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## Question

A, B, and C have a three-way duel.

The rules are as following: A shoots first, then B, then C, and so on. Each man can shoot at whoever he wants on his turn. However, when one of them gets hit, he does not shoot anymore, and no one shoots at him. The duel continues until only one man is left standing. It is well-known fact that A hits his intended target 30% of the time, C – 50%, whereas B never misses.

What is the best strategy for A?

Small hint:

This variation of the duel puzzle actually requires some calculation.

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If A shoot B first he has about 26% of winning

If A shoot C first he has only about 12% of winning

But it he shoot the floor and let B shoot C first, he has a 30% of winning.

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I'm a little confused here. If A shoots first he's go a 30% chance of hitting either B or C. Now if he shoots at C and hits him then he's dead because B goes next and has only A to shoot at, unless of course B decides to commit suicide in which case A wins. Now if A shoots at B and hits then there's a 50% chance that C will shoot him. So if C hits then A is dead. If C misses then A still has a 30% chance of hitting C and C a 50% chance of hitting A. No matter how many times they go back and forth those numbers will never change so in theory they could go on forever. Statistics don't carry on over a series they are reset everytime the situation presents itself.

Now if A shoots at B and misses he could still survive if B decides to shoot at C in which case A better hope he doesn't miss the second time or he's dead. With the fact that each person can shoot at each other or shoot themselves or shoot into the air there is no real way to come up with a good stategy for this. All of that fancy math that everyone is posting is really meaningless because if A misses on his first shot the he either gets taken out by B or B shoots into the air and C shoots at A giving A a 50% chance to live or B takes out C and gives A another shot. See without the rule of the person being shot at will shoot back at the person doing the shooting or cannot shoot at the person doing the shooting and them being able to shoot at themselves or the air then there is too much speculation involved and statistics go out the window. Plus the part about once someone is hit they can't be shot at still only give A a 30% chance of survival each turn because he only has a 30% chance to hit himself. Then assuming that he shoots at that nonleathal part and misses you can factor in his chances but you still have to figure out if B or C is going to take aim on A or each other or themselves or the air. So if A shoots at his leg and mises then he has a 0% chance if B shoots him and a 50% chance of survival if C shoots at him which would in turn mean that B either shot himself or the air because he didn't aim at A and C is still alive. And then what if B does shoot at C but decides to only wound him and what if B gets two turns and shoots both of C's hands then C can't shoot anymore so he's just a sitting duck. Then does he die later from blood loss?

I can go on and on about the posibilites because there are no rules other than A is first B is second and C is last and when one gets hit he doesn't fire or fired at. Statistics are mathematics and can be manipulated to show what you want to show.

Bottom line the best strategy for A is to just run since neither of them can shoot until he does.

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I can go on and on about the posibilites because there are no rules other than A is first B is second and C is last and when one gets hit he doesn't fire or fired at. Statistics are mathematics and can be manipulated to show what you want to show.

Bottom line the best strategy for A is to just run since neither of them can shoot until he does.

Sorry that is supposed to say statistics are NOT mathematics. They use math but no statistic is absolutely solid.

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The solution seems to be using a fishbone diagram

Start at the left and follow the arrows to their conclusion.

50% A dead (%chance of dying = .3 *.5 = 15%)

hit/

then C=>A 30% C dead (%chance of surviving= .3 *.5 *.3= 4.5%)

30% miss\ hit/

hit/ 50% A=>C 50% A dead (%chance of dying = .3 *.5 *.7 *.5 = 5.25%)

A=>B miss\ hit/

miss\ 50% A dead 70% C=>A

\ miss\

\ 50% A=>C etc. (Approx chance of surviving <5%)

\

\ 50% A dead ( %chance of dying = .7 *.5 = 35%)

\ A/

70% B chooses target 30% B dead (%chance of surviving= .7 *.5 *.3= 10.5%)

B\ hit/

50% A=> B

miss\

70% B=> A (%chance of dying= .7 *.5 *.7= 24.5%)

Total chance of surviving if A=>B = 4.5% + 10.5% + <5% = <20%

B=>A A dead (%chance of dying = .3 = 30%)

30%

hit/

A=>C

miss\

\ 50% A dead ( %chance of dying = .7 *.5 = 35%)

\ A/

70% B chooses target 30% B dead (%chance of surviving= .7 *.5 *.3= 10.5%)

B\ hit/

50% A=> B

miss\

70% B=> A (%chance of dying= .7 *.5 *.7= 24.5%)

Total chance of surviving if A=>C = 10.5%

So A is about twice as likely to be killed if he shoots at C than if he shoots at B

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The solution seems to be using a fishbone diagram

Start at the left and follow the arrows to their conclusion.

