Quote by chroot
This is not a trivial sum to compute.

You can get an approximate answer by replacing the binomial distribution by a normal distribution. The pdf of the "two scores are equal" distribution is the square of the pdf of the normal distribution.
Since [tex](e^{x^2})^2 = e^{2x^2}[/tex] this is similar to a normal distribution so you can integrate it.
The approximate probability of equal scores after n tosses is [tex]1/\sqrt{n\pi}[/tex].