CM Alternative Voting Methods

Section 7.2 Alternative Voting Methods

As we have seen, the plurality method has some downsides, and it does not use all the information on a preference table to decide the winner. In this section, we will introduce some alternative voting methods that consider the voters’ full preferences, with the hope that some of these methods may be more fair than plurality, or at least better suited to some situations.

Subsection 7.2.1 The Borda Count

Our first method of ranked voting is called the Borda Count. Its rules are fairly easy to state.

The Borda Count.

The Borda Count method of voting assigns points to candidates. Each candidate earns 1 point for every voter that ranked them last, 2 points for every voter that ranked them second-to-last, and so on. After adding up each candidate’s total points, the candidate with the most points wins.

Let’s illustrate it with an example.

Example 7.2.1.

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

Table 7.2.2.

		Number of voters
Borda points	Rankings	6	5	4	2
4	1st choice	A	D	C	B
3	2nd choice	B	B	D	C
2	3rd choice	D	C	B	D
1	4th choice	C	A	A	A

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.

Notice that the plurality winner of the election in Example 7.2.1 was Candidate A, with 6 votes out of 17 (about 35%). Candidate B had only 2 first choice votes, but many others ranked B second. In contrast, Candidate A was ranked last by everyone other than the 6 who ranked A first. We could say that B is a consensus or compromise candidate, while A is a more polarizing figure.

This behavior is common in the Borda Count, and so it is often described as a more consensus-based voting system. For this reason, the Borda Count, or some variation of it, is commonly used in awarding sports awards. Variations are used to determine the Most Valuable Player in baseball, to rank teams in NCAA sports, and to award the Heisman trophy. In contrast, it is only rarely used in elections for political office. As of 2019, it was used only in two countries: Nauru (a small Pacific island nation) and Slovenia (which only uses it to elect two members of parliament).

Example 7.2.3.

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?

Table 7.2.4.

		Number of voters
Borda points	Rankings	6	3	2
3	1st choice	F	S	M
2	2nd choice	S	M	S
1	3rd choice	M	F	F

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.

Exploration 7.2.1.

Let’s revisit the election from Exploration 7.1.1.

Table 7.2.5.

	Number of voters (in thousands)
Rankings	44	14	20	70	22	80	39
1st choice	G	G	G	M	M	B	B
2nd choice	M	B		G	B	M
3rd choice	B	M		B	G	G

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.

Solution.

The candidates’ points are as follows:

G: \(132+42+60+140+22+80+39 = 515\)
M: \(88+14+20+210+66+160+39 = 597\)
B: \(44+28+20+70+44+240+117 = 563\)

Therefore, M (McCarthy) wins the election when the Borda Count is used.

Subsection 7.2.2 The Method of Instant Runoff Voting (IRV)

Our second alternative voting method is called instant runoff voting. It is similar to the runoff elections described in Section 7.1, but instead of eliminating all but the top two candidates in the first election, it only eliminates candidates one at a time.

Instant Runoff Voting (IRV).

The method of Instant Runoff Voting (IRV) uses several rounds. Each round simulates a “runoff”, but they are all part of a single election. The rules are as follows:

For the first round, count how many first choice votes each candidate has received. Check to see if a candidate has a majority of 1st place votes. If there is such a candidate, then this candidate wins the election. If there is not a majority winner, continue to the next step.
Eliminate the candidate with the fewest first choice votes.
Use the preference lists of the voters who favored the eliminated candidate to transfer their votes to their next choice among the remaining candidates.
For the next round, count how many votes each remaining candidate has (including newly transferred votes as well as first choice votes). If a candidate has a majority, they are elected. If not, eliminate the candidate with the fewest votes in this round, and proceed to the next round.
Repeat the above steps until some candidate wins with a majority (which must happen at least by the time only two candidates remain, if there is not a tie).

Notice that, because of step 1, any candidate with a majority of first choice votes will automatically win the election. This is one notable property of IRV that is not necessarily shared by all voting methods, as we will see in the next section.

What else can we say about IRV? Let’s see how it works in practice.

Example 7.2.6.

The preference table below shows the same election we considered in Example 7.2.1. We found that Candidate B wins this election when using the Borda count. Let’s see who is the winner using IRV.

