"We are to play a version of Russian Roulette, the revolver is a standard six shooter but I will put one bullet in the gun. I spin the chambers, put the gun to my head, pull the trigger and survive. I hand you the gun and give you a choice... You may put the gun straight to your head and pull the trigger, or you may re-spin the gun before you do the same.
What is your choice and why? "
My question here is the probability of shooting myself if I don't randomly spin. The answer for this seems to be 1/5 as mentioned here.
The explanation for the same sounds fair. But, I have a different approach which gives a different answer.
I have following cases:
1. B-----
2. -B----
3. --B---
4. ---B--
5. ----B-
6. -----B
Now, I will shoot myself if the second case occurs i.e the first trigger doesn't do anything and the second trigger fires. Also, the probability of that case occurring is 1/6. So, shouldn't the probability be 1/6 instead of 1/5.
Coloring blue the chamber used by the first person and red the following chamber which would be used by the second person if it were not spun again, $O$'s for empty chambers and $X$'s for filled chambers, when initially spinning the chambers the following six outcomes are indeed equally likely:
We are told that the first player just grabbed the gun and pulled the trigger. The first player survives and then hands the gun off to the second player. We are asked... given this information what is the probability that the second player will shoot themselves.
Well... since we are told that the first player survived... we know that the original state of the revolver was not $\color{blue}{X}\color{red}{O}OOOO$ as if it were this case then the first player would have shot themselves. We can imagine then that this was not included in the list. The remaining five outcomes are still equally likely to have occurred, only one of which result in us shooting ourselves.
From the comments "Let B be the event that my partner didn't shoot himself, and let A be the event that I shoot myself. Then P(A|B)=1/5. Got it. But, the probability of case 2 occurring should be 1/6 without any conditions i.e P(A∩B)=1/6, right ? Although, I will have to compare 1/6 with 5/36 (non conditional prob) instead of 1/5 (conditional prob), right ?"
Do not lose sight of what the problem asked. The problem asks whether you should spin again or not spin. The implication being that we want to avoid shooting ourselves and the game will end immediately after the next time the trigger is pulled.
If we weren't to spin again, we will shoot ourselves with probability $Pr(A\mid B) = \frac{1}{5}$. If we were to spin again we will shoot ourselves with probability $\frac{1}{6}$ which is less than $\frac{1}{5}$. Since spinning again reduces the chance that we will shoot ourselves this is what we should choose to do (assuming that the spins were truly random). End of story.