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?

(In reply to Answer by K Sengupta)

At the outset, we observe that the probability that the first man loses on the first shot is 1/6, while the probability that he does not lose out on the  first shot is 1 - 1/6 = 5/6

In the event of the latter case, the chamber is respun and the probability of a loss at the first shot for the second man is 1/6, and the probability that he will lose out is 5/6.

Accordingly, the probability that the first man takes a second shot is (5/6)*(5/6) =25/36

This sequence will repeat indefinitely and will terminate as soon as one of the man loses out on a shot.  

Let the probability that the first man will lose be w.

Then, we must have:

(25/36)*w + 1/6 = w
or, 1/6 = 1 - (25/36)*w
or, 1/6 = 11*w/36
or, w = (1/6)*(36/11) = 6/11

Consequently, the required probability is 6/11

Edited on April 14, 2008, 4:38 pm

Posted by K Sengupta on 2008-04-14 16:24:41