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.
Yes, and Sqrt(n) (times a small constant) is a good approximation to the above for large n. See Stirling’s formula.