Probability having to do with “Russian roulette”

A shotgun has 6 slots and there is 1 bullet in it. The first person pulls the trigger and nothing was fired. The second person pulls the trigger without shuffling the chamber. What's probability that the second person dies?

Assume $P(A) = 1/6$ is the probability that the second person dies from firing the shotgun and $P(B)$ is the probability the first person doesn't die, so $P(B) = 5/6$. Then $P(B|A) = 1$ because the second person dies. Then

$$P(A|B) = \frac{P(B|A) P(A)}{P(B)} = 1/5$$

Is that right?

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Yes, that's correct. –  Chris Taylor Aug 26 '11 at 6:14
Do shotguns have slots? –  Gerry Myerson Aug 26 '11 at 6:16
Chiming in with Gerry :-). Something was lost in translation. Shotguns use shells (or rarely slugs), not bullets. I think you are talking about something that would be called a revolver, a six-shooter or just a gun as in Westerns. Also the minimum legal barrel length of a shotgun makes it impractical for Russian roulette - not that a bunch of folks playing that would necessarily care about the legality of the thing. –  Jyrki Lahtonen Aug 26 '11 at 7:03
Playing with a sawn-off shotgun would add to the frisson, by putting at more risk the players not currently holding the gun, and bystanders... –  slim Jan 4 '12 at 16:58

Yes, that's right. You could also find the same result by counting the number of slots left that might contain the bullet, each of which has equal probability to contain the bullet.

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