Fairness and Equity: Notes 8

Prepared by:

Joseph Malkevitch
Department of Mathematics and Computer Studies
York College (CUNY)
Jamaica, New York 11451



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Theory of Elections (continued)

Here is a more extended discussion of the notation that was developed by Duncan Black, a British political scientist for displaying in an easy to understand way the preferences of voters for candidates (alternatives). This notation appears in Black's book Theory of Committees and Elections.

How are we to interpret the diagram above? There are three candidates (or alternatives) whose names are A, B, and C. Nine (9) voters feel A is most preferred of the three candidates, B next preferred, and the least preferred is C. In more detail, A is preferred to B and C, and B is preferred to C. Note, that this does not necessarily mean that the voters had any negative feelings about C. However, they do prefer either A or B to C. It is possible, of course, that C is actually very hated. One can not tell this from the information that the voter provided. Black's elegant notation also allows the voters to express indifference:

Thus, in the diagram below we see that the voters faced 5 alternatives, whose names are A, B, C, D, and E. Six (6) voters had the following perceptions of the alternatives:

B was preferred to A, C, D, and E; C was preferred to A, D, and E. The six voters were indifferent between A and D, they viewed them as having the same appeal on balance. (Thus, they may have liked A more than D on some issues but D more than A on other issues so that overall, they viewed the two with indifference. Finally, they viewed all of the candidates A, B, C, and D more positively than E. Again, they may not "hate" E; they only feel of all the choices, E is the least preferred.

Now we consider how we can take into account the fact that voters rank the alternatives (candidates) in different positions to design an election decision method. The Borda Count, named for Jean de Charles Borda (who lived during the period of the French Revolution) is the best known of these point count systems where first place, second place, etc. is taken into account in deciding the winner.

Since there are three candidates below C, we give C three points, and weight this for the 8 voters who voted for this preference schedule. Thus, C gets 3(8) = 24 points. B has two candidates below him, so he gets 2(8) = 16 points. Similarly, D gets 1(8) points. Since A came in last, no candidates are below A so A gets 0(8) = 0 points.

If we have an election such as:

Election 1

Who is the plurality winner in this election? A gets 12 first place votes; B gets 10 and D gets 9, while D gets 0, since no one ranks D first. However, A does not have a majority. If we conduct a run-off election, we eliminate C and D. Now, if an election is conducted between A and B we get:

A gets 12 + 0 + 0 while B gets 0 + 10 + 10 votes, so B is the run-off winner. Note, this is a different person from the plurality winner.

Let us determine the Borda Count winner:

A gets 3(12) + 1 (10) + 0 (9) point. (Here is a more detailed analysis. In the first preference schedule A came in first. Thus, there are 3 candidates below her. She therefore gets 3(12) points because 12 voters voted for this schedule; from the second preference schedule, A came in 4th with one candidate (C) below her. Thus, A gets 1 (10) = 10. Since the last 8 voters ranked A last, she gets no points here.) So, in all, A gets 36 + 10 = 46 points.

B gets 0(12) + 3(10) + 2(9) = 48 points

C gets 1(12) + 0(10) + 3(9) = 39.

D gets 2(12) + 2(10) + 1(9) = 53.

Hence, D is the Borda Count winner. Now some may find this peculiar, that the winner of this election is a candidate who no one liked best. The reason that this happened is that no one ranked D very low either. On the average D was higher than the other candidates in his position. Here is the vote table for this election:

The meaning of the number in row C and column B, for example, is 21 voters preferred C to B. The entry in row B and column C means that 10 voters preferred B to C. Note that 21 + 10 = 31, which is the total number of people who voted in the election.

  A B C D Total
A   12 22 12 46
B 19   10 19 48
C 9 21   9 39
D 19 12 22   53

Vote Table 1

Note that the row totals in this table are the same as the Borda Count values computed in a different way earlier. This is not an accident. The row sum of the table is always the Borda count.

Election 2

Again we can compute the Borda Count. Note that for each of A and D on the left-most preference schedule there is only one candidate below each of these candidates, namely E.


A gets 1(6) + 2 (10) = 26

B gets 3(6) + 4(10) = 48 points

C gets 3(6) + 1(10) = 28 points

D gets 6(1) + 3(10) = 36 points

E gets 0(6) + 0 (10) = 0 points.

Here is the vote table for this election:

  A B C D E Total
A   0 10 0 16 26
B 16   16 16 16 48
C 6 0   6 16 28
D 10 0 10   16 36
E 0 0 0 0   0

Vote Table 2

Note that again the row sums of the vote table give the same values as the Borda Count computed directly from the number of candidates below a given candidate. Not surprisingly, since all the voters preferred A to the other candidates, the Borda Count gives B as the winner. In this case B not only is the plurality winner, B is the majority winner!

In addition to the value of the vote table for computing the Borda Count it can also be used to find the Condorcet winner of an election if there is one. The Condorcet winner of an election is that person (alternative) who can beat every other alternative in a two-way race.

From Vote Table 1 we see that B beats A in a 2-way race (19 to 12); A beats C in a 2-way race (22 to 9); C beats B by 10; B beats D (19 to 12) and D beats A (19 to 12)! Thus, there is no person who can beat all the others in a 2-way race for this election! Here is a diagram that shows the unintuitive results concerning the winners of the 2-way races for this election.

It is easy to verify that for Vote Table 2 B can beat all of the other candidates in a 2-way race. B is the Condorcet winner. Many elections have a Condorcet winner but many do not. What can be done if there is no Condorcet winner? We will leave this question on the side for the moment, as we continue to get insight into different ways to decide elections and what is appealing or not appealing about these methods.

Are you bothered by the results of Election 1 using the different methods? It is common for people in discussing democracy to use phrases such as "X has a mandate from the people" or "Let the best candidate win." Yet, we see that the candidate who wins, using the same input data, can vary! Different methods, each of which seems to have reasonable characteristics (or more precisely some verbal description about why it's attractive) can yield different winners. This leaves us with the question of whether or not any of these methods is better than any of the others. Again we will come back to this issue after seeing a few other methods.

Here are a few other interesting ways to decide elections, which for convenience we assume have no indifference on the part of the voters for the candidates:

a. Add up votes for candidates who are either highest or second highest on the voters ordinal preference schedules.

In Election 1, A gets 12, B gets 10 + 9, C gets 9, and D gets 12 + 10 = 22, and wins.

b. Add up votes for candidates who are either at the median level or higher on the preference schedules of the voters.

(In Election 1 since there are 4 candidates, median or above means top two, and the result is the same as for a. above.)

c. Coombs' Method

This method due to the psychologist Clyde Coombs is especially interesting. In the run-off procedures we have discussed so far, the run-off was based on the number of votes that a candidate got for first place. Candidates were eliminated on the basis of having received few first place votes. In Coombs' method, candidates are eliminated on the basis of how many last place votes they got. Candidates with more last place votes are eliminated first!

The method works as follows. If no candidate has a majority of first place votes, eliminate that candidate with the largest number of last place votes. Check again, to see if any candidate has a majority and repeat the procedure.

Thus, in Election 1 above, no one got a majority. Since candidate B got the largest number of last place votes, B is eliminated. In the next round, A gets 12 votes, D gets 10, and C gets 9. No one again has a majority, so now candidate C is eliminated. Now there are 12 votes for A and 19 votes for D (since C and B are not available to the 9 voters who selected them earlier). Thus, D gets a majority and wins. The winner is again a candidate who got no first place votes.