Comparison of Voting Systems: Plurality, Borda Count, Sequential Pairwise, and Hare System, Study notes of Mathematics

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Voting:

Issues, Problems, and

Systems, continued

Last time we discussed four voting systems, plurality voting, the Borda count, sequential pairwise voting, and the Hare system. We review each of these systems and recall the problems that each have.

A candidate is a Condorcet winner if he or she is the preferred candidate based on head to head competition with each candidate. A voting system satisfies the Condorcet winner criterion if the Condorcet winner, if there is one, always wins the election. We saw an example last time to show that plurality voting does not satisfy the Condorcet winner criterion.

The Borda Count

In the Borda count, voters rank order the candidates. If there are n candidates, each first place vote is worth n-1 points, each second place vote is worth n-2 points, and so on. The person who received the most points wins the election.

Sequential Pairwise

Voting

In this voting system, voters rank candidates as in the Borda count. The winner is determined by comparing pairs of candidates. The loser between a comparison is eliminated and the winner is then compared to the next candidate. The last person remaining is then the winner of the election. This system requires ordering the candidates to decide the order of comparison.

Pareto condition : if everybody prefers one candidate to another, then the latter is not elected. An example we considered last time shows that sequential pairwise voting does not satisfy this condition.

Monotonicity Criterion : if no voter were to switch his/her preference between the winner and another, then the outcome of the election would be the same. We saw that the Hare system does not satisfy this criterion.

The four systems we have considered, Plurality voting, the Borda count, Sequential Pairwise voting, and the Hare system, each violates at least one of the following conditions: the Condorcet winner condition, independence of irrelevant alternatives, the Pareto condition, and monotonicity.

How would the 1998

Minnesota

gubernatorial election

come out with other

systems?

To check this, let’s suppose the voters would rank the three candidates, Coleman, Humphreys, and Ventura, as follows. This table is not based on exit polls. I made up the table, trying to make a reasonable guess as to what would have happened. Recall that Ventura received 37% of the first place votes, while Coleman received 35% and Humphreys 28%. 20% 25% 8% 10% 20% 17% H C H C V V C H V V C H V V C H H C

H C H C V V

C H V V C H

V V C H H C

first place votes second place votes third place votes total points C 35 40 25 110 H 28 42 30 98 V 37 18 45 92

We computed Coleman’s totals by: 35 * 2 + 40 * 1 + 25 * 0 = 70 + 40 = 110. The other totals were computed similarly. Thus, with the Borda count, Coleman wins the election. first place votes second place votes third place votes total points C 35 40 25 110 H 28 42 30 98 V 37 18 45 92

Hare system: If we use the Hare system, then Humphreys is eliminated since he received the fewest first place votes, 28%, compared to 35% and 37% for the other two candidates. We then reinterpret the ballot by eliminating Humphreys. This gives 20% 25% 8% 10% 20% 17% C C V C V V V V C V C C

After reinterpreting the ballots, Coleman then has 55% first place votes to Ventura’s 45%, so Coleman would win under the Hare system. So, in all of the systems, other than plurality voting, Coleman would win the election.