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## Question

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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First, the prob of each turn containing a bullet is equal in both pistols.

example, with a 10 slot pistol, prob that 1st slot has bullet = prob that 1st slot doesnot have bullet but 2nd does = prob that 1 and 2 do not have bullets but third does.... in this case, it doesnt matter who goes first; they both have equal prob of living.

In the 11 slot pistol, the prob for bullet in each slot is the same but the one who goes first goes 6 times, while the second goes only 5 times; thus giving the second one an advantage.

So, for first question, I would choose to go second hoping that the other prisoner makes the mistake of choosing 11 slot pistol

And for the second question, if the other person has chosen to go second, I choose the 10 slot pistol

However, if the other person chooses to go first, I choose 11 slot pistol

If the other prisoner chooses 10 slot pistol, I can choose to go first or second; it doesnt make a difference

If the other person chooses to 11 slot pistol first, i choose to go second

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Player should choose 10 slot pistol and shoot first.

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In the 11 slot pistol, the prob for bullet in each slot is the same but the one who goes first goes 6 times, while the second goes only 5 times; thus giving the second one an advantage.

Assuming that "goes" means "pulls the trigger" isn't it just the opposite?

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I would rather have the 10 shot pistol, and be shot at first, because of the likelyhood of the first shot being a blank. After that you have a 4/9 chance of being hit, while your opponent has a 5/9 chance of being hit.

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[spoiler='if i understand the way you have been solving these

']the 10 bullet is 1/10 1st shot

1/9 2nd shot

11 bullet gun is 1/11 1st shot

1/10 2nd shot

and so forth.

The difference is that

pris A10 1/10,1/8, 1/6, 1/4, 1/2

Pris B10 1/9,1/7,1/5,1/3,1/1

11 shot

Pris A11 1/11, 1/9,1/7, 1/5,1/3,1/1 total of 6 shots

Pris B11 1/10,1/8, 1/6, 1/4, 1/2 total of 5 shots

But do you have to allow for probability that a previous chamber was empty? That's been and gone, matching bullets to heads seems different than matching hats to heads... isn't it just a straight 1/(whatever chambers have not yet fired)?

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To me it seems that A10 and A11 are both slightly worse off for the first 5 shots, but then A11 gets that extra shot which is a certainty if he makes it that far. Oh, is that why you have to allow for empty chambers? But would that only apply to the case of A11?

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you would want to shoot first because then you will always have a lower chance of being shot than your opponent.

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you would want to shoot first because then you will always have a lower chance of being shot than your opponent.

But if A10 has a 1/10 chance of hitting, then B has a 1/9 chance of hitting, isn't that A10(first to shoot) having the higher chance of being shot?

A11 is more confusing because of that extra shot.

Edited by chicory
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10 shot and point it at the warden. Maybe I'm just old fashioned though...

*hides*

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I don't think either question matters, and it doesn't matter which choice you make.

10-shot revolver (OP says it's like in a western. That means it's a revolver). When you get to shoot, either you have the even cylinders or you have the odd cylinders. it's 50-50 whether your cylinders have the bullet or not. Doesn't matter whether you shoot first.

11-shot revolver. Once again, the cylinder will either appear when your opponent shoots, or when you shoot. It's 50-50.

The only thing that does matter is the random choice of initial cylinder. You don't get a choice that affects that.

edit: end the game sooner, pick the 10-shot gun.

Edited by CaptainEd
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The gun doesn't matter. Just shoot first to win. Sorry second choice. They are not shooting simultaneously, so the first shooter always has the odds (he gets to shoot once before you, twice before your two, etc...) Here's the math. Hope I didn't make mistakes

Edited by bekabeh
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I take back what I said, take the 11 shot pistol and shoot first.

Like someone above said, once they put the bullet in the 10 shot pistol, spin the little bullet wheel thing, and hand the gun to someone, at that point the bullet has a 50/50 chance to be in an even chamber or an odd chamber. Shooting first or second does not change the odds b/c there is only 1 bullet and you are only using 1 gun. The odds of that bullet being in an "odd" chamber or "even" chamber does not change with who shoots first.

But the 11 shot pistol has 1 extra chamber, so 1 player gets to shoot 6 times and the other shoots 5. Regardless of the odds at each step of the game, the bullet is in 1 out of 11 chambers, and the prisoner to shoot first gets 6 chambers, and the other prisoner gets 5, which gives the first shooter a slight advantage.

Edited by genmaximus2000
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Assuming the barrels aren't spun after each shot, I believe the solution is calculated from the odds of NOT being shot

Clearly whoever shoots first will end up with the 11 shot gun (or vice versa)

1st round

1st shot odds of B not being killed is 10/11

2nd shot odds of A not being killed is 9/10

2nd round

1st shot odds of B not being killed is 10/11 * 9/10 = .818

2nd shot odds of A not being killed is 9/10 * 8/9 = .800

3rd round

1st shot odds of B NOT being killed is .818 * 8/9 = .727

2nd shot odda of A NOT being killed is .8 * 7/8 = .700

Summary of *** survival odds

--------1st shot--2nd shot

Rnd----11 slot-----10 slot

1--------.909---------.9

2--------.818---------.8

3--------.727---------.7

4--------.636---------.6

5--------.545---------.5

6--------.454---------.4

Therefore, the 2nd shooter has a 50/50 chance of killing the first shooter by the 5th round after following the first shooter's .455 chance of killing him by that round

This suggests that given a choice, the first should be to select the 10 chambered gun

Edited by Simon Legree
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I take back what I said, take the 11 shot pistol and shoot first.

