[Guest]

An eccentric billionaire knows that three people, A, B and C, who would dearly like to kill each other, are all perfect shots with a Colt 45 revolver at 25 paces. He arranges a three-way duel (a tri-el) whose rules are as follows.

(a) The contestants, each supplied with a large number of revolvers, are positioned at the apices of an equilateral triangle of side 25 paces. Each of A's revolvers is loaded with 2 live rounds and 4 blanks at random. Effectively, if he aims and fires at anybody, his chances of killing him with one shot are one third. In a likewise manner, B's revolvers are loaded with 3 live rounds and 3 blanks, while C's revolvers are loaded with 6 live rounds. Effectively, A's chances of hitting a target at which he is aiming are one third, B's one half, while C's are certainty.

(b) A is allowed to take 1 shot, after which he discards his revolver and picks another. Then B, if he is still alive, is permitted one shot, after which he discards his revolver and picks another. Then C, if alive, has his chance. The tri-el continues cyclically until there is just one survivor who is given one billion dollars (and immunitly from prosecution.)

Question 1 (easy).

Suppose C is dead and A and B are still alive. What are A's chances, if it is his shot, of eventually winning?

Question 2 (a bit more difficult.)

Is A's best strategy on his first shot to aim at C?

[Guest]

Suppose C is dead and A and B are still alive. What are A's chances, if it is his shot, of eventually winning?

1/3 +2/3 *1/2 *1/3 +(2/3)^2 *(1/2)^2 *1/3 ...

1/3 +(1/3)^2 +(1/3)^3...

(1/3)/(1-1/3) = 1/2.

[Guest]

Suppose C is dead and A and B are still alive. What are A's chances, if it is his shot, of eventually winning?

1/3 +2/3 *1/2 *1/3 +(2/3)^2 *(1/2)^2 *1/3 ...

1/3 +(1/3)^2 +(1/3)^3...

(1/3)/(1-1/3) = 1/2.

I fully concur with your method, if not your arithmetic.

Incidentally, my arithmetic seems to be deteriorating a bit at the moment. That is why I have joined this site.

[Guest]

Well...

Instinctively, as a hypothesis, I would say yes A should shoot at C, because if A kills one of the other two, then he's in a duel and it's not his turn. He'd (much) rather have B shooting at him than C. Here's my math:

Let's suppose that it is C's turn, and that everyone is alive. He will shoot at B, because that would put C in a duel against A (with A's turn), which gives him a 2/3 chance of surviving (as opposed to only 1/2 if he shoots and kills A and has a duel against B) and killing A immediately afterward.

B, knowing this, will have to shoot at C, assuming everyone is still alive (i.e. A missed on his turn). If he hits (1/2 chance), then it's a duel between A and B with A's turn, which, as stated above, is a 1/2 chance of winning for A. If B misses (another 1/2 chance), then C will kill B, giving A a 1/3 chance of killing C and winning. Thus, if A misses, and then B shoots at C, A will have a (1/2)*(1/2) + (1/2)*(1/3) = 5/12 chance of winning.

So who should A target on his first turn? Let's suppose A shoots at C. He has a 1/3 chance of killing him, and a 2/3 chance of missing. The chance that A wins a duel against B when B gets the first shot (which will happen if A manages to kill C) is:

(1/2)*(1/3) + (1/2)*(2/3)*(1/2)*(1/3) + (1/2)*(2/3)*(1/2)*(2/3)*(1/2)*(1/3) + ...

= 1/6 + (1/6)*(1/3) + (1/6)*(1/3)² + ...

= (1/6)/(1-(1/3)) = (1/6)/(2/3) = 1/4.

As stated above, if A misses, he will have a 5/12 chance of winning, so if A targets C on the first turn, then the chances of A winning are (1/4)*(1/3) + (5/12)*(2/3) = 13/36.

Let's say now that A targets B. He has a 1/3 chance of hitting, but that results in C shooting A on the very next turn, and C wins. The chance that A wins if he misses is still 5/12, so the chance that A wins if he targets B is (1/3)*0 + (2/3)*(5/12) = 5/18.

At this point, it may seem that A's best strategy is to target C. However, we've forgotten one scenario. Remember that if A misses, his chance of winning is 5/12. His chances of winning when targeting B and C are respectively 13/36 and 5/18, both of which are less than 5/12. Thus, A's best strategy is to shoot into the air, missing on purpose, which would give him a 5/12 chance of winning.

Edited August 9, 2009 by Chuck

[Guest]

I do not think that 2 is solvable because it depends on who B and C decide to shoot at (if they survive). If they both decide to gang up on A then shooting air, as suggested by Chuck, would not be a good idea.

[Guest]

I do not think that 2 is solvable because it depends on who B and C decide to shoot at (if they survive). If they both decide to gang up on A then shooting air, as suggested by Chuck, would not be a good idea.

Well, the point is, that B and C will play so that they have their best possible chance of surviving. I'm pretty sure that my solution incorporates the best strategies of B and C.

[Guest]

Well, the point is, that B and C will play so that they have their best possible chance of surviving. I'm pretty sure that my solution incorporates the best strategies of B and C.

That accords with my thinking Chuck. I first saw this in Martin Gardner's column in Scientific American, and that is what he thought too.

[Guest]

Well...

Instinctively, as a hypothesis, I would say yes A should shoot at C, because if A kills one of the other two, then he's in a duel and it's not his turn. He'd (much) rather have B shooting at him than C. Here's my math:

Let's suppose that it is C's turn, and that everyone is alive. He will shoot at B, because that would put C in a duel against A (with A's turn), which gives him a 2/3 chance of surviving (as opposed to only 1/2 if he shoots and kills A and has a duel against B) and killing A immediately afterward.