50% A dead (%chance of dying = .3 *.5 = 15%)

hit/

then C=>A 30% C dead (%chance of surviving= .3 *.5 *.3= 4.5%)

30% miss\ hit/

hit/ 50% A=>C 50% A dead (%chance of dying = .3 *.5 *.7 *.5 = 5.25%)

A=>B miss\ hit/

miss\ 50% A dead 70% C=>A

\ miss\

\ 50% A=>C etc. (Approx chance of surviving <5%)

\

\ 50% A dead ( %chance of dying = .7 *.5 = 35%)

\ A/

70% B chooses target 30% B dead (%chance of surviving= .7 *.5 *.3= 10.5%)

B\ hit/

50% A=> B

miss\

70% B=> A (%chance of dying= .7 *.5 *.7= 24.5%)

Total chance of surviving if A=>B = 4.5% + 10.5% + <5% = <20%

B=>A A dead (%chance of dying = .3 = 30%)

30%

hit/

A=>C

miss\

\ 50% A dead ( %chance of dying = .7 *.5 = 35%)

\ A/

70% B chooses target 30% B dead (%chance of surviving= .7 *.5 *.3= 10.5%)

B\ hit/

50% A=> B

miss\

70% B=> A (%chance of dying= .7 *.5 *.7= 24.5%)

Total chance of surviving if A=>C = 10.5%

So A is about twice as likely to be killed if he shoots at C than if he shoots at B

My spaces didn't show up so my fishbone diagram looked terrible.

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All of these duels are limited by the fact that you don't know the relative value to each person of his own life.

For example, A shouldn't shoot anyone, because doing so means that he has a higher chance of dying (see many previous posts for an explanation). But then comes B. Even though he hits 100% of the time, he could shoot the floor as well. This would put C in the same position as A, with the additional option of also shooting the floor. This creates a psychological problem. If you shoot someone, there is a chance you will die, but if you choose to shoot away and others do too, you can be guaranteed survival and that's not much of a duel.

This is where the value of a life comes into play. For B, shooting C means a 70% chance of survival and winning the duel, as A will get one shot, and B will finish it off if he is alive. If he shoots the floor and he knows C will also choose survival over chance of winning he has guaranteed himself survival. Is a 70% chance of winning the duel better than 100% chance of survival? Who knows, it's all up to the duelists.

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All of these duels are limited by the fact that you don't know the relative value to each person of his own life.

For example, A shouldn't shoot anyone, because doing so means that he has a higher chance of dying (see many previous posts for an explanation). But then comes B. Even though he hits 100% of the time, he could shoot the floor as well. This would put C in the same position as A, with the additional option of also shooting the floor. This creates a psychological problem. If you shoot someone, there is a chance you will die, but if you choose to shoot away and others do too, you can be guaranteed survival and that's not much of a duel.

This is where the value of a life comes into play. For B, shooting C means a 70% chance of survival and winning the duel, as A will get one shot, and B will finish it off if he is alive. If he shoots the floor and he knows C will also choose survival over chance of winning he has guaranteed himself survival. Is a 70% chance of winning the duel better than 100% chance of survival? Who knows, it's all up to the duelists.

Assume, they shoot at each other with paintballs, and they compete to win.

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Assume, they shoot at each other with paintballs, and they compete to win.

This doesn't necessarily solve the problem. I refer you to this post where things are a bit clearer. In this case, the probabilities of successfully hitting a target are 7/10, 8/10 and 9/10. In this case, no matter what, if you shoot first and hit, you are most likely going to lose the duel. Therefore, no one should shoot.

This case is less extreme, in that one player has a significantly better chance of winning than the others (namely B), but still the problem exists. The paintball idea does nothing to relieve it, it just changes the wording from "dying" to "not losing". Would I rather have a 100% chance of not losing or a 70% chance of winning? If by letting someone else shoot first I can increase my chances of winning, then why would I shoot now and lower them?

P.S. Note that in this case B has a better chance of winning if he shoots first than if C shoots first, so he must rely on the fact that C can increase his chances of winning by not shooting as well.

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Are they aware of their shooting chances?

If they are I guess A and C are stupid to enter a duel with B and therefore they probably arent capable of statistical calculations.

So in such a duel I doubt anyone would make any calculations appart from B who might consider will he shoot A or C first.

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A misses C on purpose so that B will think A isnt trying to kill him. B kills C because if B kills A, C will kill B. A kills B. THE END!

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