Jane estimates the probability of her vote tilting the presidential election at 1 in 1,000,000; Eric estimates the probability of his vote tilting the presidential election at 1 in 100,000,000. I find both of these estimates orders of magnitude too low.
Eric presumably is modeling the election by saying that with 100,000,000 voters (besides himself), there are 100,000,001 outcomes of their votes, only one of which is a tie which his vote will break. But his conclusion that the odds of deciding the election are about 1 in 100,000,000 assumes that all of these outcomes are equally probable, which is a hard-to-defend assumption.
If every other voter is flipping a fair coin to determine their vote, for example, then the probability of a tied vote is exactly 100,000,000! / [ (50,000,000!)^2 * 2^100,000,000], which is approximately 1⁄12,500. Moreover, I estimate that a solid 40% of the voters will vote Republican no matter what, and a solid 40% will vote Democrat no matter what. If the other 20,000,000 voters flip their fair coins, now the probability of a tied vote is approximately 1⁄5,600.
This model is oversimplified, of course, because factors that tend to bias individual votes (such as the current economy) will tend to bias many votes in the same direction. Still, I am much more confident in a 1-in-10,000 chance to affect the presidential election outcome than I am in 1-in-100,000,000.
I also agree with Kaj’s comment that my vote influences other people to vote, which would make the odds of affecting the outcome better still.
What is the probability your vote will make a difference? seems to be the state-of-the-art in the “deciding vote” type of reasoning. It concludes “On average, a voter in America had a 1 in 60 million chance of being decisive in the presidential election.”
I’d need to read it again, with pen and paper, to gain an understanding of why the Student-t distribution is the right thing to compute. At the very least I can say this: the probability of one’s vote tilting the election is certainly higher in very close elections (as measured beforehand by polls, say) than in an election such as Obama-McCain 2008. The article you quoted suggests the difference in probabilities is much higher than I anticipated. (Unless my calculation, which models the closest possible election, is incorrect.)
Edited to add: Okay, I’ve incorporated the probability p that the coin lands heads into the calculation. Even when p=50.05% instead of 50% (closer than any presidential election since Garfield/Hancock), the chance of one vote tilting the election drops by over four orders of magnitude. So for practical purposes, my initial calculation is irrelevant. - At least this was a good lesson in bias: this argument was easy to find, once Wei’s comment got me to consider the alternative in the first place.
Jane estimates the probability of her vote tilting the presidential election at 1 in 1,000,000; Eric estimates the probability of his vote tilting the presidential election at 1 in 100,000,000. I find both of these estimates orders of magnitude too low.
Eric presumably is modeling the election by saying that with 100,000,000 voters (besides himself), there are 100,000,001 outcomes of their votes, only one of which is a tie which his vote will break. But his conclusion that the odds of deciding the election are about 1 in 100,000,000 assumes that all of these outcomes are equally probable, which is a hard-to-defend assumption.
If every other voter is flipping a fair coin to determine their vote, for example, then the probability of a tied vote is exactly 100,000,000! / [ (50,000,000!)^2 * 2^100,000,000], which is approximately 1⁄12,500. Moreover, I estimate that a solid 40% of the voters will vote Republican no matter what, and a solid 40% will vote Democrat no matter what. If the other 20,000,000 voters flip their fair coins, now the probability of a tied vote is approximately 1⁄5,600.
This model is oversimplified, of course, because factors that tend to bias individual votes (such as the current economy) will tend to bias many votes in the same direction. Still, I am much more confident in a 1-in-10,000 chance to affect the presidential election outcome than I am in 1-in-100,000,000.
I also agree with Kaj’s comment that my vote influences other people to vote, which would make the odds of affecting the outcome better still.
What is the probability your vote will make a difference? seems to be the state-of-the-art in the “deciding vote” type of reasoning. It concludes “On average, a voter in America had a 1 in 60 million chance of being decisive in the presidential election.”
I’d need to read it again, with pen and paper, to gain an understanding of why the Student-t distribution is the right thing to compute. At the very least I can say this: the probability of one’s vote tilting the election is certainly higher in very close elections (as measured beforehand by polls, say) than in an election such as Obama-McCain 2008. The article you quoted suggests the difference in probabilities is much higher than I anticipated. (Unless my calculation, which models the closest possible election, is incorrect.)
Edited to add: Okay, I’ve incorporated the probability p that the coin lands heads into the calculation. Even when p=50.05% instead of 50% (closer than any presidential election since Garfield/Hancock), the chance of one vote tilting the election drops by over four orders of magnitude. So for practical purposes, my initial calculation is irrelevant. - At least this was a good lesson in bias: this argument was easy to find, once Wei’s comment got me to consider the alternative in the first place.