2.10: Copeland’s Method (Pairwise Comparisons)
So far none of our voting methods have satisfied the Condorcet Criterion. The Copeland Method specifically attempts to satisfy the Condorcet Criterion by looking at pairwise (one-to-one) comparisons.
In this method, each pair of candidates is compared, using all preferences to determine which of the two is more preferred. The more preferred candidate is awarded 1 point. If there is a tie, each candidate is awarded \(\frac{1}{2}\) point. After all pairwise comparisons are made, the candidate with the most points, and hence the most pairwise wins, is declared the winner.
Variations of Copeland’s Method are used in many professional organizations, including election of the Board of Trustees for the Wikimedia Foundation that runs Wikipedia.
Consider our vacation group example from the beginning of the chapter. Determine the winner using Copeland’s Method.
\(\begin{array}{|l|l|l|l|l|}
\hline & 1 & 3 & 3 & 3 \\
\hline 1^{\text {st }} \text { choice } & \mathrm{A} & \mathrm{A} & \mathrm{O} & \mathrm{H} \\
\hline 2^{\text {nd }} \text { choice } & \mathrm{O} & \mathrm{H} & \mathrm{H} & \mathrm{A} \\
\hline 3^{\text {rd }} \text { choice } & \mathrm{H} & \mathrm{O} & \mathrm{A} & \mathrm{O} \\
\hline
\end{array}\)
Solution
We need to look at each pair of choices, and see which choice would win in a one-to-one comparison. You may recall we did this earlier when determining the Condorcet Winner. For example, comparing Hawaii vs Orlando, we see that 6 voters, those shaded below in the first table below, would prefer Hawaii to Orlando. Note that Hawaii doesn’t have to be the voter’s first choice – we’re imagining that Anaheim wasn’t an option. If it helps, you can imagine removing Anaheim, as in the second table below.
Based on this, in the comparison of Hawaii vs Orlando, Hawaii wins, and receives 1 point.
Comparing Anaheim to Orlando, the 1 voter in the first column clearly prefers Anaheim, as do the 3 voters in the second column. The 3 voters in the third column clearly prefer Orlando. The 3 voters in the last column prefer Hawaii as their first choice, but if they had to choose between Anaheim and Orlando, they'd choose Anaheim, their second choice overall. So, altogether 1+3+3=7 voters prefer Anaheim over Orlando, and 3 prefer Orlando over Anaheim. So, comparing Anaheim vs Orlando: 7 votes to 3 votes: Anaheim gets 1 point.
All together,
\(\begin{array} {ll} {\text{Hawaii vs Orlando: }} & {6\text{ votes to }4\text{ votes: Hawaii gets }1\text{ point}} \\ {\text{Anaheim vs Orlando:}} & {7\text{ votes to }3\text{ votes: Anaheim gets }1\text{ point}} \\ {\text{Hawaii vs Anaheim: }} & {6\text{ votes to }4\text{ votes: Hawaii gets }1\text{ point}} \end{array}\)
Hawaii is the winner under Copeland’s Method, having earned the most points.
Notice this process is consistent with our determination of a Condorcet Winner.
Consider the advertising group’s vote we explored earlier. Determine the winner using Copeland’s method.
\(\begin{array}{|l|l|l|l|l|l|l|}
\hline & 3 & 4 & 4 & 6 & 2 & 1 \\
\hline 1^{\text {st }} \text { choice } & \text { B } & \text { C } & \text { B } & \text { D } & \text { B } & \text { E } \\
\hline 2^{\text {nd }} \text { choice } & \text { C } & \text { A } & \text { D } & \text { C } & \text { E } & \text { A } \\
\hline 3^{\text {rd }} \text { choice } & \text { A } & \text { D } & \text { C } & \text { A } & \text { A } & \text { D } \\
\hline 4^{\text {th }} \text { choice } & \text { D } & \text { B } & \text { A } & \text { E } & \text { C } & \text { B } \\
\hline 5^{\text {th }} \text { choice } & \text { E } & \text { E } & \text { E } & \text { B } & \text { D } & \text { C } \\
\hline
\end{array}\)
Solution
With 5 candidates, there are 10 comparisons to make:
\( \begin{array} {ll} {\text{A vs B: }11\text{ votes to }9\text{ votes}} & {\text{A gets }1\text{ point}} \\ {\text{A vs C: }3\text{ votes to }17\text{ votes}} & {\text{C gets }1\text{ point}} \\ {\text{A vs D: }10\text{ votes to }10\text{ votes }} & {\text{A gets }\frac{1}{2}\text{ point, D gets }\frac{1}{2}\text{ point}} \\ {\text{A vs E: }17\text{ votes to }3\text{ votes}} & {\text{A gets }1\text{ point}} \\ {\text{B vs C: }10\text{ votes to }10\text{ votes}} & {\text{B gets }\frac{1}{2}\text{ point, C gets }\frac{1}{2}\text{ point}} \\ {\text{B vs D: }9\text{ votes to }11\text{ votes}} & {\text{D gets }1\text{ point}} \\ {\text{B vs E: }13\text{ votes to }7\text{ votes}} & {\text{B gets }1\text{ point}} \\ {\text{C vs D: }9\text{ votes to }11\text{ votes}} & {\text{D gets }1\text{ point}} \\ {\text{C vs E: }17\text{ votes to }3\text{ votes}} & {\text{C gets }1\text{ point}} \\ {\text{D vs E: }17\text{ votes to }3\text{ votes}} & {\text{D gets }1\text{ point}} \end{array}\)
Totaling these up:
A gets \(2\frac{1}{2}\) points
B gets \(1\frac{1}{2}\) points
C gets \(2\frac{1}{2}\) points
D gets \(3\frac{1}{2}\) points
E gets \(0\) points
Using Copeland’s Method, we declare D as the winner.
Notice that in this case, D is not a Condorcet Winner. While Copeland’s method will also select a Condorcet Candidate as the winner, the method still works in cases where there is no Condorcet Winner.
Consider again the election from Try it Now 1. Find the winner using Copeland’s method. Since we have some incomplete preference ballots, we’ll have to adjust. For example, when comparing M to B, we’ll ignore the 20 votes in the third column which do not rank either candidate.
\(\begin{array}{|l|l|l|l|l|l|l|l|}
\hline & 44 & 14 & 20 & 70 & 22 & 80 & 39 \\
\hline 1^{\text {st }} \text { choice } & \mathrm{G} & \mathrm{G} & \mathrm{G} & \mathrm{M} & \mathrm{M} & \mathrm{B} & \mathrm{B} \\
\hline 2^{\text {nd }} \text { choice } & \mathrm{M} & \mathrm{B} & & \mathrm{G} & \mathrm{B} & \mathrm{M} & \\
\hline 3^{\text {rd }} \text { choice } & \mathrm{B} & \mathrm{M} & & \mathrm{B} & \mathrm{G} & \mathrm{G} & \\
\hline
\end{array}\)
- Answer
-
Using Copeland’s Method:
Looking back at our work from Try it Now #2, we see
G vs M: \(44+14+20 = 78\) prefer G, \(70+22+80=172\) prefer M: M preferred – 1 point
G vs B: \(44+14+20+70=148\) prefer G, \(22+80+39 = 141\) prefer B: G preferred – 1 point
M vs B: \(44+70+22=136\) prefer M, \(14+80+39=133\) prefer B: M preferred – 1 point
M earns 2 points; G earns 1 point. M wins under Copeland’s method.