Re: probability of n ties in a race to three (roshambo)
The number of ties before a decisive throw follows a geometric distribution with mean 1/2. There is a 50% chance that a best-of-three-decisive-throws match will last 2 decisive throws, and a 50% chance that it lasts 3. So, the distribution is 50% of a convolution of 2 geometric distributions of mean 1/2 plus 50% of a convolution of 3 geometric distributions of mean 1/2.
P(n) = 1/2 (n+1 C 1) (1/3)^n (2/3)^(2) + 1/2 (n+2 C 2) (1/3)^n (2/3)^3.
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