For my state (Texas), a simple binomial model suggests my chance of being the critical vote is about 1 in 10^51121.
Are you sure? The simplest possible binomial model for Texas (which has 26 million people) gives a probability of 0.016% that exactly 13 million people will vote for each candidate, and therefore your vote will be critical. What exactly are you calculating?
Edit: I forgot to take into account the probability that Texas will matter to the election. But there are 2^49 < 10^15 equally likely outcomes for the other 49 states; if even one of them results in Texas being a pivotal state, then we only get a factor of 10^-15, which is still very far away from your estimate.
Edit: I guess you’re assuming that Texans have a roughly 55% chance of voting Republican, which does give roughly the probability in your comment (although at that point we can also take into account the number of people that voted in the last election, which brings the probability up to better than 10^-25000.) I guess that’s a reasonable thing to take into account; there are of course objections people could make, but I see where your math is coming from.
Edit: Here’s a fancier model. Let P be the probability that someone votes Republican in an election. We have a good estimate of what P was in previous elections; suppose the change in P is normally distributed. Then we find that with a 0.08% chance, P will be between 49.99% and 50.01%, at which point the correction factor from the 50% binomial distribution is negligible. Given that P is in this range, the chance your vote will be pivotal for Texas is better than 0.02%, so the overall probability is about 1 in 5 million. Of course, then there’s the probability that the Texas vote changes the election to take into account.
I expected Texas voter turnout for a Presidential election to be about 8,075,000. Assuming everyone votes either for Obama or Romney, averaging the polls gives a probability for each vote of about 0.415 for Obama and 0.595 for Romney. That story fits a binomial distribution, and my vote would be critical if the votes were split evenly.
binopdf(0.5*8075000,8075000,0.415) evaluated to approximately 10^-51120, and at that point I just upped the exponent one rather than trying to figure out the electoral college details.
Are you sure? The simplest possible binomial model for Texas (which has 26 million people) gives a probability of 0.016% that exactly 13 million people will vote for each candidate, and therefore your vote will be critical. What exactly are you calculating?
Edit: I forgot to take into account the probability that Texas will matter to the election. But there are 2^49 < 10^15 equally likely outcomes for the other 49 states; if even one of them results in Texas being a pivotal state, then we only get a factor of 10^-15, which is still very far away from your estimate.
Edit: I guess you’re assuming that Texans have a roughly 55% chance of voting Republican, which does give roughly the probability in your comment (although at that point we can also take into account the number of people that voted in the last election, which brings the probability up to better than 10^-25000.) I guess that’s a reasonable thing to take into account; there are of course objections people could make, but I see where your math is coming from.
Edit: Here’s a fancier model. Let P be the probability that someone votes Republican in an election. We have a good estimate of what P was in previous elections; suppose the change in P is normally distributed. Then we find that with a 0.08% chance, P will be between 49.99% and 50.01%, at which point the correction factor from the 50% binomial distribution is negligible. Given that P is in this range, the chance your vote will be pivotal for Texas is better than 0.02%, so the overall probability is about 1 in 5 million. Of course, then there’s the probability that the Texas vote changes the election to take into account.
I expected Texas voter turnout for a Presidential election to be about 8,075,000. Assuming everyone votes either for Obama or Romney, averaging the polls gives a probability for each vote of about 0.415 for Obama and 0.595 for Romney. That story fits a binomial distribution, and my vote would be critical if the votes were split evenly.
binopdf(0.5*8075000,8075000,0.415) evaluated to approximately 10^-51120, and at that point I just upped the exponent one rather than trying to figure out the electoral college details.