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I place two bullets into a revolver with six cartridges in random positions. I spin it and shoot once; nothing comes out. If you do not want me to shoot, would you rather I pull the trigger again and shoot (thus, shooting the adjacent bullet) or spin the barrel again and shoot?

I know that if I re-spin, the probability of shooting is $2/6 = 1/3$. But, if I don't spin again, then I'm equally likely to be at one of the empty cartridges. But, the bullets don't necessarily have to be adjacent. I don't know what the probability is in this case. Can someone help me?

J. W. Tanner
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  • See also https://math.stackexchange.com/questions/849347/russian-roulette-should-a-player-pull-the-trigger-or-spin-the-cylinder?rq=1 – Michael Lugo Feb 07 '19 at 21:39

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If you spin the barrel again, you indeed have a probability of $\frac{2}{6} = \frac{1}{3}$ of shooting me in the next turn. If you don't spin again, the previous empty position cannot be reached, and should therefore be discarded. Since there are five slots left for two bullets, the probability becomes $\frac{2}{5}$. Therefore, I would like you to spin again!

jvdhooft
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