6.1: Voting Methods
Every couple of years or so, voters go to the polls to cast ballots for their choices for mayor, governor, senator, president, and so on. Then, election officials count the ballots and declare a winner. But how do the election officials determine who the winner is?
If there are only two candidates, then there is no problem figuring out the winner. The candidate with more than 50% of the votes wins. This is known as the majority . So the candidate with the majority of the votes is the winner.
Majority rule means that the candidate or choice receiving more than 50% of all the votes is the winner.
But what happens if there are three candidates, no one receives more than 50%, and so no candidate receives a majority? The answer to that question depends on where you live.
Some places decide that the person with the most votes wins, even if they don’t have a majority. There are problems with this in that someone could be liked by 35% of the people, but is disliked by 65% of the people. So you have a winner that the majority doesn’t like. Other places conduct runoff elections where the top two candidates have to run again, and then the winner is chosen from the runoff election. There are some problems with this method. First, it is very costly for the candidates and the election office to hold a second election. Second, you don’t know if you will have the same voters voting in the second election, and so the preferences of the voters in the first election may not be taken into account.
Preference Voting
What can be done to have a better election that has someone liked by more voters yet doesn't require a runoff election? A ballot method that can fix this problem is known as a preference ballot . In this method, voters choose not only their favorite candidate but also their second favorite, their third favorite, and so on. You also may hear this referred to as ranked choice voting .
Ballots in which voters choose not only their favorite candidate, but voters actually order all of the candidates from their most favorite down to their least favorite.
To understand how a preference ballot works and how to determine the winner, we will look at an example.
Suppose an election is held to determine which bag of candy will be opened. The choices (candidates) are Hershey’s Miniatures (M), Nestle Crunch (C), and Mars’ Snickers (S). Each voter is asked to complete a ballot by marking their first, second, and third place choices.
Here are the ballots cast by the 18 voters.
| Voter | Anne | Bob | Chloe | Dylan | Eli | Fred |
|---|---|---|---|---|---|---|
| 1 st choice | C | M | C | M | S | S |
| 2 nd choice | S | S | M | C | M | M |
| 3 rd choice | M | C | S | S | C | C |
| Voter | George | Hiza | Isha | Jacy | Kalb | Lan |
|---|---|---|---|---|---|---|
| 1 st choice | S | S | S | M | C | M |
| 2 nd choice | M | M | M | C | M | C |
| 3 rd choice | C | C | C | S | S | S |
| Voter | Makya | Nadira | Ochen | Paki | Quinn | Riley |
|---|---|---|---|---|---|---|
| 1 st choice | S | S | C | C | S | S |
| 2 nd choice | M | M | M | M | M | M |
| 3 rd choice | C | C | S | S | C | C |
Now we must count the ballots. It isn’t as simple as just counting how many voters like each choice of candy. We have to look at how many liked the candidate in first place, second place, and third place. So there needs to be a better way to organize the results. This is known as a preference schedule.
A preference schedule is a table used to organize the results of all the preference ballots in an election.
Create a preference schedule for the ballots in the the Candy Election.
Solution
We are interested in counting how many voters chose each ordering. Looking at the preference ballots from Example 1A, you can see that
- Dylan, Jacy, and Lan all voted for the same order: M, then C, then S.
- Bob is the only voter with the order: M, then S, then C.
- Chloe, Kalb, Ochen, and Paki chose the order C, then M, then S.
- Anne is the only voter who voted C, then S, then M.
- The other 9 voters selected the order S, then M, then C.
- No voter selcted the order S, then C, then M, so this will not be included in the preference table.
We can summarize and present this information in the preference schedule.
| Number of voters | 3 | 1 | 4 | 1 | 9 |
|---|---|---|---|---|---|
| 1 st choice | M | M | C | C | S |
| 2 nd choice | C | S | M | S | M |
| 3 rd choice | S | C | S | M | C |
Methods of Determining a Winner
Once voters have voted, it is time to determine a winner. There are several different methods that be used to determine a winner of an election. The ones we will discuss here are
- Plurality Method,
- Borda Count method,
- Plurality with Elimination Method, and
- Pairwise Comparisons Method.
Each of these methods will be illustrated using the Candy Election ballots.
