7.1: Voting Methods
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Every couple of years or so, voters go to the polls to cast ballots for their choices for mayor, governor, senator, president, etc. Then the 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: This concept means that the candidate (choice) receiving more than 50% of the vote is the winner.
But what happens if there are three candidates, and no one receives the majority? That 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.
So 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.
Preference Ballots: Ballots in which voters choose not only their favorite candidate, but they actually order all of the candidates from their most favorite down to their least favorite.
Note: Preference Ballots are transitive: If a voter prefers choice A to choice B and also prefers choice B to choice C, then the voter must prefer choice A to choice C.
To understand how a preference ballot works and how to determine the winner, we will look at an example.
Example \(\PageIndex{1}\): Preference Ballot for the Candy Election
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 fill in the following ballot, by marking their first, second, and third place choices.
Figure \(\PageIndex{1}\): Preference Ballot for the Candy Election
Each voter fills out the above ballot with their preferences, and what follows is the results of the election.
Voter | Anne | Bob | Chloe | Dylan | Eli | Fred |
1st choice | C | M | C | M | S | S |
2nd choice | S | S | M | C | M | M |
3rd choice | M | C | S | S | C | C |
Voter | George | Hiza | Isha | Jacy | Kalb | Lan |
1st choice | S | S | S | M | C | M |
2nd choice | M | M | M | C | M | C |
3rd choice | C | C | C | S | S | S |
Voter | Makya | Nadira | Ochen | Paki | Quinn | Riley |
1st choice | S | S | C | C | S | S |
2nd choice | M | M | M | M | M | M |
3rd 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 candidate. You 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.
Preference Schedule: A table used to organize the results of all the preference ballots in an election.
Example \(\PageIndex{2}\): Preference Schedule for the Candy Election
Using the ballots from Example \(\PageIndex{1}\), we can count how many people liked each ordering. Looking at Table \(\PageIndex{2}\), you may notice that three voters (Dylan, Jacy, and Lan) had the order M, then C, then S. Bob is the only voter with the order M, then S, then C. Chloe, Kalb, Ochen, and Paki had the order C, M, S. Anne is the only voter who voted C, S, M. All the other 9 voters selected the order S, M, C. Notice, no voter liked the order S, C, M. We can summarize this information in a table, called the preference schedule.
Number of voters | 3 | 1 | 4 | 1 | 9 |
1st choice | M | M | C | C | S |
2nd choice | C | S | M | S | M |
3rd choice | S | C | S | M | C |
Methods of Counting Ballots:
Now that we have organized the ballots, how do we determine the winner? There are several different methods that can be used. The easiest, and most familiar, is the Plurality Method.
Plurality Method: The candidate with the most first-place votes wins the election.
Example \(\PageIndex{3}\): The Winner of the Candy Election—Plurality Method
Using the preference schedule in Table \(\PageIndex{3}\), find the winner using the Plurality Method.
From the preference schedule you can see that four (3 + 1) people choose Hershey’s Miniatures as their first choice, five (4 + 1) picked Nestle Crunch as their first choice, and nine picked Snickers as their first choice. So Snickers wins with the most first-place votes, although Snickers does not have the majority of first-place votes.
There is a problem with the Plurality Method. Notice that nine people picked Snickers as their first choice, yet seven chose it as their third choice. Thus, nine people may be happy if the Snickers bag is opened, but seven people will not be happy at all. So let’s look at another way to determine the winner.
The Borda Count Method (Point System): 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 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.
Example \(\PageIndex{4}\): The Winner of the Candy Election—Borda Count Method
Using the preference schedule in Table \(\PageIndex{3}\), find the winner using the Borda Count Method.
Number of voters | 3 | 1 | 4 | 1 | 9 |
1st choice | M | M | C | C | S |
2nd choice | C | S | M | S | M |
3rd choice | S | C | S | M | C |
The third choice receives one point, second choice receives two points, and first choice receives three points. There were three voters who chose the order M, C, S. So M receives 3*3 = 9 points for the first-place, C receives 3*2 = 6 points, and S receives 3*1 = 3 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 |
1st choice |
M 9 |
M 3 |
C 12 |
C 3 |
S 27 |
2nd choice |
C 6 |
S 2 |
M 8 |
S 2 |
M 18 |
3rd choice |
S 3 |
C 1 |
S 4 |
M 1 |
C 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.
So who is the winner? With one method Snicker’s wins and with another method Hershey’s Miniatures wins. The problem is that it all depends on which method you use. Therefore, you need to decide which method to use before you run the election.
