EXAMPLE 2 Finding a Successful Manipulation with the Borda Count

Suppose there are two voters and four candidates, and suppose the true preferences of the voters are reflected in the following ballots:

Using the Borda count with point values 3, 2, 1, 0 (or by counting the number of occurrences of other candidates below the one in question, as described in Section 9.3), we see that the Borda scores of the four candidates are as follows:

Thus, is the winner of this election using the Borda count. Voter 2 can indeed change the outcome of the election, but only in a pointless way. That is, the winning alternative is her top choice, and so it is quite impossible for her to do any better than this!

Voter 1, however, is quite unhappy with being the winner, and she would like to change her ballot so as to make either or the winner (as she prefers both of these to ). Let's first see if Voter 1 can change her ballot so as to make the winner. Voter 1 certainly cannot increase 's Borda score of 3 from the original election. But had a Borda score of 4 and the most Voter 1 can do is to reduce this by 1 (by moving to the bottom of her ballot). Hence, there is no way that Voter 1 can change her ballot to make the unique winner.

Voter 1, however, can make the unique winner. To do this, she first moves to the top of her ballot (an obvious thing to do because she is trying to make the winner). That raises 's Borda score to 4. But still has a Borda score of 4, so she must move to the bottom of her ballot, reducing the Borda score of to 3. Now we must check that neither nor have a Borda score of 4, but this is easily done. Hence, Voter 1 can submit a disingenuous ballot and achieve a result she prefers to that of the original election. Hence, this example also serves to show that the Borda count is manipulable.