Two men are playing Russian roulette using a pistol with six chambers.
A single bullet is used and the chamber is spun after every turn.

What is the probability that the first man will lose?

The probability that the first man will lose is 54/99. To get this answer you must sum an infinite series. The first term is 1/6, the probability that he will lose on the first round. The next term will be (5/6)*(5/6)*(1/6), the probability that he will not lose on his first shot times the probability that his opponent will not lose on his first shot times the probablity that he will lose on his second shot.
Continuing in this way, his overall chances of losing are 1/6 + (5/6)^2*(1/6) + (5/6)^4*(1/6) + . . . + (5/6)^(n-1)*(1/6) = 54/99.
He has a slightly greater chance of losing than his opponent as he must take his chances first. This assumes that one man must lose and that they do not quit after one round.

Posted by wonshot on 2003-11-10 10:25:44