Plurality Method
The easiest, and the most common method in the United States, is the Plurality Method . In this method, the choice with the most first-preference votes is declared the winner. Only the first choice matters in the vote counting, and all other preferences are ignored.
The choice with the most first-place votes is declared the winner.
Use the preference schedule from the Candy Election to find the winner using the Plurality Method.
| Number of voters | 3 | 1 | 4 | 1 | 9 |
|---|---|---|---|---|---|
| 1 st choice | M | M | C | C | S |
| 2 nd choice | C | S | M | S | M |
| 3 rd choice | S | C | S | M | C |
Solution
From the preference schedule you can see that
- \(3 + 1 = 4\) people choose Hershey’s Miniatures as their first choice,
- \(4 + 1 = 5\) people picked Nestle Crunch as their first choice, and
- \(9\) people picked Snickers as their first choice.
So, Snickers wins with the most first-place votes, although Snickers does not have the majority. A candy would need more than 50% of the 18 votes, or more than 9 votes, to win a majority.
There is a problem with the Plurality Method. Notice that 9 people picked Snickers as their first choice, yet 7 chose it as their third choice. Thus, 9 people may be happy if the Snickers bag is opened, 7 voters will not be happy at all. So let’s look at another way to determine the winner that might be more pleasing to the entire group of voters.
Borda Count Method
Borda Count is another voting method, named for Jean-Charles de Borda, who developed the system in 1770. In this method, points are assigned to candidates based on their ranking. The point values for all ballots are totaled, and the candidate with the largest point total is the winner.
Each place on a preference ballot is assigned points. Last place receives one point, next-to-last place receives two points, and so on. Thus, if there are N candidates, then first-place receives N points. Now, multiply the point value for each place by the number of voters at the top of the column of the preference schedule to find the points each candidate wins in a column. Lastly, total up all the points for each candidate. The candidate with the most points wins.
Use the preference schedule from the Candy Election to find the winner using the Borda Count Method.
| Number of voters | 3 | 1 | 4 | 1 | 9 |
|---|---|---|---|---|---|
| 1 st choice | M | M | C | C | S |
| 2 nd choice | C | S | M | S | M |
| 3 rd choice | S | C | S | M | C |
Solution
Third choice receives one point, second choice receives two points, and first choice receives three points.
- There were 3 voters who chose the order M, C, S. So, M receives \(3 \cdot 3 = 9\) points for first place, C receives \(3 \cdot 2 = 6\) points, and S receives \(3 \cdot 1 = 3\) points for those ballots.
- There was 1 voter who chose the order M, S, C. So, M receives \(1 \cdot 3 = 3\) points for first place, C receives \(1 \cdot 2 = 2\) points, and S receives \(1 \cdot 1 = 1\) points for those ballots.
- There were 4 voters who chose the order C, M, S. So, C receives \(4 \cdot 3 = 12\) points for first place, M receives \(4 \cdot 2 = 8\) points, and S receives \(4 \cdot 1 = 4\) points for those ballots.
- The same process is conducted for the other columns. The table below summarizes the points that each candy received.
| Number of voters | 3 | 1 | 4 | 1 | 9 |
|---|---|---|---|---|---|
|
1 st choice 3 points |
M \(3 \cdot 3 = 9\) |
M \(1 \cdot 3 = 3\) |
C \(4 \cdot 3 = 12\) |
C \(1 \cdot 3 = 3\) |
S \(9 \cdot 3 = 27\) |
|
2 nd choice 2 points |
C \(3 \cdot 2 = 6\) |
S \(1 \cdot 2 = 2\) |
M \(4 \cdot 2 = 8\) |
S \(1 \cdot 2 = 2\) |
M \(9 \cdot 2 = 18\) |
|
3 rd choice 1 point |
S \(3 \cdot 1 = 3\) |
C \(1 \cdot 1 = 1\) |
S \(4 \cdot 1 = 1\) |
M \(1 \cdot 1 = 1\) |
C \(9 \cdot 1 = 9\) |
Adding up these points gives,
M: \(9 + 3 + 8 + 1 + 18 = 39\) C: \(6 + 1 + 12 + 3 + 9 = 31\) S: \(3 + 2 + 4 + 2 + 27 = 38\)
Thus, Hershey’s Miniatures wins using the Borda Count Method with 39 points.
