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Copeland's method or Copeland's pairwise aggregation method is a Condorcet method in which candidates are ordered by the number of pairwise victories, minus the number of pairwise defeats.[1]
Proponents argue that this method is easily understood by the general populace, which is generally familiar with the sporting equivalent. In many round-robin tournaments, the winner is the competitor with the most victories. It is also easy to calculate.
When there is no Condorcet winner (i.e. when there are multiple members of the Smith set), this method often leads to ties. For example, if there is a three-candidate majority rule cycle, each candidate will have exactly one loss, and there will be an unresolved tie between the three.
Critics argue that it also puts too much emphasis on the quantity of pairwise victories and defeats rather than their magnitudes.[citation needed]
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Imagine that Tennessee is having an election on the location of its capital. The population of Tennessee is concentrated around its four major cities, which are spread throughout the state. For this example, suppose that the entire electorate lives in these four cities and that everyone wants to live as near to the capital as possible.
The candidates for the capital are:
The preferences of the voters would be divided like this:
| 42% of voters (close to Memphis) |
26% of voters (close to Nashville) |
15% of voters (close to Chattanooga) |
17% of voters (close to Knoxville) |
|---|---|---|---|
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To find the Condorcet winner, every candidate must be matched against every other candidate in a series of imaginary one-on-one contests. In each pairing, each voter will choose the city physically closest to their location. In each pairing the winner is the candidate preferred by a majority of voters. When results for every possible pairing have been found they are as follows:
| Comparison | Result | Winner |
|---|---|---|
| Memphis vs Nashville | 42 v 58 | Nashville |
| Memphis vs Knoxville | 42 v 58 | Knoxville |
| Memphis vs Chattanooga | 42 v 58 | Chattanooga |
| Nashville vs Knoxville | 68 v 32 | Nashville |
| Nashville vs Chattanooga | 68 v 32 | Nashville |
| Knoxville vs Chattanooga | 17 v 83 | Chattanooga |
The wins and losses of each candidate sum as follows:
| Candidate | Wins | Losses | Wins - Losses |
|---|---|---|---|
| Memphis | 0 | 3 | -3 |
| Nashville | 3 | 0 | 3 |
| Knoxville | 1 | 2 | -1 |
| Chattanooga | 2 | 1 | 1 |
Nashville, with no defeats, is a Condorcet winner and, with the greatest number of net wins, is a Copeland winner.
In an election with five candidates competing for one seat, the following votes were cast using a ranked voting method (100 votes with four distinct sets):
| 31: A>E>C>D>B | 30: B>A>E | 29: C>D>B | 10: D>A>E |
The results of the 10 possible pairwise comparisons between the candidates are as follows:
| Comparison | Result | Winner | Comparison | Result | Winner |
|---|---|---|---|---|---|
| A v B | 41 v 59 | B | B v D | 30 v 70 | D |
| A v C | 71 v 29 | A | B v E | 59 v 41 | B |
| A v D | 61 v 39 | A | C v D | 60 v 10 | C |
| A v E | 71 v 0 | A | C v E | 29 v 71 | E |
| B v C | 30 v 60 | C | D v E | 39 v 61 | E |
The wins and losses of each candidate sum as follows:
| Candidate | Wins | Losses | Wins - Losses |
|---|---|---|---|
| A | 3 | 1 | 2 |
| B | 2 | 2 | 0 |
| C | 2 | 2 | 0 |
| D | 1 | 3 | -2 |
| E | 2 | 2 | 0 |
No Condorcet winner (candidate who beats all other candidates in pairwise comparisons) exists. Candidate A is the Copeland winner, with the greatest number of wins minus losses.
As a Condorcet completion method, Copeland requires a Smith set containing at least five candidates to give a clear winner unless two or more candidates tie in pairwise comparisons.
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