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2 prisoners are on a death row. The warden gives them a chance to live. He shows them a gun that has a cylindrical barrel with slots for bullets (kind of like a six-shooter in old western movies except this gun has more bullet slots). He will slip a bullet into a random position in the gun barrel, and then have the prisoners take turn shooting at each other using the same gun. As you know, the first shot is most likely a blank, since there's only 1 bullet in the gun, thus the chance that the bullets fires on the first shot is low. However, as the prisoners take turn pulling the trigger, the chance that the bullet fires goes up accordingly. The person still standing after the gun fires wins his freedom. If the bullet fires, assume the prisoner the gun is aimed at will die.

On the day of the game, the warden show them two guns- one with 10 bullet slots, and one with 11 bullet slots. He then says that the prisoners will pick from the following two options. If a prisoner picks one option, the other options falls to his fellow prisoner by default.

A) pick a gun (10-bullet slot or 11 bullet slot ) to use during the game

B) choose to go first or second.

1) Suppose that you are prisoner 1, and you are required to pick first. What option do you pick, A or B, and what choices within that option?

2) Suppose that you are prisoner 2, and you get to pick second. Your son-of-a-gun prison-mate made the same choices as the answer in part 1, what would you pick now?

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Am I way off here? For me this seems reasonable. Feedback please.

I still believe this is easier than it seems. Let me explain.

On the left side of the gun you see the cylinder and on the right side of the gun you see the

cylindrical barrel still being just a regular barrel. But in the riddle it is written cylindrical barrel

to confuse the readers.

handheld%20gun.jpg

With that being said. The prisoner should always choose the options leading to his fellow prisoner being shot first, because the gun will always shoot the first time, since the bullet is in the cylindrical barrel, still being just a regular barrel.

If I read your answer right, you're saying that the OP is really a trick question, where 10-shot and 11-shot don't matter. You never know with these puzzlemakers, it could be :-)

Personally, I suspect that the OP was intended to be a real question, where 10-shot vs. 11-shot might matter as well as who shoots first. So I interpreted it as meaning the warden puts a bullet in a chamber, spins the chambers (the cylinder, I guess), and hands the gun to the first shooter.

Therefore, I'm convinced by chicory that going first with the 11-shot is best, and going last with the 11-shot is worst, with any position with the 10-shot being intermediate. So, if I could choose first, I'd avoid the 11-shot (as my opponent would choose to shoot first). I'd either choose the 10-shot, and guarantee the intermediate chances, or choose to go first (and hope my opponent would accidentally choose the 11-shot).

Edited by CaptainEd
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Am I way off here? For me this seems reasonable. Feedback please.

I still believe this is easier than it seems. Let me explain.

On the left side of the gun you see the cylinder and on the right side of the gun you see the

cylindrical barrel still being just a regular barrel. But in the riddle it is written cylindrical barrel

to confuse the readers.

handheld%20gun.jpg

With that being said. The prisoner should always choose the options leading to his fellow prisoner being shot first, because the gun will always shoot the first time, since the bullet is in the cylindrical barrel, still being just a regular barrel.

I meant for the puzzle to be a statistical exercise (i.e. chance of dying the same for A and B on 10-shot gun). But good job for thinking outside the box. I know much more about statistics than I do about gun, as your post just showed.

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