tie. Candidate A, Candidate B, and Candidate C are all tied with 1 point each.
Despite its strengths, the pairwise comparison method is not without drawbacks. In some elections, as we see in Example #7.8.4, no candidate defeats all others in head‑to‑head contests. This situation, known as the Condorcet paradox, can produce circular preferences (for example, A beats B, B beats C, but C beats A). When this occurs, the method may rely on point totals that depend on how ties and wins are scored. Additionally, because the method requires many comparisons, it can become cumbersome when there are many candidates.
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Consider the following preference table with Candidates A, B, & C:
Number of Voters 7 6 6 5 3 First Choice C B A B A Second Choice A A B C C Third Choice B C C A B Use the plurality-with-elimination method to determine a winner.
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A group of 41 people is choosing where to eat lunch during their meeting. The four fast food delivery options are: Board & Brew, Naugles, T.K. Burgers, or In-N-Out Burgers. Each voter ranks the restaurants from most preferred to least preferred.
Number of Voters 11 9 8 6 4 3 First Choice In-N-Out T.K. Burgers Naugles Board & Brew T.K.Burgers In-N-Out Second Choice Naugles Naugles T.K.Burgers T.K. Burgers In-N-Out T.K. Burgers Third Choice T.K. Burgers Board & Brew In-N-Out In-N-Out Naugles Board & Brew Fourth Choice Board & Brew In-N-Out Board & Brew Naugles Board & Brew Naugles Use the pairwise comparison method to determine a winner.
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A 24-member committee is selecting a chairperson. The 3 candidates are David (D), Edward (E), and Fernando (F). Each committee member completely ranked the candidates on a separate ballot. The preference schedule is listed below.
| Number of Voters | 8 | 6 | 6 | 4 |
|---|---|---|---|---|
| First Choice | D | E | F | E |
| Second Choice | E | F | D | D |
| Third Choice | F | D | E | F |
Use the pairwise comparison method to determine a winner.
- Answers
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- Candidate A = 2 points, Candidate B = 1 point, Candidate C = 0 points. Candidate A wins.
- In-N-Out = 2 pts; T.K. Burgers = 2 pts; Naugles = 2 pts; Board & Brew = 0 pts. In‑N‑Out, T.K. Burgers, and Naugles are tied with 2 points each. No single pairwise comparison winner exists.
- David = 1.5 points, Edward = 1, Fernando = 0.5 points. David wins.