Using the same estimate as the OP for n (1.56x10^23) gives p=3.02x10^-22. Still fantastically low.
Related question. If there are exactly 2N people who vote in a binary election (ie: for presidential candidates) and they have an even 50/50% chance of voting either way, how do I compute the odds that they will have a even split? This is a generous estimate for the probability my vote will matter.
Check out: http://en.wikipedia.org/wiki/Binomial_distribution
I think that this is way too low. Shouldn't it be the quite large number
1 - \prod_{i = 1}^n (1 - (i - 1)/52!)
(a la the birthday paradox)?
Mathematica overflowed when I tried to compute this by brute force. The next best thing I can think of is to use the exponential approximation
1 - x ≈ e^{-x},
\prod_{i = 1}^n (1 - (i - 1)/52!)
≈ \prod_{i = 1}^n e^{-(i - 1)/52!}
= e^{-n(n - 1)/52!}
≈ 1 - n(n - 1)/52!.
\frac1 2\sum_{i = 1}^n [(i - 1)/52!]^2
≈ n^3/(2(52!)^2),