Learning Outcomes
- Determine the winner of an election using Copeland’s method
- Evaluate the fairness of an election determined by Copeland’s method
Study Strategy
When learning new vocabulary and processes it often takes more than a careful reading of the text to gain understanding. Remember to use flashcards for vocabulary, writing the answers out by hand before checking to see if you have them right. When learning new processes, writing them out by hand as you read through them will help you simultaneously memorize and gain insight into the process.
Pro-tip: Write out each of the examples in this section using paper and pencil, trying each of the steps as you go, until you feel you could explain it to another person.
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.
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 ½ 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
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 |
Here is the same example presented in a video.
Example
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 |
Watch the same example from above being worked out in this video.
Try It
Consider again the election from earlier. 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.
| 44 | 14 | 20 | 70 | 22 | 80 | 39 | |
| 1st choice | G | G | G | M | M | B | B |
| 2nd choice | M | B | G | B | M | ||
| 3rd choice | B | M | B | G | G |
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
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 gets ½ point, B gets ½ point |
| A vs C: 14 votes to 6 votes: | A gets 1 point |
| A vs D: 5 votes to 15 votes: | D gets 1 point |
| B vs C: 4 votes to 16 votes: | C gets 1 point |
| B vs D: 15 votes to 5 votes: | B gets 1 point |
| C vs D: 11 votes to 9 votes: | C gets 1 point |
Totaling:
| A has 1 ½ points | B has 1 ½ points |
| C has 2 points | D has 1 point |
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 | A gets ½ point, B gets ½ point |
| 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 ½ points | B has ½ point |
| C has 1 point |
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 the last example, the IIA Criterion was violated.
Watch this video to see the example from above worked out again,
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.
