Election sums. posted by Richard Seymour

Guest post by 'christian h':

Those pesky preference votes...

First, the conclusion: A Ken (1st) - Third Party (2nd) vote is mathematically equivalent to a Third Party (1st) - Ken (2nd) vote as far as the eventual outcome is concerned. However, the Third Party (1st) - Ken (2nd) option is preferable as any second preference votes of Third Party won't be counted.

Here's how it works.

It's election time in Londongrad. Running for mayor are Dr. Evil (Boris, to his admirers), Mini Me (aka Ken) and Austin Powers (fusion of left-of-Ken candidates). The voting system is "preference voting", also known as "instant run-off voting." Voter X really prefers Austin Powers, but she absolutely doesn't want Dr. Evil to win. What should she do?

The system works as follows: every voter can assign two votes, a 1st and 2nd preference vote. It is legal to leave the 2nd preference blank, but it isn't possible to only vote for a 2nd preference. The votes have to go to different candidates. After polls close, all 1st preference votes are counted. If one candidate obtains more than 50% of those - that is, more votes than both his opponents together - he is declared the winner. Otherwise, all 1st preference votes of the candidate with the fewest votes are discarded and the corresponding 2nd preference votes counted instead. since only two candidates are left in the race, one of them now is guaranteed to win.

Assume there are 100 eligible voters in Londongrad. On election day, 78 come out to vote - the rest are watching football. Of those 78, 40 vote Dr. Evil 1st preference, 37 vote Mini Me, and one (that's X) votes Austin Powers. The next day, the website socialistsplitters.com accuses X of throwing the vote to Dr. Evil. Are they right? No. If X had voted Mini Me 1st preference, Dr. Evil still had 40 votes - more than half. In formulas, if E, M and A denote 1st preference votes for Dr. Evil, Mini Me and Austin respectively, Dr. Evil wins outright if and only if E > A + M. Only the sum of A and M matters, not the individual totals.

... phew, bad dream! Turns out, Dr. Evil got only 38 of the 1st preference votes, Mini Me 37, and Austin Powers got 3 (one of them cast by X). Now Austin has the fewest 1st preference votes, so they are discarded; instead, the 2nd preference votes on those ballots are now added to the totals of Dr. Evil and Mini Me. If at least two Austin-voters did their duty and voted Mini Me with their 2nd preference, Mini Me has 37+2 = 39 total votes to Dr. Evil's 38, and wins. Only if X and her comrades inexplicably decided to leave 2nd preference blank will Dr. Evil walk away victorious.

In formulas, if e, m and denote second preference votes of the Austin Powers voters, for Dr. Evil and Mini Me respectively, Dr. Evil will now win if E + e > M + m.

To recap, Dr. Evil will win if

(a) either E > M + A

(b) or M + A >= E and E + e > M + m. and E + e > M + m.

Since M + A is at least M + m and no Austin voter will vote Dr. Evil with 2nd preference (that is, e = 0), this simplifies to give that Dr. Evil will win if and only if E > M + m. That is, the only number that matters is M + m - a 2nd preference vote for Mini Me is equally as good as a 1st preference vote.

Labels: gla elections, left list, london, mayor, respect