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.
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.