So who is the winner? With one method, Snicker’s wins. With another method, Hershey’s Miniatures wins. It all depends on which method you use. Therefore, you need to decide which method to use before you run the election.
Plurality with Elimination Method
The Plurality with Elimination Method should only be used when the Plurality Method winner does not have a majority. The choice with the least first-place votes is then eliminated from the election, and any votes for that candidate are redistributed to the voters’ next choice. This continues until a choice has a majority (over 50%). This is similar to the idea of holding runoff elections, but since every voter’s order of preference is recorded on the ballot, the runoff can be computed without requiring a second costly election. A version of the Plurality with Elimination Method is used by the International Olympic Committee to select host nations.
Eliminate the candidate with the least amount of 1st place votes and re-distribute their votes amongst the other candidates according to their 2nd place preference. Repeat this process until a winner is decided. At any time during this process if a candidate has a majority of first-place votes, then that candidate is the winner.
Use the preference schedule from the Candy Election to find the winner using the Plurality with Elimination Method.
| Number of voters | 3 | 1 | 4 | 1 | 9 |
|---|---|---|---|---|---|
| 1 st choice | M | M | C | C | S |
| 2 nd choice | C | S | M | S | M |
| 3 rd choice | S | C | S | M | C |
Solution
This isn’t the most exciting example, since there are only three candidates, but the process is the same whether there are three or many more.
First, look at how many first-place votes there are: M has \(3+1 =4\) votes, C has \(4+1=5\), and S has 9 votes. There are a total of 18 votes and majority would be 10 votes. No candidate has received a majority, but M has the fewest number of votes so M is eliminated from the preference schedule.
| Number of voters | 3 | 1 | 4 | 1 | 9 |
|---|---|---|---|---|---|
| 1 st choice |
|
|
C | C | S |
| 2 nd choice | C | S |
|
S |
|
| 3 rd choice | S | C | S |
|
C |
After M is removed from the election and the next preferences are used, the preference schedule becomes
| Number of voters | 3 | 1 | 4 | 1 | 9 |
|---|---|---|---|---|---|
| 1 st choice | C | S | C | C | S |
| 2 nd choice | S | C | S | S | C |
And then we can combine all the same preferences down into a simpler table:
| Number of voters | 8 | 10 |
|---|---|---|
| 1 st choice | C | S |
| 2 nd choice | S | C |
So C has 8 first-place votes, and S has 10 first-place votes. S has achieved a majority. Thus, Snicker's wins using the Plurality with Elimination Method.
Pairwise Comparisons Method
The final method we will examine is the Pairwise Comparisons Method . In this method, each pair of candidates is compared, using all preferences to determine which of the two is "more preferred." The more preferred candidate is awarded 1 point. If there is a tie, each candidate is awarded \(1/2\) point. After all pairwise comparisons are made, the candidate with the most points is declared the winner.
Compare each candidate to the other candidates in one-on-one match-ups. Give the winner of each pairwise comparison a point. Give each candidate \(1/2\) point if there is a tie. The candidate with the most points wins.
Use the preference schedule from the Candy Election to find the winner using the Pairwise Comparisons Method.
| Number of voters | 3 | 1 | 4 | 1 | 9 |
|---|---|---|---|---|---|
| 1 st choice | M | M | C | C | S |
| 2 nd choice | C | S | M | S | M |
| 3 rd choice | S | C | S | M | C |
Solution
Taking only two candidates at a time, there are 3 one-on-one matchups possible: M vs. C, M vs. S, and S vs. C.
If you only have an election between M and C, then M wins the 3 votes in the first column, the 1 vote in the second column, and the 9 votes in the last column. C receives the rest of the votes. That means that M has \(3+1+9=13\) votes while C has \(4+1 =5\) votes. So, M wins when compared to C. M gets 1 point.
If you only compare M and S (the next one-on-one match-up), then M wins the 3 votes in the first column, the 1 vote in the second column, and the 4 votes in the third column. S receives the rest of the votes. That means that M has \(3+1+4=8\) votes while S has \(1+9 =10\) votes. So, S wins when compared to M. S gets 1 point.
