A Russian gangster kidnaps you. He puts two bullets in consecutive order in an empty six-round revolver, spins it, points it at your head and shoots. *click* You're still alive. He then asks you, do you want me to spin it again and fire or pull the trigger again. For each option, what is the probability that you'll be shot?

1/4 for pulling the trigger. 1/3 for spinning and pulling the trigger.

Consider where the bullets could be so they are consecutive: (BBxxxx), (xBBxxx), (xxBBxx), (xxxBBx), (xxxxBB), (BxxxxB). You have a 2/3 chance of getting shot on pull number 1. If you respin you still have that same chance, assuming the spin is random. Now assume the first was a blank. Only 4 of the cases above will work, and in one of them you'll get shot on a second pull. So your chances of getting shot are 1/4 on the second pull without a respin Don't respin! The two consecutive bullets makes this an interesting case, but what it means is that after a first blank you know *neither* bullet was in that position. This tips the probabilities just enough to make the right answer counterintuitive. The answer is just the opposite if there is only 1 bullet. In that case you should respin.

Hmmm.. I think the odds are 2/5 for pulling the trigger or 2/6 (which is the same as 1/3) for spinning and then pulling. There's 5 chambers left and 2 bullets, where would 1/4 come from?

'1/4 for pulling the trigger. 1/3 for spinning and pulling the trigger.' =candidate is right Probabilities for 'next shot'=kill 1/4 (no spin) or 3/12 2/6 (spin) or 4/12 or 1/3 so, for (no spin) the probability of getting shot is lower (1/4, because first pull = empty, means he hit 1 of 4 empty ones. As he puts the bullets in consecutive orders, 3/4 empty will be followed by another empty one. 1/4 will be followed by a bullet) (2/6, if he spins, the probability stays the same, because he can hit again the same empty one)

First pull is blank. Since the two bullets have to be in order, out of the five slots left, there are only four ways to arrange the bullets. Only if the next two are bullets will you be shot.

There are only 4 ways to arrange the revolver with consecutive bullets so that the first shot is nothing. Out of those 4 there is only one way to arrange the revolver so the second shot kills. So for the 2 consecutive hammer pulls(granted the first trigger pull is empty) you have a 1/4 chance of getting shot. As for pulling the trigger and spinning the chamber and pulling again. That is a selection with replacement. The likelihood that the second shot will fire is the same as the likelihood that the first shot will fire. 2/6 or 1/3. Now the likelihood that you would pull the trigger on an empty chamber, spin chamber, pull the trigger on a loaded chamber is (4/6)*(2/6)=2/9

1/4 for pulling the trigger again and 1/2 for spinning. You will have more chances in case of spinning.

How about asking the gangster why he kidnapped you in the first place and what he wants? If killing me is the answer, well, then just do it. Don't mess around and then try to get some DNA on you so if and when they find your body, they have evidence. If they come back with some weird response, talk your way out of it. I'd rather see if I can work my way out of that vs betting on a Russian's weapon and my inherent math skills.

The odds are 1 in 3 for spinning it again and pulling the trigger. The odds are 1 in 6 for pulling the trigger again without spinning it. So, of course, it is "better" odds that he should have him pull the trigger again without spinning it.

This strikes me as being very similar to the Monty Hall Problem. Here's my attempt to answer this question (would love any feedback on whether I'm right). The fact that the two bullets are in adjacent chambers is important. It lets you know that you can rephrase the problem in the following way: "Assuming the gangster just fired one shot and the chamber was empty, what are the odds that the current chamber will also be empty?" This is like saying "What are the odds that the previous shot was *not* fired from the final chamber in a series of empty chambers?" Let's calculate those odds now. There were originally 2 adjacent chambers *with* bullets, and 4 adjacent chambers *without* bullets. We know that the shot which was just fired had to be one of the 4 *without* bullets, which means there are now 3 chambers without bullets and 2 with bullets. To rephrase the problem yet again, what are the odds that the next chamber does *not* have a bullet, given the last chamber did not have one? Furthermore, are those odds better than the odds of a random re-shuffling of the chambers, considering the current relative odds of each scenario? We know there were originally 4 adjacent empty chambers followed by a non-empty one, and we used one of those empties. This means the likelihood of the current chamber being empty are 3/1 if we *don't* spin. Next, let's calculate what the odds would be if we *do* spin the chamber. For this, let's assume that if we re-spin the chamber, our odds of landing on another empty chamber are random depending on the relative quantity of empty-to-full chambers (i.e. 3 empty / 2 full). In reality, the randomness will be highly dependent on how many rotations it makes. But for this example, let's ignore that. Since 3/1 odds are better than 3/2 odds, I recommend *not* spinning the chamber.

It matters how the bullets are loaded which is why the question stipulates that they are in adjacent chambers. If the bullets are not adjacent it is better to spin. Spinning gives a 2/6 chance of being shot and not spinning gives a 2/4 chance. With adjacent bullets, spinning produces the same odds as above, 2/6, but not spinning improves the odds to 1/4. Interview candidate is right.

2/3 for pulling the trigger again and 1/2 for spinning. You will have more chances in case of spinning.