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.
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.