Finally, if you compare C and S (the last one-on-one-match-up), then C wins the 3 votes in the first column, the 4 votes in the third column, and the 1 vote in the fourth column. S receives the rest of the votes. That means that C has \(3+4+1=8\) votes while S has \(1+9=10\) votes. So, S wins when compared to C. S gets 1 point.
The table summarizes the results.
| Match-Up 1 | Match-Up 2 | Match-Up 3 |
|---|---|---|
| M vs. C | M vs. S | S vs. C |
| 13 to 5 | 8 to 10 | 10 to 8 |
| Winner of Match-Up 1: M | Winner of Match-Up 2: S | Winner of Match-Up 3: S |
- M: 1 point
- S: 2 points
- C: 0 points
Thus, Snickers wins the election using the Pairwise Comparisons Method.
The problem with this method is that many overall elections (not just the one-on-one match-ups) will end in a tie, so you need to have a tie-breaker method designated before beginning the tabulation of the ballots. Another problem is that if there are more than three candidates, the number of pairwise comparisons that need to be analyzed becomes unwieldy. So, how many pairwise comparisons are there?
In Example 1F, there were 3 one-on-one comparisons when there were 3 candidates. You may think that means the number of pairwise comparisons is the same as the number of candidates, but that is not correct. Let’s see if we can come up with a way of knowing how many one-on-one matchups will be needed from the number of candidates.
Suppose you have 4 candidates called A, B, C, and D. A is to be matched up with B, C, and D (3 comparisons). B is to be compared with C and D, but has already been compared with A (2 comparisons). C needs to be compared with D, but has already been compared with A and B (1 more comparison). Therefore, the total number of one-on-one match-ups is \(3+2+1=6\) comparisons that need to be made with 4 candidates.
What about 5 or 6, or more candidates? Looking at 5 candidates, the first candidate needs to be matched-up with 4 other candidates, the second candidate needs to be matched-up with 3 other candidates, the third candidate needs to be matched-up with 2 other candidates, and the fourth candidate needs to only be matched-up with the last candidate for 1 more match-up. Thus, the total is \(4+3+2+1=10\) pairwise comparisons when there are 5 candidates. For 6 candidates, you would have \(5+4+3+2+1=15\) pairwise comparisons to do. For small numbers of candidates, it isn't hard to add these numbers up, but for large numbers of candidates there is a shortcut for adding numbers together.
It turns out that the following formula is true: \((n-1)+ (n-2)+ \cdots + 3 + 2 + 1 = \frac{n(n-1)}{2}\). So, if there are \(n\) candidates, then there are \(\frac{n(n-1)}{2}\) pairwise comparisons that can be made. For example, in an election with 10 candidates, there are \(9+8+7+6+5+4+3+2+1= \frac{10(10-1)}{2} = \frac{10(9)}{2} = 45\) pairwise comparisons.
As you can see from, the winner of an election can change depending on which voting method is used. The remainder of this section provides more examples using each of the four voting methods we have discussed.
A vacation club is trying to decide which destination to visit this year: Hawaii (H), Orlando (O), or Anaheim (A). Ten voters' individual ballots are shown below. Summarize the results into a preference schedule and decide the winner using the Plurality Method.
AOH, AHO, OHA, HAO, AHO, OHA, HAO, OHA, HAO, AHO
Solution
- 1 voter has the preference AOH,
- 3 voters have the preference AHO,
- 3 voters have the preference OHA, and
- 3 voters have the preference HAO.
The preference schedule for this election is
\(\begin{array}{|c|c|c|c|}
\hline & 1 & 3 & 3 & 3 \\
\hline 1^{\text {st }} \text { choice } & \mathrm{A} & \mathrm{A} & \mathrm{O} & \mathrm{H} \\
\hline 2^{\text {nd }} \text { choice } & \mathrm{O} & \mathrm{H} & \mathrm{H} & \mathrm{A} \\
\hline 3^{\text {rd }} \text { choice } & \mathrm{H} & \mathrm{O} & \mathrm{A} & \mathrm{O} \\
\hline
\end{array}\)
For the plurality method, we only care about the first-choice options. Totaling them up,
- A: 1+3 = 4 first-choice votes,
- O: 3 first-choice votes, and
- H: 3 first-choice votes.
Anaheim is the winner using the Plurality Method. Notice that Anaheim won with 4 out of 10 votes, 40% of the votes, which is a plurality but not a majority of the votes. Anaheim did not receive a majority of the vote because 4 is not more than 50% of 10 votes.