The Plurality with Elimination Method (Sequential Runoffs): Eliminate the candidate with the least amount of 1st place votes and re-distribute their votes amongst the other candidates. Repeat this process until you find a winner. Note: At any time during this process if a candidate has a majority of first-place votes, then that candidate is the winner. |
Example \(\PageIndex{5}\): The Winner of the Candy Election—Plurality with Elimination Method
Using the preference schedule in Table \(\PageIndex{3}\), find the winner using the Plurality with Elimination Method.
Number of voters | 3 | 1 | 4 | 1 | 9 |
1st choice | M | M | C | C | S |
2nd choice | C | S | M | S | M |
3rd choice | S | C | S | M | C |
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. So look at how many first-place votes there are. M has , C has , and S has 9. So M is eliminated from the preference schedule.
Number of voters | 3 | 1 | 4 | 1 | 9 |
1st choice | M | M | C | C | S |
2nd choice | C | S | M | S | M |
3rd choice | S | C | S | M | C |
So the preference schedule becomes:
Number of voters | 3 | 1 | 4 | 1 | 9 |
1st choice | C | S | C | C | S |
2nd choice | S | C | S | S | C |
And then we can condense it down to:
Number of voters | 8 | 10 |
1st choice | C | S |
2nd choice | S | C |
So C has eight first-place votes, and S has 10. So S wins.
The Method of Pairwise Comparisons: Compare each candidate to the other candidates in one-on-one match-ups. Give the winner of each pairwise comparison a point. The candidate with the most points wins.
Example \(\PageIndex{6}\): The Winner of the Candy Election—Pairwise Comparisons Method
Using the preference schedule in Table \(\PageIndex{3}\), find the winner using the Pairwise Comparisons Method.
Number of voters | 3 | 1 | 4 | 1 | 9 |
1st choice | M | M | C | C | S |
2nd choice | C | S | M | S | M |
3rd choice | S | C | S | M | C |
If you only have an election between M and C (the first one-on-one match-up), then M wins the three votes in the first column, the one vote in the second column, and the nine votes in the last column. That means that M has thirteen votes while C has five. So M wins when compared to C. M gets one point.
If you only compare M and S (the next one-on-one match-up), then M wins the first three votes in column one, the next one vote in column two, and the four votes in column three. M has eight votes and S has 10 votes. So S wins compared to M, and S gets one point.
Comparing C to S, C wins the three votes in column one, the four votes in column three, and one vote in column four. C has eight votes while S has 10 votes. So S wins compared to C, and S gets one point.
To summarize, M has one point, and S has two points. Thus, S wins the election using the Method of Pairwise Comparisons.
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
S: 2
C: 0
Thus, S wins the election.
Note: If any one given match-up ends in a tie, then both candidates receive ½ point each for that match-up.
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 \(\PageIndex{6}\), there were three one-on-one comparisons when there were three 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 formula for the number of candidates. Suppose you have four candidates called A, B, C, and D. A is to be matched up with B, C, and D (three comparisons). B is to be compared with C and D, but has already been compared with A (two comparisons). C needs to be compared with D, but has already been compared with A and B (one more comparison). Therefore, the total number of one-on-one match-ups is comparisons that need to be made with four candidates. What about five or six or more candidates? Looking at five candidates, the first candidate needs to be matched-up with four other candidates, the second candidate needs to be matched-up with three other candidates, the third candidate needs to be matched-up with two other candidates, and the fourth candidate needs to only be matched-up with the last candidate for one more match-up. Thus, the total is pairwise comparisons when there are five candidates.
Now, for six candidates, you would have pairwise comparisons to do. Continuing this pattern, if you have N candidates then there are pairwise comparisons. 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 the numbers together. It turns out that the following formula is true: . Thus, for 10 candidates, there are pairwise comparisons. So you can see that in this method, the number of pairwise comparisons to do can get large quite quickly.
Now that we have reviewed four different voting methods, how do you decide which method to use? One question to ask is which method is the fairest? Unfortunately, there is no completely fair method. This is based on Arrow’s Impossibility Theorem.
Arrow's Impossibility Theorem: No voting system can satisfy all four fairness criteria in all cases.
This brings up the question, what are the four fairness criteria? They are guidelines that people use to help decide which voting method would be best to use under certain circumstances. They are the Majority Criterion, Condorcet Criterion, Monotonicity Criterion, and Independence of Irrelevant Alternatives Criterion.
Fairness Criteria:
The Majority Criterion (Criterion 1): If a candidate receives a majority of the 1st-place votes in an election, then that candidate should be the winner of the election.
