Given that we stop after the first failure technically this should be a negative binomial distribution (http://en.wikipedia.org/wiki/Negative_binomial_distribution) with p = 0.979 and r = 1. The mean is p / (1-p) = 46.62. Also there is a p ^ 75 = 0.204 chance that someone of same skill will achieve a better score.
Not true. Assuming both players move completely random (both choose either 1 or 0 with 50% probability each) and independently (past actions don't influence the current event) player 1 has a 2:1 advantage, i.e. he will win with probability 2/3. This is because player 1 can win after an even number of rounds (including 0) with 50% probability. So his total chance of winning is 1/2 * (1 + 1/4 + 1/16 + 1/64 + ...) = 1/2 * 4/3 = 2/3.
