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Let's say you play Russian roulette with a $6$-chamber gun and there is only one bullet in it. Your friend spins and pulls the trigger, they're still alive, and then they give the gun to you and you need to make a decision: spin or don't spin.

If you spin then your chances of dying are $1/6$. But if you don't spin then it's a bit more complicated:

If $B$ is the event that your friend survived, then $P(B) = 5/6$. Then assume $A$ is the event that you survive. There are $5$ chambers left and a bullet in one of them, so your chances of dying are $P(A \cap B) = 1/5$. Then $P(A|B) = (1/5)/(5/6) = 6/25 $.

So $1/6$ chance of death if you spin, $6/25$ chance of death if you don't spin. Therefore it's better if you spin.

Is that correct? I'm trying to find the solution with the best probability of surviving.

Greg Martin
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Jonathan
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3 Answers3

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In the don't-spin scenario: Your argument that there are 5 chambers left and a bullet in one of them after your friend survives is actually an argument that $P(A|B)=1/5$, not that $P(A\cap B)=1/5$. In fact, $P(A\cap B)=2/3$, since there is a 4 in 6 chance that the bullet is not in either of the first two chambers. (You also define $A$ to be the event that you survive, but your calculation treats $A$ as the event that you die.)

Greg Martin
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  • This answer, like the question, is inconsistent about what $A$ and $B$ represent. Note that by definition, $P(A\cap B) = P(A\mid B) P(B),$ so the statements $P(A\mid B)=1/5$ and $P(A\cap B)=2/3$ imply that $P(B) = 10/3.$ Something is mislabeled. – David K May 26 '25 at 15:15
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Maybe I am being too simplistic, but isn't it just
spin : $1/6$
don't spin : $1/5$

As long as you spin, there is always a chance you might survive. If you never spin, someone will eventually die $(1/6 \rightarrow 1/5 \rightarrow 1/4 \rightarrow ...\rightarrow 1/1)$. There is no doubts that spinning is the way to go.

Benji Altman
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stack reader
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Your best probability of surviving is not to play :).

As the other two answers imply, you're making it too complicated. If your friend survives, the bullet is in one of the remaining five chambers, with equal probability: each has probability $1/5$. If you don't spin, you die with probability $1/5$. If you spin, you're back to the initial distribution, where you die with probability $1/6$. Spin!

But if you want to use the probability notation, let's be precise. Let $B$ be the event that the bullet isn't initially in the first chamber. Let $A$ be the event that the bullet isn't initially in the second chamber. Then your friend survives with $P(B)=5/6$; and if your friend survives and you don't spin, then you survive with $$ P(A | B) = \frac{P(A \cap B)}{P(B)}=\frac{2/3}{5/6}=\frac{4}{5}. $$

mjqxxxx
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