B, knowing this, will have to shoot at C, assuming everyone is still alive (i.e. A missed on his turn). If he hits (1/2 chance), then it's a duel between A and B with A's turn, which, as stated above, is a 1/2 chance of winning for A. If B misses (another 1/2 chance), then C will kill B, giving A a 1/3 chance of killing C and winning. Thus, if A misses, and then B shoots at C, A will have a (1/2)*(1/2) + (1/2)*(1/3) = 5/12 chance of winning.

So who should A target on his first turn? Let's suppose A shoots at C. He has a 1/3 chance of killing him, and a 2/3 chance of missing. The chance that A wins a duel against B when B gets the first shot (which will happen if A manages to kill C) is:

(1/2)*(1/3) + (1/2)*(2/3)*(1/2)*(1/3) + (1/2)*(2/3)*(1/2)*(2/3)*(1/2)*(1/3) + ...

= 1/6 + (1/6)*(1/3) + (1/6)*(1/3)² + ...

= (1/6)/(1-(1/3)) = (1/6)/(2/3) = 1/4.

As stated above, if A misses, he will have a 5/12 chance of winning, so if A targets C on the first turn, then the chances of A winning are (1/4)*(1/3) + (5/12)*(2/3) = 13/36.

Let's say now that A targets B. He has a 1/3 chance of hitting, but that results in C shooting A on the very next turn, and C wins. The chance that A wins if he misses is still 5/12, so the chance that A wins if he targets B is (1/3)*0 + (2/3)*(5/12) = 5/18.

At this point, it may seem that A's best strategy is to target C. However, we've forgotten one scenario. Remember that if A misses, his chance of winning is 5/12. His chances of winning when targeting B and C are respectively 13/36 and 5/18, both of which are less than 5/12. Thus, A's best strategy is to shoot into the air, missing on purpose, which would give him a 5/12 chance of winning.

This is precisely what Martin Gardner stated when he printed the problem in his Scientific American column.

[Guest]

If C is dead in the first round then only A & B are left.

The probability of A winning is:

1/3 * (1 - 1/2) = 1/6

[Guest]

for question 1:

If C is dead in the first round then only A & B are left.

The probability of A winning is: 1/3 * (1 - 1/2) = 1/6

for question 2:

if he aims at B:

his chances of being killed by B is (1 - 1/3) * 1/2 * 1/2 = 1/6

his chances of being killed by C is (1 - 0) * 1/1 * 1/2 = 1/2

so his overall chance of surviving first round is 1/2 * 1/6 = 1/12

if he aims at C:

his chances of being killed by B is (1 - 0) * 1/2 * 1/2 = 1/4

his chances of being killed by C is (1 - 1/3) * 1/1 * 1/2 = 1/3

so his overall chance of surviving first round is 1/4 * 1/3 = 1/12

in conclusion it doesn't matter his chances of surviving first round is the same.

[Guest]

Reading Chuck's post, I realize I was approaching this problem incorrectly.

I read the part where each member takes a shot and then "he discards his revolver and picks another." For some reason, I didn't assume that it was another revolver from his own pile, but a choice of anybody's revolver. With C having a certain kill, I was originally assuming both players would pick up one of C's many fully loaded guns.

[Guest]

I think everyone has overlooked the human factor in this riddle. Although it may be in the best interest of B and C to shoot at each other once it's their turn, their emotions may take over. If I'm C or B and person A shoots a gun at me (intending to kill me) but I survive, I will sure as hell try to kill A the first chance I get. In the case where A shoots into the air, I will still do my best to kill him/her because, assuming we're outside, the bullet could come back down and hit me.

[superprismatic]

An eccentric billionaire knows that three people, A, B and C, who would dearly like to kill each other, are all perfect shots with a Colt 45 revolver at 25 paces. He arranges a three-way duel (a tri-el) whose rules are as follows.

(a) The contestants, each supplied with a large number of revolvers, are positioned at the apices of an equilateral triangle of side 25 paces. Each of A's revolvers is loaded with 2 live rounds and 4 blanks at random. Effectively, if he aims and fires at anybody, his chances of killing him with one shot are one third. In a likewise manner, B's revolvers are loaded with 3 live rounds and 3 blanks, while C's revolvers are loaded with 6 live rounds. Effectively, A's chances of hitting a target at which he is aiming are one third, B's one half, while C's are certainty.

(b) A is allowed to take 1 shot, after which he discards his revolver and picks another. Then B, if he is still alive, is permitted one shot, after which he discards his revolver and picks another. Then C, if alive, has his chance. The tri-el continues cyclically until there is just one survivor who is given one billion dollars (and immunitly from prosecution.)

Question 1 (easy).

Suppose C is dead and A and B are still alive. What are A's chances, if it is his shot, of eventually winning?

Question 2 (a bit more difficult.)

Is A's best strategy on his first shot to aim at C?

A should waste his first shot. Then B will shoot at C because C is the greater threat of A and C. If B shoots a blank, then C will kill B (again, because B is the greater threat of A and B). Then A will have the first at C. But if B shoots a live round, C will be dead and A will have first crack at B.

[Guest]

Thank you for clearing this up. It didn't say in the OP that B and C also do an optimum stratergy so I wasn't sure.

[bonanova]

with different probabilities but same strategy.

Locking this thread for now.

If there are differences from previous puzzle, explain them to me in a PM, and I'll be happy to unlock.

- bn