EXAMPLE 2 Condorcet’s Method
Suppose we have four candidates (, , , and , with these initials chosen for a soon-to-be-revealed reason) and the following sequence of preference list ballots, where the heading of “6” indicates that 6 of the 15 voters hold the ballot with over over over , the heading of “5” indicates that 5 of the 15 voters hold the ballot with over over over , and so on.
411
| Number of Voters (15) | ||||
|---|---|---|---|---|
| Rank | 6 | 5 | 3 | 1 |
| First | ||||
| Second | ||||
| Third | ||||
| Fourth | ||||
We claim that is the winner in this election if we use Condorcet’s method. Let’s check the one-on-one scores for each possible pair of opponents:
versus : is over on of the ballots, while the reverse is true on of the ballots. Thus, defeats by a score of 8 to 7.
versus : is over on of the ballots, while the reverse is true on 3 of the ballots. Thus, defeats by a score of 12 to 3.
versus : is over on of the ballots, while the reverse is true on 1 of the ballots. Thus, defeats by a score of 14 to 1.
This shows that is the winner using Condorcet’s method.