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 Quit While You're Ahead (Posted on 2003-11-10)
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?

 See The Solution Submitted by DJ Rating: 3.5556 (9 votes)

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 Solution | Comment 1 of 23
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.
 Posted by Popstar Dave on 2003-11-10 07:08:00

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