Like someone above said, once they put the bullet in the 10 shot pistol, spin the little bullet wheel thing, and hand the gun to someone, at that point the bullet has a 50/50 chance to be in an even chamber or an odd chamber. Shooting first or second does not change the odds b/c there is only 1 bullet and you are only using 1 gun. The odds of that bullet being in an "odd" chamber or "even" chamber does not change with who shoots first.

But the 11 shot pistol has 1 extra chamber, so 1 player gets to shoot 6 times and the other shoots 5. Regardless of the odds at each step of the game, the bullet is in 1 out of 11 chambers, and the prisoner to shoot first gets 6 chambers, and the other prisoner gets 5, which gives the first shooter a slight advantage.

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Just a thought...

The first shot is always the safest, 10 or 11. So No shouldn't you let your opponent start?

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Just a thought...

The first shot is always the safest, 10 or 11. So No shouldn't you let your opponent start?

Safety is not the issue. The first, third, fifth, seventh, ninth, (and eleventh, for the 11-shot) chambers belong to one player, the second, fourth, sixth, eighth, and tenth chambers belong to the other.

I like Genmax's argument that the first player with the 11-shot has more chambers that have a chance of having a bullet than the second player, so there's a slightly better chance of winning by going first with 11-shot.

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What genmax said makes most sense.

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...He will slip a bullet into a random position in the gun barrel...

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?

He slips the bullet into the barrel, meaning that it will shoot the first time. Not randomly.

With that being said, I suppose the answer is obvious.

Oh, and by the way. This is the answer.

Reason for edit: Typo (Probably still is).

Edited by TheRene
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I don't think either question matters, and it doesn't matter which choice you make.

10-shot revolver (OP says it's like in a western. That means it's a revolver). When you get to shoot, either you have the even cylinders or you have the odd cylinders. it's 50-50 whether your cylinders have the bullet or not. Doesn't matter whether you shoot first.

11-shot revolver. Once again, the cylinder will either appear when your opponent shoots, or when you shoot. It's 50-50.

The only thing that does matter is the random choice of initial cylinder. You don't get a choice that affects that.

edit: end the game sooner, pick the 10-shot gun.

marvelous

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i think the mistake that some people are making is the chance to shoot on the first shot (lets use the 10 gun for this example) is 1/10. The chance of the second shot is 1/9, but thats once you get to that shot. You only get there 9/10 of the time so the probability of someone getting shot on the second shot is 9/10 *1/9=1/10. same applies for every shot.

i think the question is you get to choose the gun or which time you go so if i understand correctly or in a correct manner

if you get to choose the gun choose the 10 shooter. because if you chose the 11 your opponent will choose to shoot first gaining an advantage. this way he can't

if you get to pick which turn to shoot, if its the 10 shooter. I would pick to go second because its a lil of a psych advantage i think (and theres nothing like not only killing a person but also psyching him out), but there is no statistical advantage. if its the 11 shooter shoot first

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i think the mistake that some people are making is the chance to shoot on the first shot (lets use the 10 gun for this example) is 1/10. The chance of the second shot is 1/9, but thats once you get to that shot. You only get there 9/10 of the time so the probability of someone getting shot on the second shot is 9/10 *1/9=1/10. same applies for every shot.

i think the question is you get to choose the gun or which time you go so if i understand correctly or in a correct manner

if you get to choose the gun choose the 10 shooter. because if you chose the 11 your opponent will choose to shoot first gaining an advantage. this way he can't

if you get to pick which turn to shoot, if its the 10 shooter. I would pick to go second because its a lil of a psych advantage i think (and theres nothing like not only killing a person but also psyching him out), but there is no statistical advantage. if its the 11 shooter shoot first

Thank you for explaining that

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10 shot and point it at the warden. Maybe I'm just old fashioned though...

*hides*

That warden is pretty mean and canny. I doubt if he is available to be shot at.

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I don't think either question matters, and it doesn't matter which choice you make.

10-shot revolver (OP says it's like in a western. That means it's a revolver). When you get to shoot, either you have the even cylinders or you have the odd cylinders. it's 50-50 whether your cylinders have the bullet or not. Doesn't matter whether you shoot first.

11-shot revolver. Once again, the cylinder will either appear when your opponent shoots, or when you shoot. It's 50-50.

The only thing that does matter is the random choice of initial cylinder. You don't get a choice that affects that.

edit: end the game sooner, pick the 10-shot gun.

That is an interesting way of looking at it. So it is rather like a 50:50 chance of rolling an even number when you toss a single die?

But then , wouldn't the 11 shots affect that? Since it would be like tossing a 7 sided die instead of a 6 sided die, the odds are no longer 50-50 that you will roll an odd number. One prisoner will be rolling with the 7 sided die instead of the 6 sided die.

Statistics is definitely not intuitive for me. I am convinced by almost every different argument I read!

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I think that the best case scenario would be having the A10 and shooting first. Your first shot is a blank. Your opponent then shoots a blank at you. You will always be one shot ahead of your opponent. Also, because the A11 shoots more times than the A10, you will be done shooting before he/she shoots every round. Your chances of living are higher if you just think logic, no statistics.

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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..."

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

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.

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

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