The Condorcet Criterion (Criterion 2): If there is a candidate that in a head-to-head comparison is preferred by the voters over every other candidate, then that candidate should be the winner of the election. This candidate is known as the Condorcet candidate.
The Monotonicity Criterion (Criterion 3): If candidate X is a winner of an election and, in a re-election, the only changes in the ballots are changes that favor X, then X should remain a winner of the election.
The Independence of Irrelevant Alternatives Criterion (Criterion 4): If candidate X is a winner of an election and one (or more) of the other candidates is removed and the ballots recounted, then X should still be a winner of the election.
Example \(\PageIndex{7}\): Condorcet Criterion Violated
Suppose you have a vacation club trying to figure out where it wants to spend next year’s vacation. The choices are Hawaii (H), Anaheim (A), or Orlando (O). The preference schedule for this election is shown below in Table \(\PageIndex{9}\).
Number of voters | 1 | 3 | 3 | 3 |
1st choice | A | A | O | H |
2nd choice | O | H | H | A |
3rd choice | H | O | A | O |
Using the Plurality Method, A has four first-place votes, O has three first-place votes, and H has three first-place votes. So, Anaheim is the winner. However, if you use the Method of Pairwise Comparisons, A beats O (A has seven while O has three), H beats A (H has six while A has four), and H beats O (H has six while O has four). Thus, Hawaii wins all pairwise comparisons against the other candidates, and would win the election. In fact Hawaii is the Condorcet candidate. However, the Plurality Method declared Anaheim the winner, so the Plurality Method violated the Condorcet Criterion.
Example \(\PageIndex{8}\): Monotonicity Criterion Violated
Suppose you have a voting system for a mayor. The resulting preference schedule for this election is shown below in Table \(\PageIndex{10}\).
Number of voters | 37 | 22 | 12 | 29 |
1st choice | Adams | Brown | Brown | Carter |
2nd choice | Brown | Carter | Adams | Adams |
3rd choice | Carter | Adams | Carter | Brown |
Using the Plurality with Elimination Method, Adams has 37 first-place votes, Brown has 34, and Carter has 29, so Carter would be eliminated. Carter’s votes go to Adams, and Adams wins. Suppose that the results were announced, but then the election officials accidentally destroyed the ballots before they could be certified, so the election must be held again. Wanting to “jump on the bandwagon,” 10 of the voters who had originally voted in the order Brown, Adams, Carter; change their vote to the order of Adams, Brown, Carter. No other voting changes are made. Thus, the only voting changes are in favor of Adams. The new preference schedule is shown below in Table \(\PageIndex{11}\).
Number of voters | 47 | 22 | 2 | 29 |
1st choice | Adams | Brown | Brown | Carter |
2nd choice | Brown | Carter | Adams | Adams |
3rd choice | Carter | Adams | Carter | Brown |
Now using the Plurality with Elimination Method, Adams has 47 first-place votes, Brown has 24, and Carter has 29. This time, Brown is eliminated first instead of Carter. Two of Brown’s votes go to Adams and 22 of Brown’s votes go to Carter. Now, Adams has 47 + 2 = 49 votes and Carter has 29 + 22 = 51 votes. Carter wins the election. This doesn’t make sense since Adams had won the election before, and the only changes that were made to the ballots were in favor of Adams. However, Adams doesn’t win the re-election. The reason that this happened is that there was a difference in who was eliminated first, and that caused a difference in how the votes are re-distributed. In this example, the Plurality with Elimination Method violates the Monotonicity Criterion.
Example \(\PageIndex{9}\): Majority Criterion Violated
Suppose a group is planning to have a conference in one of four Arizona cities: Flagstaff, Phoenix, Tucson, or Yuma. The votes for where to hold the conference are summarized in the preference schedule shown below in Table \(\PageIndex{12}\).
Number of voters | 51 | 25 | 10 | 14 |
1st choice | Flagstaff | Phoenix | Yuma | Tucson |
2nd choice | Phoenix | Yuma | Phoenix | Phoenix |
3rd choice | Tucson | Tucson | Tucson | Yuma |
4th choice | Yuma | Flagstaff | Flagstaff | Flagstaff |
If we use the Borda Count Method to determine the winner then the number of Borda points that each candidate receives are shown in Table \(\PageIndex{13}\).
