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 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 $$(e^{-x^2})^2 = e^{-2x^2}$$ this is similar to a normal distribution so you can integrate it.

The approximate probability of equal scores after n tosses is $$1/\sqrt{n\pi}$$.