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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.2011-08-26
  • 9
    Do shotguns have slots?2011-08-26
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    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.2011-08-26
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    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...2012-01-04

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