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electoral calculus
Nov. 3rd, 2014 11:04 amIf you're like me, you may be wondering whether to vote for a third party candidate who will almost certainly lose, or for the lesser of two evils. One factor I'm considering for this is the likelihood that mine is the deciding vote. If the chance of this is better than 1/n, where n is the number of voters in the election, then it seems to me that I'm better off voting for the lesser of two evils. If it's worse, then I might as well vote my conscience.
The statistics behind this comes from modeling the behavior of the voters, and from modeling the accuracy of the polls. It turns out that the latter effect is vastly more significant than the former, and roughly speaking if a tie vote is within the 90% error bars for the aggregate polling then your undecided vote carries more than average weight in the head-to-head race, and if it's outside the error bars then it carries less.
I was a little surprised to see that Nate Silver was giving much chance of winning to the particular lesser evil I'm considering. But as long as that's true, I think the chances of a one vote margin are too high to ignore.
The statistics behind this comes from modeling the behavior of the voters, and from modeling the accuracy of the polls. It turns out that the latter effect is vastly more significant than the former, and roughly speaking if a tie vote is within the 90% error bars for the aggregate polling then your undecided vote carries more than average weight in the head-to-head race, and if it's outside the error bars then it carries less.
I was a little surprised to see that Nate Silver was giving much chance of winning to the particular lesser evil I'm considering. But as long as that's true, I think the chances of a one vote margin are too high to ignore.
