# [Math] Russian Roulette and conditional probability

bayes-theoremprobability

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, he's still alive, and then he gives 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.

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.)