This seems counterintuitive, because we imagine that the votes of everyone else are “locked in” somehow, and that we’re only deciding whether to add ours to the pile- in which case, the only way that it could matter is in the event that it makes or breaks an exact tie.
Furthermore, even if you think of everybody else’s vote as locked in the odds of you being the deciding vote (assuming everybody else votes randomly and you don’t know how) is 1:sqrt{N} which is a lot better than your odds in Erewhon. Furthermore, in a close election with partisan voters and undecideds your odds are even better.
How did you calculate those odds? Shouldn’t it be the binomial distribution where n=number of voters, k=n/2, and p=.5, meaning the probability of you being the deciding vote is (.5^n)*(n choose n/2). Still better than Erewhon, though.
I’m not particularly confident that I’m right about this though, so please explain if I’m wrong.
Shouldn’t it be the binomial distribution where n=number of voters, k=n/2, and p=.5, meaning the probability of you being the deciding vote is (.5^n)*(n choose n/2).
Yes, and Sqrt(n) (times a small constant) is a good approximation to the above for large n. See Stirling’s formula.
Furthermore, even if you think of everybody else’s vote as locked in the odds of you being the deciding vote (assuming everybody else votes randomly and you don’t know how) is 1:sqrt{N} which is a lot better than your odds in Erewhon. Furthermore, in a close election with partisan voters and undecideds your odds are even better.
How did you calculate those odds? Shouldn’t it be the binomial distribution where n=number of voters, k=n/2, and p=.5, meaning the probability of you being the deciding vote is (.5^n)*(n choose n/2). Still better than Erewhon, though.
I’m not particularly confident that I’m right about this though, so please explain if I’m wrong.
Yes, and Sqrt(n) (times a small constant) is a good approximation to the above for large n. See Stirling’s formula.