The Cornell Daily Sun published on its front page last week, the results of the student trustee election.

http://cornellsun.com/section/news/content/2009/04/22/students-elect-asa-craig-%E2%80%9911-serve-board-trustees

I quickly skimmed through the article, but what caught my eye was the paragraph:

“Out of the 3,423 ballots cast for the 10 candidates, Craig was ranked first on 651 of those ballots. Using the Hare-Clark system, the candidate with the least amount of votes is systematically eliminated. The eliminated candidates’ votes are then transferred to other candidates depending on the rankings of voters’ preferences. In the end, Craig defeated Raymond Mensah ’11 with a final count of 1,578 votes.”

What is the Hare-Clark system? I did a quick google search, and what turned out was a very in-depth article by Antony Green, an election commentator for ABC. (http://www.abc.net.au/elections/tas/2006/guide/hareclark.htm) (Caution! His article is extremely long!) But here’s a simpler example I found on Wikipedia which is fun and easy to read:

An example

Suppose a food election is conducted to determine what to serve at a party. There are 5 candidates, 3 of which will be chosen. The candidates are: Oranges, Pears, Chocolate, Strawberries, and Sweets. The 20 guests at the party have the preferences marked on their ballots in the two tables below (the first is a numerical representation, the second is pictorial). In the following table only some of the second preferences and none of the lower preferences are shown because they happen to not be needed in the count (a different set of votes could be constructed where first, second and third preferences of some voters must be considered).

# of Guests	x x x x	x x	x x x x x x x x	x x x x	x	x
1st Preference
2nd Preference

First, the quota is calculated. Using the Droop quota, with 20 voters and 3 winners to be found, the number of votes required to be elected is:

When ballots are counted the election proceeds as follows:

Candidate:
Round 1	x x x x	x x	x x x x x x x x x x x x	x	x	Round 1: Chocolate is declared elected, since Chocolate has more votes than the quota
Round 2	x x x x	x x	x x x x x x	x x x x x	x x x	Round 2: Chocolate’s surplus votes transfer proportionately to Strawberry and Sweets according to the Chocolate voters’ second choice preferences. However, even with the transfer of this surplus no candidate has reached the quota. Therefore Pear, who has the fewest votes, is eliminated.
Round 3	x x x x x x		x x x x x x	x x x x x	x x x	Round 3: Pear’s votes transfer to their second preference, Oranges, causing Orange to reach the quota and be elected. Orange barely meets the quota, and therefore has no surplus to transfer.
Round 4	x x x x x x		x x x x x x	x x x x x	x x x	Round 4: Neither of the remaining candidates meets the quota, so Sweets are eliminated. Strawberry is the only remaining candidate and so wins the final seat.

Result: The winners are Chocolate, Oranges and Strawberries. (Of course!)

There are several problems with this system of voting. This system of voting may not be fair because it is not monotonic. In class, we learned about how a fair election should be Pareto efficient. A non-monotonic voting system cannot be Pareto-efficient. See the example below for an explanation of how this system of voting might be unfair:

Suppose 100 people vote for 3 candidates, and a majority of 51 is needed to win the election. In the first election, votes are cast as follows:

Number of votes	1st Preference	2nd Preference
39	Andrea	Belinda
35	Belinda	Cynthia
26	Cynthia	Andrea

Cynthia is eliminated, thus transferring votes to Andrea, who is elected with a majority. She then serves a full term, and does such a good job that she persuades ten of Belinda’s supporters to change their votes to her at the next election.

This election looks thus:

Number of votes	1st Preference	2nd Preference
49	Andrea	Belinda
25	Belinda	Cynthia
26	Cynthia	Andrea

Because of the votes Belinda loses, she is eliminated first this time, and her second preferences are transferred to Cynthia, who now wins 51 to 49. In this case Andrea’s preferential ranking increased between elections - more electors put her first - but this increase in support appears to have caused her to lose. Counterintuitively, it was the increase in support for Andrea that hurt her. (Source: http://en.wikipedia.org/wiki/Monotonicity_criterion)

We learned in class that the only aggregation procedure that satisfies the Pareto Principle, Independence of Irrelevant Alternatives and produces a complete and transitive group ranking for all collections of individual ranks is dictatorship (the Arrow’s Impossibility Theorem). Thus, it seems inevitable that our system of voting for the student trustee election has certain exploitable flaws in it.

However, I feel that as a student who could have voted in this election, not enough emphasis was given to the fact that the voting system used was the Hare-Clark system. Any voting system can work well only if voters are educated sufficiently about the system of voting. Indeed, the Hare-Clark system which is still in use in Tasmania has been used since the 1900s, and so voters are familiar with the system, and so know well how they ought to vote.

Also, I wonder if it is fair to adopt this system of voting in the student trustee election, since, as I quote from Antony Green, “Hare-Clark works particularly well in Tasmania because of the state’s highly stable population, the clear regional divisions reflected in electoral boundaries and long experience of using Hare-Clark. Voting is based very strongly on personal knowledge of candidates.” Clearly, the student population at Cornell is not stable, and it would be fair to say that most people do not have much personal knowledge about many of the candidates. This could thus potentially lead to election results that do not reflect the wants of the students.