EXAMPLE 9 Shortcomings of the Hare System

Suppose we have the following sequence of preference list ballots, where, as before, the heading of “5” indicates that 5 of the 13 voters hold the ballot with over over , the heading of “4” indicates that 4 of the 13 voters hold the ballot with over over , and so forth.

Candidates and have only 4 first-place votes (while has 5). Thus, and are eliminated in the first round, and wins the election.

Now, suppose that the voter in the last column moves Candidate up on his list. Let’s look at the new election. Notice that, even though won the last election, the only change we are making in ballots for the new election is one that is favorable to . The ballots for the new election are as follows:

422

If we apply the Hare system again, only is eliminated in round 1, as he or she has 3 first-place votes, as opposed to 4 for and 6 for . Thus, after this round, the ballots are as follows:

We now have on top of 6 lists and on top of 7 lists. Thus, at stage 2, (our previous winner!) is eliminated and is the winner of this new election.