A single bullet is randomly put into a 6-chamber revolver. 2 players (A and B) take turns pulling the trigger with the barrel pointed at themself. Each time the trigger is pulled, the barrel rotates to the next chamber (I think this is how revolvers work?). What is the probability of surviving if you're player A or B?
So my intuition tells me that each player has equal probability of surviving. Player $A$ will be assigned to chambers 1,3,5, and player $B$ will be assigned to chambers 2,4,6. The bullet is equally likely to be in any of these chambers, so each player has a 1/2 chance, I think.
I wanted to prove this with actual equations using conditional probability for practice, but it seems that I am doing something wrong in my work below.
There are 3 rounds, where a round is defined by players $A$ and $B$ pulling the trigger in sequence. Let $A_1, A_2, A_3, B_1, B_2, B_3$ denote the events that players A and B survive rounds 1,2,3.
First round
$$
P(A_1) = \frac{5}{6} \\
P(A_1^c) = \frac{1}{6} \\
P(B_1) = P(B_1|A_1)P(A_1) + P(B_1|A_1^c)P(A_1^c) \\
= \frac{4}{5}\frac{5}{6} + 1*\frac{1}{6} \\
= \frac{5}{6} = P(A_1)
$$
So players $A$ and $B$ have equal probability of surviving the first round.
Second round
$$
P(A_2) = P(A_2|A_1, B_1)P(A_1, B_1) + P(A_2|A_1, B_1^c)P(A_1, B_1^c) + \text{zero terms} \\
= \frac{3}{4}\frac{2}{3} + \frac{1}{6} = \frac{2}{3} \\
\\
P(B_2) = P(B_2|A_1, B_1, A_2)P(A_1,B_1,A_2) + P(B_2|A_1, B_1, A_2^c)P(A_1,B_1,A_2^c) + \text{zero terms} \\
= \frac{2}{3}\frac{1}{2} + 1*\frac{1}{6} = \frac{1}{2}
$$
Hmmm, the 2 players don't have equal probability of surviving the 2nd round.
Third round
$$
P(A_3) = P(A_3|A_1,B_1,A_2,B_2)P(A_1,B_1,A_2,B_2) +
P(A_3|A_1,B_1,A_2,B_2^c)P(A_1,B_1,A_2,B_2^c) + \text{zero terms} \\
= \frac{1}{2}\frac{1}{3} + 1*\frac{1}{6} = \frac{1}{3} \\
\\
P(B_3) = P(B_3|A_1,B_1,A_2,B_2,A_3)P(A_1,B_1,A_2,B_2,A_3) +
P(B_3|A_1,B_1,A_2,B_2,A_3^c)P(A_1,B_1,A_2,B_2,A_3^c) + \text{zero terms}\\
=P(B_3|A_1,B_1,A_2,B_2,A_3^c)P(A_1,B_1,A_2,B_2,A_3^c) \\
= \frac{1}{6}
$$
So it seems that something is wrong with my calculations in rounds 2 and 3 because the 2 probabilities aren't equal? In addition, how do I go about computing the probability that $A$ or $B$ survives all 3 rounds, i.e., $P(A_1, A_2, A_3)$ and $P(B_1, B_2, B_3)$. I understand 1 is the complement of the other, but I'm unsure how to get either one. It seems to calculate $P(A_1, A_2, A_3)$, I would have to do something like:
$$
P(A_1,A_2,A_3) + P(B_1,B_2,B_3) = 1 \\
P(A_1, A_2, A_3) = P(A_1, A_2, A_3|B_1^c)P(B_1^c) + P(A_1, A_2, A_3|B_2^c)P(B_2^c) + P(A_1, A_2, A_3|B_3^c)P(B_3^c) = P(B_1^c) + P(B_2^c) + P(B_3^c) = \frac{1}{6} + \frac{1}{2} + \frac{5}{6} > 1 \\
P(B_1, B_2, B_3) = P(B_1, B_2, B_3|A_1^c)P(A_1^c) + P(B_1, B_2, B_3|A_2^c)P(A_2^c) + P(B_1, B_2, B_3|A_3^c)P(A_3^c) = P(A_1^c) + P(A_2^c) + P(A_3^c) = \frac{1}{6} + \frac{1}{3} + \frac{2}{3} > 1 \\
$$
But this is obviously wrong. Is this the correct way to get $P(A_1, A_2, A_3)$ and $P(A_1, A_2, A_3)$?
Best Answer
For the second round, you must also include the term $$ P(B_2|A_1^c, B_1, A_2^c)P(A_1^c,B_1,A_2^c) = 1\cdot P(A_1^c) = \frac16, $$ which is not zero. When this term is included, $P(B_2) = P(A_2) = \frac23.$
In the third round, your calculation of $P(A_3)$ should include the term $$ P(A_3|A_1,B_1^c,A_2,B_2^c)P(A_1,B_1^c,A_2,B_2^c) =\frac16$$
while your calculation of $P(B_3)$ should include the terms $$ P(B_3|A_1,B_1,A_2,B_2,A_3^c)P(A_1,B_1,A_2,B_2,A_3^c) =\frac16$$ and $$ P(B_3|A_1,B_1,A_2,B_2,A_3^c)P(A_1,B_1,A_2,B_2,A_3^c) =\frac16.$$
When you include these missing terms, $P(A_3) = P(B_3) = \frac12,$ which is the correct probability that each player survives at the end.
Actually the probabilities "in real life" should be $P(A_3) = \frac13,$ $P(B_3) = \frac23,$ because if the first five chambers are empty then when $A$ hands the revolver to $B$ to try the final chamber, $B$ shoots $A$. Avid readers of MAD Magazine would know this. A variation on this theme was a plot element of The Deer Hunter.