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?
As the chamber is spun after every turn the probability of a shot being fired on any turn is (1/6) the probability of a miss is (5/6).
Player 1 could lose on the first turn. This has a probability of (1/6).
If not, they could lose by missing on the first turn, then the second player also misses, then the first player loses on their second turn. This has a probability of (5/6)²(1/6).
If this still doesn't happen then they could lose by both players missing on their first two turns then player one loses on their third turn. This has a probability of (5/6)^4(1/6).
As this pattern continues indefinately, the probability of player one losing is the total of the infinite series:
(1/6) + (5/6)^2(1/6) + (5/6)^4(1/6) + ... + (5/6)^n(1/6)
This equates to 6/11 or a 54.545% chance of the first player losing.
Solution
