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 |
Answer:
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.
|
1 |
3 |
3 |
3 |
1st choice |
A |
A |
O |
H |
2nd choice |
O |
H |
H |
A |
3rd choice |
H |
O |
A |
O |
|
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 [latex]1+3+3=7[/latex] 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,
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.
Here is the same example presented in a video.
https://youtu.be/FftVPk7dqV0?list=PL1F887D3B8BF7C297
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 |
Answer:
With 5 candidates, there are 10 comparisons to make:
A vs B: 11 votes to 9 votes |
A gets 1 point |
A vs C: 3 votes to 17 votes |
C gets 1 point |
A vs D: 10 votes to 10 votes |
A gets ½ point, D gets ½ point |
A vs E: 17 votes to 3 votes |
A gets 1 point |
B vs C: 10 votes to 10 votes |
B gets ½ point, C gets ½ point |
B vs D: 9 votes to 11 votes |
D gets 1 point |
B vs E: 13 votes to 7 votes |
B gets 1 point |
C vs D: 9 votes to 11 votes |
D gets 1 point |
C vs E: 17 votes to 3 votes |
C gets 1 point |
D vs E: 17 votes to 3 votes |
D gets 1 point |
Totaling these up:
A gets 2½ points
B gets 1½ points
C gets 2½ points
D gets 3½ 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.
Watch the same example from above being worked out in this video.
https://youtu.be/sWdmkee5m_Q?list=PL1F887D3B8BF7C297
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 |
|
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.