Section 11.5 Copeland's Method
Objectives
Find the winner of an election using Copeland's method
Understand whether an election has violated the independence of irrelevant alternatives criterion
Subsection 11.5.1 Copeland's Method
Definition 11.5.1. Copeland’s Method.
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.
Example 11.5.2.
Consider our vacation group example from the beginning of the chapter. Determine the winner using Copeland’s Method.
| 1 | 3 | 3 | 3 | |
|---|---|---|---|---|
| 1st choice | A | A | O | H |
| 2nd choice | O | H | H | A |
| 3rd choice | H | O | A | O |
We need to look at each pair of choices, and see which choice would win in a one-to-one comparison.
Hawaii vs Orlando
Hawaii vs Anaheim
Anaheim vs Orlando
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 table below.
| 1 | 3 | 3 | 3 | |
|---|---|---|---|---|
| 1st choice | O | H | ||
| 2nd choice | O | H | H | |
| 3rd choice | H | O | O |
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.
To sumarize:
Hawaii vs Orlando: 6 votes to 4 votes: Hawaii gets 1 point
Anaheim vs Orlando: 7 votes to 3 votes: Anaheim gets 1 point
Hawaii vs Anaheim: 6 votes to 4 votes: Hawaii gets 1 point
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.
Example 11.5.5.
Consider the advertising group’s vote we explored earlier. Determine the winner using Copeland’s method.
| 3 | 4 | 4 | 6 | 2 | 1 | |
|---|---|---|---|---|---|---|
| 1st Choice | B | C | B | D | B | E |
| 2nd Choice | C | A | D | C | E | A |
| 3rd Choice | A | D | C | A | A | D |
| 4th Choice | D | B | A | E | C | B |
| 5th Choice | E | E | E | B | D | C |
With 5 candidates, there are 10 comparisons to make:
A vs B: 11 votes to 9 votes: A gets 1 pt
A vs C: 3 votes to 17 votes: C gets 1 pt
A vs D: 10 votes to 10 votes: Both A and D get \(\frac{1}{2}\) pt
A vs E: 17 votes to 3 votes: A gets 1 pt
B vs C: 10 votes to 10 votes: Both B and C get \(\frac{1}{2}\) pt
B vs D: 9 votes to 11 votes: D gets 1 pt
B vs E: 13 votes to 7 votes: B gets 1 pt
C vs D: 9 votes to 11 votes: D gets 1 pt
C vs E: 17 votes to 3 votes: C gets 1 pt
D vs E: 17 votes to 3 votes: D gets 1 pt
Totaling these up:
A = \(2\frac{1}{2}\)pts
B = \(1\frac{1}{2}\)pts
C = \(2\frac{1}{2}\)pts
D = \(3\frac{1}{2}\)pts
E = \(0\)pts
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.
Subsection 11.5.2 What's Wrong with Copeland's Method?
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 11.5.7.
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:
| 5 | 5 | 6 | 4 | |
|---|---|---|---|---|
| 1st Choice | D | A | C | B |
| 2nd Choice | A | C | B | D |
| 3rd Choice | C | B | D | A |
| 4th Choice | B | D | A | C |
Making the comparisons:
A vs B: 10 votes to 10 votes: A and B both get \(\frac{1}{2}\) pt
A vs C: 14 votes to 6 votes: A gets 1 pt
A vs D: 5 votes to 15 votes: D gets 1 pt
B vs C: 4 votes to 16 votes: C gets 1 pt
B vs D: 15 votes to 5 votes: B gets 1 pt
C vs D: 11 votes to 9 votes: C gets 1 pt
Totaling:
A has \(1\frac{1}{2}\) pts
B has \(1\frac{1}{2}\) pts
C has \(2\) pts
D has \(1\) pt
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:
| 5 | 5 | 6 | 4 | |
|---|---|---|---|---|
| 1st Choice | A | A | C | B |
| 2nd Choice | C | C | B | A |
| 3rd Choice | B | B | A | C |
A vs B: 10 votes to 10 votes: Both A and B get \(\frac{1}{2}\) pt
A vs C: 14 votes to 6 votes: A gets \(1\) point
B vs C: 4 votes to 16 votes: C gets \(1\) point
Totaling:
A has \(1\frac{1}{2}\) points
C has \(1\) point
B has \(\frac{1}{2}\) point
Suddenly Anna is the winner! This leads us to another fairness criterion.
Definition 11.5.10. 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 previous example, 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.