probability of n ties in a race to three (roshambo)

[pzhon]

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.

[Siegmund]

You can replace the phrase "convolution of n geometric distributions" with "negative binomial distribution" (and "with mean 1/2" by "with probability of success 2/3") to make it look a little less scary.

But it's the same calculation.