Three candidates are running in an election for County Executive: Goings (G), McCarthy (M), and Bunney (B). The preference schedule is shown below. Which candidate wins using the Plurality Method? Does the winning candidate receive a majority of first place votes?
\(\begin{array}{|c|c|c|c|c|}
\hline & 54 & 24 & 70 & 22 & 119 \\
\hline 1^{\text {st }} \text { choice } & \text { G } & \text { G } & \text { M } & \text { M } & \text { B } \\
\hline 2^{\text {nd }} \text { choice } & \text { M } & \text { B } & \text { G } & \text { B } & \text { M } \\
\hline 3^{\text {rd }} \text { choice } & \text { B } & \text { M } & \text { B } & \text { G } & \text { G } \\
\hline
\end{array}\)
- Answer
-
Using the Plurality method,
- G gets \(54+24 = 78\) first-choice votes,
- M gets \(70+22 = 92\) first-choice votes, and
- B gets \(119\) first-choice votes.
Bunney (B) wins under the Plurality Method. There are a total of \(78 + 92 + 119 = 289\) votes, but 119 is only about 41% of the votes. So, Bunney did not receive a majority.
A group of mathematicians are getting together for a conference. The members are coming from four cities: Seattle, Tacoma, Puyallup, and Olympia. Their approximate locations on a map are shown to the right. The votes for where to hold the conference are summarized in the preference schedule.
Where should the conference be held if the decision is made using the Borda Count Method?
\(\begin{array}{ |c|c|c|c| }
\hline & 51 & 25 & 10 & 14 \\
\hline 1^{\text {st }} \text { choice } & \text { Seattle } & \text { Tacoma } & \text { Puyallup } & \text { Olympia } \\
\hline 2^{\text {nd }} \text { choice } & \text { Tacoma } & \text { Puyallup } & \text { Tacoma } & \text { Tacoma } \\
\hline 3^{\text {rd }} \text { choice } & \text { Olympia } & \text { Olympia } & \text { Olympia } & \text { Puyallup } \\
\hline 4^{\text {th }} \text { choice } & \text { Puyallup } & \text { Seattle } & \text { Seattle } & \text { Seattle } \\
\hline
\end{array}\)
Solution
For each of the 51 ballots in the first column of the preference table, Puyallup will be given 1 point, Olympia 2 points, Tacoma 3 points, and Seattle 4 points. Multiplying the points per vote times the number of votes allows us to calculate points awarded. We proceed through the remaining 25, 10, and 14 ballots in the table.
\(\begin{array}{ |c|c|c|c| }
\hline & 51 & 25 & 10 & 14 \\
\hline 1^{\text {st choice }} & \text { Seattle } & \text { Tacoma } & \text { Puyallup } & \text { Olympia } \\
4 \text { points } & 4 \cdot 51=204 & 4 \cdot 25= 100 & 4 \cdot 10=40 & 4 \cdot 14=56 \\
\hline 2^{\text {nd choice }} & \text { Tacoma } & \text { Puyallup } & \text { Tacoma } & \text { Tacoma } \\
3 \text { points } & 3 \cdot 51=153 & 3 \cdot 25=75 & 3 \cdot 10=30 & 3 \cdot 14=42 \\
\hline 3^{\text {rd }} \text { choice } & \text { Olympia } & \text { Olympia } & \text { Olympia } & \text { Puyallup } \\
2 \text { points } & 2 \cdot 51=102 & 2 \cdot 25=50 & 2 \cdot 10=20 & 2 \cdot 14=28 \\
\hline 4^{\text {th }} \text { choice } & \text { Puyallup } & \text { Seattle } & \text { Seattle } & \text { Seattle } \\
1 \text { point } & 1 \cdot 51=51 & 1 \cdot 25=25 & 1 \cdot 10=10 & 1 \cdot 14=14 \\
\hline
\end{array}\)
Adding up the points for each city,
- Seattle: \(204 + 25 + 10 + 14 = 253\) points,
- Tacoma: \(153 + 100 + 30 + 42 = 325\) points,
- Puyallup: \(51 + 75 + 40 + 28 = 194\) points, and
- Olympia: \(102 + 50 + 20 + 56 = 228\) points.
