Loaded revolver puzzle.

Question

This was a puzzle asked in one of the interviews. It goes like : There are 3 consecutive bullets in a revolver barrel (total out of 6), so 3 are empty. Now you roll the barrel so you don't know which one is at the current position. You press the trigger and a bullet is fired. What's the probability of getting a bullet again if you press the trigger again? I think the answer to above should be 2/3 because only 2 out of 3 cases would lead to a bullet-firing scenario.

The follow up question was : If on pressing the trigger again, a bullet is fired, what's the probability of getting a 3rd bullet once the trigger is pressed again. My answer for the follow up is 1/2.

Would appreciate if someone could confirm and correct me (in case my answers are wrong). Thanks.

Answer 1

I agree with your reasoning. Let B denote an unfired round and X a fired round. The event which occurred was either XBB, BXB or BBX, all equally likely. In two of these events, namely XBB and BXB the second round will also be fired so the odds are $\frac{2}{3}$. When the second round is fired you know that the event was either XXB or BXX, both equally likely. So the odds that the third event will be XXX is $\frac{1}{2}$.

Answer 2

I believe your answer is correct yes. For two bullets to be consecutively fired there are only two positions you could have originally landed on. Either you landed on the first bullet out of the three consecutively lined up or you landed on the second bullet. Landing on any other position would mean two consecutive shots could not have been fired. From these two positions the next shot will either be a bullet or not so chance of getting shot is $\frac{1}{2}$.