2.11: What’s Wrong with Copeland’s Method?

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As already noted, Copeland’s Method does satisfy the Condorcet Criterion. It also satisfies the Majority Criterion and the Monotonicity Criterion. So is this the perfect method? Well, in a word, no.

Example 11

A committee is trying to award a scholarship to one of four students, Anna (A), Brian (B), Carlos (C), and Dimitry (D). The votes are shown below:

\(\begin{array}{|l|l|l|l|l|}
\hline & 5 & 5 & 6 & 4 \\
\hline 1^{\text {st }} \text { choice } & \text { D } & \text { A } & \text { C } & \text { B } \\
\hline 2^{\text {nd }} \text { choice } & \text { A } & \text { C } & \text { B } & \text { D } \\
\hline 3^{\text {rd }} \text { choice } & \text { C } & \text { B } & \text { D } & \text { A } \\
\hline 4^{\text {th }} \text { choice } & \text { B } & \text { D } & \text { A } & \text { C } \\
\hline
\end{array}\)

Solution

Making the comparisons:

\(\begin{array} {ll} {\text{A vs B: }10\text{ votes to }10\text{ votes}} & {\text{A gets }\frac{1}{2}\text{ point, B gets }\frac{1}{2}\text{ point}} \\ {\text{A vs C: }14\text{ votes to }6\text{ votes:}} & {\text{A gets }1\text{ point}} \\ {\text{A vs D: }5\text{ votes to }15\text{ votes:}} & {\text{D gets }1\text{ point}} \\ {\text{B vs C: }4\text{ votes to }16\text{ votes:}} & {\text{C gets }1\text{ point}} \\ {\text{B vs D: }15\text{ votes to }5\text{ votes:}} & {\text{B gets }1\text{ point}} \\ {\text{C vs D: }11\text{ votes to }9\text{ votes:}} & {\text{C gets }1\text{ point}} \end{array} \)

Totaling:

\(\begin{array} {ll} {\text{A has }1\frac{1}{2}\text{ points}} & {\text{B has }1 \frac{1}{2}\text{ points}} \\ {\text{C has }2\text{ points}} & {\text{D has }1\text{ point}} \end{array} \)

So Carlos is awarded the scholarship. However, the committee then discovers that Dimitry was not eligible for the scholarship (he failed his last math class). Even though this seems like it shouldn’t affect the outcome, the committee decides to recount the vote, removing Dimitry from consideration. This reduces the preference schedule to:

\(\begin{array}{|l|l|l|l|l|}
\hline & 5 & 5 & 6 & 4 \\
\hline 1^{\text {st }} \text { choice } & \mathrm{A} & \mathrm{A} & \mathrm{C} & \mathrm{B} \\
\hline 2^{\text {nd }} \text { choice } & \mathrm{C} & \mathrm{C} & \mathrm{B} & \mathrm{A} \\
\hline 3^{\text {rd }} \text { choice } & \mathrm{B} & \mathrm{B} & \mathrm{A} & \mathrm{C} \\
\hline
\end{array}\)

Totaling:

\(\begin{array} {ll} {\text{A has }1 \frac{1}{2}\text{ points}} & {\text{B has }\frac{1}{2}\text{ point}} \\ {\text{C has }1\text{ point}} & { } \end{array} \)

Suddenly Anna is the winner! This leads us to another fairness criterion.

The Independence of Irrelevant Alternatives (IIA) Criterion

If a non-winning choice is removed from the ballot, it should not change the winner of the election.

Equivalently, if choice A is preferred over choice B, introducing or removing a choice C should not cause B to be preferred over A.

In the election from Example 11, the IIA Criterion was violated.

This anecdote illustrating the IIA issue is attributed to Sidney Morgenbesser:

After finishing dinner, Sidney Morgenbesser decides to order dessert. The waitress tells him he has two choices: apple pie and blueberry pie. Sidney orders the apple pie. After a few minutes the waitress returns and says that they also have cherry pie at which point Morgenbesser says "In that case I'll have the blueberry pie."

Another disadvantage of Copeland’s Method is that it is fairly easy for the election to end in a tie. For this reason, Copeland’s method is usually the first part of a more advanced method that uses more sophisticated methods for breaking ties and determining the winner when there is not a Condorcet Candidate.