Number of voters | 51 | 25 | 10 | 14 |
1st choice 4 points |
Flagstaff 204 |
Phoenix 100 |
Yuma 40 |
Tucson 56 |
2nd choice 3 points |
Phoenix 153 |
Yuma 75 |
Phoenix 30 |
Phoenix 42 |
3rd choice 2 points |
Tucson 102 |
Tucson 50 |
Tucson 20 |
Yuma 28 |
4th choice 1 point |
Yuma 51 |
Flagstaff 25 |
Flagstaff 10 |
Flagstaff 14 |
The totals of all the Borda points for each city are:
Phoenix:
Yuma:
Tucson:
Phoenix wins using the Borda Count Method. However, notice that Flagstaff actually has the majority of first-place votes. There are 100 voters total and 51 voters voted for Flagstaff in first place (51/100 = 51% or a majority of the first-place votes). So, Flagstaff should have won based on the Majority Criterion. This shows how the Borda Count Method can violate the Majority Criterion.
Example \(\PageIndex{10}\): Independence of Irrelevant Alternatives Criterion Violated
A committee is trying to award a scholarship to one of four students: Anna (A), Brian (B), Carlos (C), and Dmitri (D). The votes are shown below.
Number of voters | 5 | 5 | 6 | 4 |
1st choice | D | A | C | B |
2nd choice | A | C | B | D |
3rd choice | C | B | D | A |
4th choice | B | D | A | C |
Using the Method of Pairwise Comparisons:
A vs B: 10 votes to 10 votes, A gets ½ point and B gets ½ point
A vs C: 14 votes to 6 votes, A gets 1 point
A vs D: 5 votes to 15 votes, D gets 1 point
B vs C: 4 votes to 16 votes, C gets 1 point
B vs D: 15 votes to 5 votes, B gets 1 point
C vs D: 11 votes to 9 votes, C gets 1 point
So A has 1½ points, B has 1 point, C has 2 points, and D has 1 point. So Carlos is awarded the scholarship.
Now suppose it turns out that Dmitri didn’t qualify for the scholarship after all. Though it should make no difference, the committee decides to recount the vote. The preference schedule without Dmitri is below.
Number of voters | 10 | 6 | 4 |
1st choice | A | C | B |
2nd choice | C | B | A |
3rd choice | B | A | C |
Using the Method of Pairwise Comparisons:
A vs B: 10 votes to 10 votes, A gets ½ point and B gets ½ point
A vs C: 14 votes to 6 votes, A gets 1 point
B vs C: 4 votes to 16 votes, C gets 1 point
So A has 1½ points, B has ½ point, and C has 1 point. Now Anna is awarded the scholarship instead of Carlos. This is an example of The Method of Pairwise Comparisons violating the Independence of Irrelevant Alternatives Criterion.
In summary, every one of the fairness criteria can possibly be violated by at least one of the voting methods as shown in Table \(\PageIndex{16}\). However, keep in mind that this does not mean that the voting method in question will violate a criterion in every election. It is just important to know that these violations are possible.
Plurality | Borda Count | Plurality with Elimination | Pairwise Comparisons | |
---|---|---|---|---|
Majority Criterion | * | Violation Possible | * | * |
Condorcet Criterion | Violation Possible | Violation Possible | Violation Possible | * |
Monotonicity Criterion | * | * | Violation Possible | * |
Independence of Irrelevant Alternatives Criterion | Violation Possible | Violation Possible | Violation Possible | Violation Possible |
* The indicated voting method does not violate the indicated criterion in any election.
Insincere Voting:
This is when a voter will not vote for whom they most prefer because they are afraid that the person they are voting for won’t win, and they really don’t want another candidate to win. So, they may vote for the person whom they think has the best chance of winning over the person they don’t want to win. This happens often when there is a third party candidate running. As an example, if a Democrat, a Republican, and a Libertarian are all running in the same race, and you happen to prefer the Libertarian candidate. However, you are afraid that the Democratic candidate will win if you vote for the Libertarian candidate, so instead you vote for the Republican candidate. You have voted insincerely to your true preference.
Approval Voting
Since there is no completely fair voting method, people have been trying to come up with new methods over the years. One idea is to have the voters decide whether they approve or disapprove of candidates in an election. This way, the voter can decide that they would be happy with some of the candidates, but would not be happy with the other ones. A possible ballot in this situation is shown in Table \(\PageIndex{17}\):
Candidate | Approve | Disapprove |
---|---|---|
Smith | X | |
Baker | X | |
James | X | |
Paulsen | X |
This voter would approve of Smith or Paulsen, but would not approve of Baker or James.
In this type of election, the candidate with the most approval votes wins the election.
One issue with approval voting is that it tends to elect the least disliked candidate instead of the best candidate. Another issue is that it can result in insincere voting as described above.
As a reminder, there is no perfect voting method. Arrow proved that there never will be one. So make sure that you determine the method of voting that you will use before you conduct an election.