Contemporary MathematicsContemporary Mathematics at Nebraska

Example7.13

Find the Borda Count winner of the election whose preference table is below.

Notice that we have added a column to the left of the preference schedule indicating how many Borda points are awarded to each ranking. Since there are 4 candidates, a first-place ranking is worth 4 points.

To find the score of Candidate A, we look at the four times A appears on the preference table. The first column says that 6 people ranked A first, giving A \(6 \times 4=24\) points. Everyone else ranked A last, giving A 1 point each. This adds up to \(5+4+2=11\) points. Therefore, A has a total of \(24+11=35\) points.

We can calculate the Borda points of the other candidates similarly, as follows:

• B: \(6 \times 3 + 5 \times 3 + 4 \times 2 + 2 \times 4 =49\)
• C: \(6 \times 1 + 5 \times 2 + 4 \times 4 + 2 \times 3 =38\)
• D: \(6 \times 2 + 5 \times 4 + 4 \times 3 + 2 \times 2 =48\)

Thus, Candidate B wins this election when the Borda Count is used.

Example7.15

The UNL Dairy Store is taking a survey of customers' favorite ice cream flavors. Their preferences are shown in the table below. The three options are Black Walnut Fudge (F), Scarlet & Cream (S), and Bavarian Mint (M). Which flavor would the Borda Count pick as the overall favorite?

First notice how the number of points assigned to each ranking has changed. The rules for the Borda count state that every last choice vote gets 1 point, and then we count going up. Hence, when there are three candidates, a 3rd choice vote gets 1 point, a 2nd choice vote gets 2 points, and a 1st choice vote gets 3 points. Comparing this to the example with four candidates, note that the number of points awarded for a first choice vote will always be the same as the number of candidates.

We can now calculate the Borda Count winner as usual, and find:

• F: \(6 \times 3 + 3 \times 1 + 2 \times 1 =23\)
• S: \(6 \times 2 + 3 \times 3 + 2 \times 2 =25\)
• M: \(6 \times 1 + 3 \times 2 + 2 \times 3 =18\)

Therefore, Scarlet & Cream wins with 25 points.

Exploration7.2

Let's revisit the election from Exploration7.1.

Find the winner of this election using the Borda Count.

Note: When voters do not rank every candidate, we will treat them as if the candidates who they didn't rank are all in last place. Thus, for example, the 20 voters who ranked only G first will give 3 points each to G, but also give 1 point each to B and M, as if both those candidates were simultaneously ranked last.

Example7.18

The preference table below shows the same election we considered in Example7.13. We found that Candidate B wins this election when using the Borda count. Let's see who is the winner using IRV.

There are three rounds of IRV, shown in the table below. As we can see, the winner of this election is different with IRV than with the Borda count.

Example7.20The 2009 mayoral election in Burlington, VT

In 2009, the city of Burlington, Vermont used IRV to elect their mayor. The results of this election are shown below, in a simplified form to include only the three main candidates. ³More complete details can be found at the Wikipedia article. These three candidates were Bob Kiss, Kurt Wright, and Andy Montroll.

Using this information, we can find the winner of this election using IRV as follows.

Thus, Bob Kiss won this election using instant runoff voting.

Notice that, in this example, the voters who ranked Montroll first had a variety of second choice candidates. When Montroll was eliminated, these votes were transferred to each of these different second choices, rather than all to one candidate.

Additionally, notice that the 5% of voters who did not rank any candidates other than Montroll did not have their votes transferred to any other candidate, since they did not indicate any second choice. Essentially they are saying that, if Montroll does not win, they have no preference between the other candidates. ⁴Such votes are referred to as exhausted ballots. It may seem strange that their votes were not part of the final tally. However, their votes did count; they just counted only for a candidate that happened to lose.

This example shows how IRV can give a different winner than plurality. The plurality winner in this election was Kurt Wright. However, more votes were transferred from Andy Montroll to Bob Kiss when Montroll was eliminated, so Kiss, rather than Wright, was the winner.

As mayor, Kiss became involved in a scandal and, partially due to his unpopularity, IRV was repealed by a narrow majority of voters in 2010. However, they replaced it with a form of ordinary runoff voting that would still have elected Kiss as mayor had it been used in 2009. This election and its aftermath is discussed on Wikipedia.

Example7.22

Suppose there is an election with the following preference table:

Let's find the winner of this election using the method of pairwise comparisons. First, we need to list all possible pairs of candidates. Since there are three candidates, there are three possible pairs to run against each other: A vs B, A vs C, and B vs C. The following table shows, for each pair, who would win if the election were only between that pair, and how many points each candidate receives as a result.

Notice that we do not ignore a ballot just because the first choice on that ballot is not for one of the two candidates in the pair we are considering. For instance, when considering the pair A vs B, the third column of 40 voters have listed C as their first choice. However, we still give these 40 votes to one of the two candidates, in this case candidate B because B is higher than A on the preference list. These voters prefer candidate B over candidate A, and would therefore vote for candidate B in an election between just candidates A and B. So these 40 votes are given to B, causing B to win against A.

Having determined the outcome of each pair, we now tally the points. A earns a total of 0.5 points, B earns a total of 1 point, and C earns a total of 1.5 points. Therefore, C wins this election if we use the method of pairwise comparisons.

Example7.24The 2009 mayoral election in Burlington, VT

Let's return to the election from Example7.20, shown again below.

Using this information, we can find the winner of this election using the method of pairwise comparisons as follows.

We see that Montroll wins two pairs for 2 points, while Kiss and Wright each won one pair for 1 point each. Thus, Montroll would win this election using the method of pairwise comparisons.