Under the Borda Count Method, Tacoma is the winner of this vote. Note that, however, if this election were decided using the Plurality Method, Seattle would win with the most votes.
Refer to the election in Try It Now 1. The preference schedule is shown below. Which candidate wins using the Borda Count Method?
\(\begin{array}{|c|c|c|c|c|}
\hline & 54 & 24 & 70 & 22 & 119 \\
\hline 1^{\text {st }} \text { choice } & \text { G } & \text { G } & \text { M } & \text { M } & \text { B } \\
\hline 2^{\text {nd }} \text { choice } & \text { M } & \text { B } & \text { G } & \text { B } & \text { M } \\
\hline 3^{\text {rd }} \text { choice } & \text { B } & \text { M } & \text { B } & \text { G } & \text { G } \\
\hline
\end{array}\)
- Answer
-
Borda points are calculated as shown in the table.
\(\begin{array}{|c|c|c|c|c|}
\hline & 54 & 24 & 70 & 22 & 119 \\
\hline 1^{\text {st }} \text { choice } & \text{G} & \text{G} & \text{M} & \text{M} & \text{B} \\
3 \text { points } &54 \cdot 3 = 162 & 24 \cdot 3 = 72 & 70 \cdot 3 = 210 & 22 \cdot 3 = 66 & 119 \cdot 3 = 357 \\
\hline 2^{\text {nd }} \text { choice } & \text{M} & \text{B} & \text{G} & \text{B} & \text{M} \\
2 \text { points } &54 \cdot 2 = 108 & 24 \cdot 2 = 48 & 70 \cdot 2 = 140 & 22 \cdot 2 = 44 & 119 \cdot 2 = 238 \\
\hline 3^{\text {rd }} \text { choice } & \text{B} & \text{M} & \text{B} & \text{G} & \text{G} \\
1 \text { point } &54 \cdot 1 = 54 & 24 \cdot 1 = 24 & 70 \cdot 1 = 70 & 22 \cdot 1 = 22 & 119 \cdot 1 = 119 \\
\hline
\end{array}\)- G: \(162+72+140+22+119 = 515\) points
- M: \(108+24+210+66+238 = 646\) points
- B: \(54+48+70+44+357 = 573\) points
McCarthy (M) is the winner using Borda Count.
Consider the preference schedule below, in which a company’s advertising team is voting on five different advertising slogans, called A, B, C, D, and E here for simplicity. The choice of slogan will be made using the Plurality with Elimination Method.
\(\begin{array}{|c|c|c|c|c|c|}
\hline & 3 & 4 & 4 & 6 & 2 & 1 \\
\hline 1^{\text {st }} \text { choice } & \mathrm{B} & \mathrm{C} & \mathrm{B} & \mathrm{D} & \mathrm{B} & \mathrm{E} \\
\hline 2^{\text {nd }} \text { choice } & \mathrm{C} & \mathrm{A} & \mathrm{D} & \mathrm{C} & \mathrm{E} & \mathrm{A} \\
\hline 3^{\text {rd }} \text { choice } & \mathrm{A} & \mathrm{D} & \mathrm{C} & \mathrm{A} & \mathrm{A} & \mathrm{D} \\
\hline 4^{\text {th }} \text { choice } & \mathrm{D} & \mathrm{B} & \mathrm{A} & \mathrm{E} & \mathrm{C} & \mathrm{B} \\
\hline 5^{\text {th }} \text { choice } & \mathrm{E} & \mathrm{E} & \mathrm{E} & \mathrm{B} & \mathrm{D} & \mathrm{C} \\
\hline
\end{array}\)
Solution
If this was a plurality election, slogan B would win with 9 first-choice votes, compared to 0 for slogan A, 4 for slogan C, 6 for slogan D, and 1 for slogan E. There are total of \(3+4+4+6+2+1 = 20\) votes. A majority would be 11 votes. No choice yet has a majority, so we proceed to elimination rounds.
Round 1 : We make the first elimination. Choice A has the fewest first-place votes (0), so we remove that choice from the preference table.
