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AlephZero
AlephZero is offline
#8
May30-07, 07:12 AM
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Quote Quote by chroot View Post
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].