Here is my answer to the puzzle that was posted here.

Label the chambers {A,B,C,D,E,F}.

Let chambers {A,B} contain bullets.

Spin the barrel. Our expectation distribution of the current chamber is X={1/6, 1/6, 1/6, 1/6, 1/6, 1/6}.

Fire the gun. We get an observation z that the chamber was empty. We want to update our expectation to P(X|z).

P(X|z) = P(X)*P(z|X)

(where the ‘equals’ sign really means proportional)

P(z|X) is the likelihood function that we would observe our specific value of z for each value of X.

P(z|x) = {0, 0, 1, 1, 1, 1}

P(X|z) = (0, 0, 1/4, 1/4, 1/4, 1/4}

With this belief distribution we can clearly see that there is only one possible chamber that is adjacent to a bullet. The chance we are on that chamber is 1/4. If they spin, then chance we land on a bullet is 2/6=1/3. Since 1/4 < 1/3 you should ask that they just pull the trigger again.