\(\begin{array}{|c|c|c|c|c|c|}
\hline & 3 & 4 & 4 & 6 & 2 & 1 \\
\hline 1^{\text {st }} \text { choice } & \mathrm{B} & \mathrm{C} & \mathrm{B} & \mathrm{D} & \mathrm{B} & \mathrm{E} \\
\hline 2^{\text {nd }} \text { choice } & \mathrm{C} & & \mathrm{D} & \mathrm{C} & \mathrm{E} & \\
\hline 3^{\text {rd }} \text { choice } & & \mathrm{D} & \mathrm{C} & & & \mathrm{D} \\
\hline 4^{\text {th }} \text { choice } & \mathrm{D} & \mathrm{B} & & \mathrm{E} & \mathrm{C} & \mathrm{B} \\
\hline 5^{\text {th }} \text { choice } & \mathrm{E} & \mathrm{E} & \mathrm{E} & \mathrm{B} & \mathrm{D} & \mathrm{C} \\
\hline
\end{array}\)
We then shift everyone’s choices up to fill the gaps.
\(\begin{array}{|c|c|c|c|c|c|}
\hline & 3 & 4 & 4 & 6 & 2 & 1 \\
\hline 1^{\text {st }} \text { choice } & \mathrm{B} & \mathrm{C} & \mathrm{B} & \mathrm{D} & \mathrm{B} & \mathrm{E} \\
\hline 2^{\text {nd }} \text { choice } & \mathrm{C} & \mathrm{D} & \mathrm{D} & \mathrm{C} & \mathrm{E} & \mathrm{D} \\
\hline 3^{\text {rd }} \text { choice } & \mathrm{D} & \mathrm{B} & \mathrm{C} & \mathrm{E} & \mathrm{C} & \mathrm{B} \\
\hline 4^{\text {th }} \text { choice } & \mathrm{E} & \mathrm{E} & \mathrm{E} & \mathrm{B} & \mathrm{D} & \mathrm{C} \\
\hline
\end{array}\)
After the first elimination, slogan B has 9 first-choice votes, compared to 4 for slogan C, 6 for slogan D, and 1 for slogan E. There is still no choice with a majority, so we eliminate again.
Round 2 : We make our second elimination. Choice E has the fewest first-place votes, so we remove that choice, shifting everyone’s options to fill the gaps.
\(\begin{array}{|c|c|c|c|c|c|}
\hline & 3 & 4 & 4 & 6 & 2 & 1 \\
\hline 1^{\text {st }} \text { choice } & \mathrm{B} & \mathrm{C} & \mathrm{B} & \mathrm{D} & \mathrm{B} & \mathrm{D} \\
\hline 2^{\text {nd }} \text { choice } & \mathrm{C} & \mathrm{D} & \mathrm{D} & \mathrm{C} & \mathrm{C} & \mathrm{B} \\
\hline 3^{\text {rd }} \text { choice } & \mathrm{D} & \mathrm{B} & \mathrm{C} & \mathrm{B} & \mathrm{D} & \mathrm{C} \\
\hline
\end{array}\)
Now slogan B has 9 first-choice votes, slogan C has 4 votes, and slogan D has 7 votes. Still no choice has reached a majority of 11, so we eliminate again.
Round 3 : We make our third elimination. Slogan C has the fewest votes and is eliminated.
\(\begin{array}{|c|c|c|c|c|c|}
\hline & 5 & 4 & 4 & 6 & 1 \\
\hline 1^{\text {st }} \text { choice } & \text { B } & \text { D } & \text { B } & \text { D } & \text { D } \\
\hline 2^{\text {nd }} \text { choice } & \text { D } & \text { B } & \text { D } & \text { B } & \text { B } \\
\hline
\end{array}\)
Now, slogan B has 9 first-place votes while slogan D has 11. Slogan D has now gained a majority and is declared the winner using Plurality with Elimination.
Refer to the election in Try It Now 1. The preference schedule is shown below. Which candidate wins using the Plurality with Elimination Method?
\(\begin{array}{|c|c|c|c|c|}
\hline & 54 & 24 & 70 & 22 & 119 \\
\hline 1^{\text {st }} \text { choice } & \text { G } & \text { G } & \text { M } & \text { M } & \text { B } \\
\hline 2^{\text {nd }} \text { choice } & \text { M } & \text { B } & \text { G } & \text { B } & \text { M } \\
\hline 3^{\text {rd }} \text { choice } & \text { B } & \text { M } & \text { B } & \text { G } & \text { G } \\
\hline
\end{array}\)
- Answer
-
Initial Votes: G has 78 first-place votes, M has 92 first-place votes, and B has 119 first place votes.