Table 7.2.7.

	Number of voters
Rankings	6	5	4	2
1st choice	A	D	C	B
2nd choice	B	B	D	C
3rd choice	D	C	B	D
4th choice	C	A	A	A

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.

Round 1:	Votes for A: 6	D: 5	C: 4	B: 2
	B is eliminated, and 2 votes are transferred to those voters’ second choice, C
Round 2:	Votes for A: 6	D: 5	C: 6
	D is eliminated, and 5 votes are transferred to those voters’ third choice, C
	(since their second choice, B, was already eliminated)
Round 3:	Votes for A: 6		C: 11
	Candidate C wins the election.

In Example 7.2.6, each candidate was the first choice of a single group of voters. This made the process of transferring votes in each round fairly easy. In the following example, this is not the case.

Example 7.2.8. The 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.

More complete details can be found at the Wikipedia article.

These three candidates were Bob Kiss, Kurt Wright, and Andy Montroll.

Table 7.2.9.

	Percent of voters
Rankings	34	37	15	9	5
1st choice	K	W	M	M	M
2nd choice	M	M	K	W
3rd choice	W	K	W	K

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

Round 1:	Percent of votes for K: 34	W: 37	M: 15+9+5=29
	M is eliminated, and votes are allocated to their different second choices.
Round 2:	K: 34+15=49	W: 37+9=46
	K wins the election.

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.

Such 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

en.wikipedia.org/wiki/2009_Burlington_mayoral_election

Of all the voting systems in this chapter, IRV is the second-most common in real elections (after plurality). In 2016, the state of Maine adopted IRV for most statewide elections. It is also used for local elections in various parts of the US. Internationally, IRV has been used for about a century in Australia (to elect their House of Representatives) and Ireland (to elect their president).

You may encounter different names for IRV in the news. Many in the US call it simply “ranked-choice voting”, even though it is only one of several voting methods in which voters rank their choices. In Australia it is called “preferential voting”, while in Britain it is referred to as the “alternative vote”.

Subsection 7.2.3 The Method of Pairwise Comparisons

The final voting method we will discuss is the method of pairwise comparisons. It is like a round robin tournament: we compare how candidates perform one-on-one. It attempts to prevent a situation where most voters would have preferred a different candidate to win, though that isn’t always possible to avoid.

The Method of Pairwise Comparisons.

The Method of Pairwise Comparisons has the following steps:

List all possible pairs of candidates.
For each pair, determine who would win if the election were only between those two candidates.
1. To do so, we must look at all the voters. Each voter casts their ballot for whichever candidate they ranked higher of the pair being considered (even if they didn’t rank that candidate first).
2. The winner of each pair is given 1 point, and the loser, 0 points. If it is a tie, they both get \(\frac{1}{2}\) point.
After finding each pairwise winner, add up the points for each candidate. The candidate with the most points is the winner.

Note that this method results in ties a little more often than some of the other methods.

Let’s use an example to see how this works in practice.

Example 7.2.10.

Suppose there is an election with the following preference table:

Table 7.2.11.

	Number of Votes
Rankings	23	17	40
1st choice	A	B	C
2nd choice	C	A	B
3rd choice	B	C	A

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.

Pair		Percent of voters	Winner/points
1	A	23	Lose = 0
	vs B	17+40=57	Win = 1
2	A	23+17=40	Tie = 0.5
	vs C	40	Tie = 0.5
3	B	17	Lose = 0
	vs C	23+40=63	Win = 1

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.

Example 7.2.12. The 2009 mayoral election in Burlington, VT.

Let’s return to the election from Example 7.2.8, shown again below.

Table 7.2.13.

	Percent of voters
Rankings	34	37	15	9	5
1st choice	K	W	M	M	M
2nd choice	M	M	K	W
3rd choice	W	K	W	K

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

Pair		Percent of voters	Winner/points
1	Kiss	34+15=49	Win=1
	versus Wright	37+9=46	Lose=0
2	Kiss	34	Lose=0
	versus Montroll	37+15+9+5=66	Win=1
3	Wright	37	Lose=0
	versus Montroll	34+15+9+5=63	Win=1

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.

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