[gregable]

Nice, I came to the comments to see if someone knew how to calculate this.

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.

[wnoise]

The exact answer is (2n choose n) * (1/2)^(2n). This is approximately sqrt(1/Pi n) as n grows large, with error O(n^(-3/2))

[flourpower]

That's the same as the probability that 2N flips of a fair coin result in exactly heads: (2N choose N)(.5^N)(.5^N).

Check out: http://en.wikipedia.org/wiki/Binomial_distribution