Russian roulette should a player pull the trigger or spin the cylinder Two men plays Russian roulette. In revolver there are 2 bullets in consecutive chambers( 2 bullets are in 2 chambers next to each other).
One man spun the cylinder, pulled the trigger and he is fine, so the chamber was empty. What should the second man do: pull the trigger or spin the cylinder.
My answer: Pull the trigger
Explanation:
If the man spins the cylinder there is 2 full chambers with the bullet, 4 empty chambers. The probability of drawing empty chamber is $\frac{4}{6} = \frac{2}{3}$
If the man just pull the trigger we have to remember that after 3 empty chambers there is empty chamber and after 1 empty chamber there is full chamber so the probability of that the next chamber is empty is $\frac{3}{4}$
$\frac{3}{4} \gt \frac{2}{3}$
Question: Am I right?
 A: Let's model the gun has a list of 6 letters ABCDEF where EF have the bullets. The first person survives, so we must be in positions ABCD. Thus, there is a 1/4 chance of death if we pull the trigger because we can end up in positions BCDE and E is the only one that kills us. 
If we spin the wheel, then we have a 2/6=1/3 chance of death.
However, the strategic part (given that one of you must die, hurting your opponent is just as good as surviving for yourself) is that suppose I pull the trigger and survive. Now, I am in position CDE, so my opponent can now pick between a 1/3 chance (spinning) and a 1/3 chance (triggering). If he chooses to trigger again, the gun can only be in positions CD, so I have a 1/2 with trigger, and 1/3 with spinning.
Assuming optimal play for both parties, player one survives (otherwise, nothing interesting), player two triggers*, player one triggers (because spinning helps player two), player two spins (because 1/3 is better than 1/2), then player one triggers*.
Note that the * parts are the same with opposite players. So basically the roles reverse. I am also of the opinion that player one has an advantage in long term, but is probably overshadowed by immediate chance to die. I will think about that some other time. 
A: Your logic is correct, but your computation is incorrect; $\frac{4}{6}$ is not equal to $\frac{1}{3}$.
A: Just another way to look at this problem: when you pull the trigger and nothing happened you can conclude 2 things:


*

*You lost one 'good' spot (empty chamber) 

*You know that the 2nd consecutive chamber is irrelevant 


We know that if you spin the roulette your death probability is 1/3 
And if we factor 'pull the trigger' approach factored with the 2 items above you have 6-2=4 spots. So death probability is 1/4. 
Since you want a lower probability of death you should choose 'pull the trigger approach' 