G is eliminated, leaving only M and B.
\(\begin{array}{|c|c|c|c|c|}
\hline & 54 & 24 & 70 & 22 & 119 \\
\hline 1^{\text {st }} \text { choice } & \text { M } & \text { B } & \text { M } & \text { M } & \text { B } \\
\hline 2^{\text {nd }} \text { choice } & \text { B } & \text { M } & \text { B } & \text { B } & \text { M } \\
\hline
\end{array}\)Now, M has 146 votes and B has 143 votes. Candidate M is the winner and has gained a majority over Candidate B.
A club is holding elections for president and will use the Pairwise Comparisons Method to decide the winner. There are four candidates (labeled A, B, C, and D for convenience). The preference schedule for the election is shown below.
\(\begin{array}{ |c|c|c|c|c|}
\hline & 14 & 10 & 8 & 4 & 1 \\
\hline 1^{\text {st }} \text { choice } & \text { A } & \text { C } & \text { D } & \text { B } & \text { C } \\
\hline 2^{\text {nd }} \text { choice } & \text { B } & \text { B } & \text { C } & \text { D } & \text { D } \\
\hline 3^{\text {rd }} \text { choice } & \text { C } & \text { D } & \text { B } & \text { C } & \text { B } \\
\hline 4^{\text {th }} \text { choice } & \text { D } & \text { A } & \text { A } & \text { A } & \text { A } \\
\hline
\end{array}\)
Solution
With 4 candidates, there are \(3 + 2 + 1 =6\), or \(\frac{4(3)}{2}\), comparisons to make:
\(\begin{array} {ll} {\text{A vs B: }14\text{ prefer A, and }10+8+4+1 = 23\text{ prefer B }} & {\text{B gets }1\text{ point}} \\ {\text{A vs C: }14\text{ prefer A, and }10+8+4+1 = 23\text{ prefer C }} & {\text{C gets }1\text{ point}} \\ {\text{A vs D: }14\text{ prefer A, and }10+8+4+1 = 23\text{ prefer D }} & {\text{D gets }1\text{ point}} \\ {\text{B vs C: }14+4=18\text{ prefer B, and }10+8+1=19\text{ prefer C}} & {\text{C gets }1\text{ point}} \\ {\text{B vs D: }14+10+4=28\text{ prefer B, and }8+1=9\text{ prefer D}} & {\text{B gets }1\text{ point}} \\{\text{C vs D: }14+10+1=25\text{ prefer C, and }8+4=12\text{ prefer D}} & {\text{C gets }1\text{ point}} \\ \end{array}\)
So, A gets 0 points, B gets 2 points, C gets 3 points, and D gets 1 point.
Using the Pairwise Comparison Method, we declare C as the winner.
Refer to the election in Try It Now 1. The preference schedule is shown below. Which candidate wins using the Pairwise Comparisons Method?
\(\begin{array}{ |c|c|c|c|c|}
\hline & 54 & 24 & 70 & 22 & 119 \\
\hline 1^{\text {st }} \text { choice } & \text { G } & \text { G } & \text { M } & \text { M } & \text { B } \\
\hline 2^{\text {nd }} \text { choice } & \text { M } & \text { B } & \text { G } & \text { B } & \text { M } \\
\hline 3^{\text {rd }} \text { choice } & \text { B } & \text { M } & \text { B } & \text { G } & \text { G } \\
\hline
\end{array}\)
- Answer
-
\(\begin{array} {ll} {\text{G vs M: }78\text{ prefer G, and }211\text{ prefer M }} & {\text{M gets }1\text{ point}} \\ {\text{G vs B: }148\text{ prefer G, and }141\text{ prefer B }} & {\text{G gets }1\text{ point}} \\ {\text{M vs B: }146\text{ prefer M, and } 143\text{ prefer B }} & {\text{M gets }1\text{ point}} \\ \end{array}\)
So, G gets 1 point, M gets 2 points, and B gets 0 points.
Using the Pairwise Comparison Method, we declare M as the winner.