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6 votes (50%)
4 votes (33.33%)
2 votes (16.66%)

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Dween
Dween
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September 29th, 2010 at 5:21:59 AM permalink
On the British show QI, comedians discuss puzzles, debate points, and answer questions, all in the name of fun and frivolity. It's about as much of a game show as Whose Line Is It Anyway?, and from clips I have seen, very entertaining. (Take the hint, BBC America, bring this show to U.S. soil!)

This clip from the show talks about Game Theory, and specifically, a "Truel", or 3-way duel. It may not be exactly the situation being discussed, but it is similar enough to possibly shed some insight, and humourous besides.
-Dween!
Wizard
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September 29th, 2010 at 6:31:03 AM permalink
Quote: Dween

This clip from the show talks about Game Theory, and specifically, a "Truel", or 3-way duel. It may not be exactly the situation being discussed, but it is similar enough to possibly shed some insight, and humourous besides.



That is a different puzzle. Again, there is a truel between A, B, and C. A has a 10% chance to hit his target, B 60%, and C 90%. A gets to go first. Who should he aim at? I assume B goes next, then C. Each person has one bullet.
It's not whether you win or lose; it's whether or not you had a good bet.
weaselman
weaselman
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September 29th, 2010 at 7:52:21 AM permalink
Does a shooter have a chance to see the outcome of the previous shot before he has to aim and shoot, or are all decisions made and fixed at the beginning?
"When two people always agree one of them is unnecessary"
Nareed
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September 29th, 2010 at 8:21:40 AM permalink
In a standoff between three armed men, it may be best to play defense rather than offense. Wear body armor and duck, then shoot whoever remains standing.

But of course if you knew you were going to wind up in that situation, that's plan B.
Donald Trump is a fucking criminal
MathExtremist
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September 29th, 2010 at 8:24:14 AM permalink
Quote: Wizard

That is a different puzzle. Again, there is a truel between A, B, and C. A has a 10% chance to hit his target, B 60%, and C 90%. A gets to go first. Who should he aim at? I assume B goes next, then C. Each person has one bullet.



That's also a different puzzle than the previous "perfect shot" scenario where all of them had 100% chances. If they're still perfect logicians and the shooters have to aim at *someone*, then based on those chances above A should play Russian roulette. He has a 90% chance of surviving, which is better than if A shoots toward anyone else. Knowing this, the other two will shoot at each other. This assumes the goal is to live, vs. to be the last standing. A's chances of being the last standing are 0 if he aims at himself...

Edit: now form the mental image of a 3-way gunfight where all the contestants draw at the same time, and where two draw on each other and the third puts the gun to his head.
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice." -- Girolamo Cardano, 1563
weaselman
weaselman
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September 29th, 2010 at 8:36:49 AM permalink
Quote: MathExtremist

That's also a different puzzle than the previous "perfect shot" scenario where all of them had 100% chances. If they're still perfect logicians and the shooters have to aim at *someone*, then based on those chances above A should play Russian roulette. He has a 90% chance of surviving, which is better than if A shoots toward anyone else.



If the other two are aiming at each other, then A's chances to survive are 100% if he shoots any one of them instead of himself. If any of them is aiming at A, then his chances will still be improved by shooting at the guy aiming at him.
So, despite how intriguing it sounds, I don't think it is actually in A's best interest to play Russian roulette in this situation.
"When two people always agree one of them is unnecessary"
ChesterDog
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September 29th, 2010 at 3:19:09 PM permalink
Quote: TheNightfly

...A, B and C all draw in less than one second. There is no way for anoyone to have a clear advantage beyond a fraction of a second - not nearly enough time for either of the other two gunslingers to make a decision AFTER anyone else has moved. Gunfighter A is MARGINALLY faster than the other two and is MARGINALLY less accurate than the other two....
...in a real gunfight, no one would have the time to notice that someone pulling their gun would be firing to miss. ...



Let's assume that the gunfighters want to kill someone and don't want to be killed themselves, and that when the shooting starts, noone has the time to see who is shooting at whom. Also, assume that if a bullet hits someone, it kills him instantly before he can fire his gun.

B knows that he's the slowest to shoot his gun, so he knows his probability of not dying does not depend on who he tries to kill; so he's equally likely to try to kill A or C.

Shooter A must ask himself who is the bigger threat to him, C or B; it's B because B is more accurate, and A and C are equally likely to be B's target. (But, more importantly, see next line.)

Shooter C knows he's slower than A, so C's probability of dying could not be reduced by his trying to kill A, so C would try to kill B. (Shooter A would understand this reasoning, so would not plan on trying to shoot C.)

So A and C each plan on trying to kill B, and B picks A or C at random to try to kill.

A and C have equal survival probabilities. And their probabilities are higher than B's unless A's and C's shooting accuracies were considerably lower than B's.
GenWyzgy
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September 29th, 2010 at 3:47:53 PM permalink
I picked Gunman C, and here's why (so you can at least pick apart one take on the original puzzle)

Since gunman A knows he's fastest and gunman B is most accurate, but slowest; gunman A shoots at gunman B

Since gunman B knows he's the slowest, he cannot beat either A or C to the draw. Therefore, he shoots at gunman A, since A is the most likely to have missed with his quicker shot.

Gunman C could shoot at either A or B, but since neither of them aimed at him, he survives.
Wizard
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September 29th, 2010 at 5:17:21 PM permalink
Quote: weaselman

If the other two are aiming at each other, then A's chances to survive are 100% if he shoots any one of them instead of himself. If any of them is aiming at A, then his chances will still be improved by shooting at the guy aiming at him.
So, despite how intriguing it sounds, I don't think it is actually in A's best interest to play Russian roulette in this situation.



Let's assume that if it gets to C's turn and all three are still alive, then C will choose randomly who to shoot at. So consider the possibility that A shoots at B or C, and misses, and then B shoots at C, and misses. There is a chance that C will choose to shoot at A. So the odds of death are more than 0% if you aim at B or C. So I pose the question, what A's probability of surival under these options:

1. Deliberately misses.
2. Aims at himself.
3. Aims at B.
4. Aims at C.

If A aims at himself, his odds of success are still 10%. I admit that seems silly, especially if he can deliberately miss with 100% chance, but just take that on faith.
It's not whether you win or lose; it's whether or not you had a good bet.
MathExtremist
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September 29th, 2010 at 5:38:53 PM permalink
Quote: weaselman

If the other two are aiming at each other, then A's chances to survive are 100% if he shoots any one of them instead of himself. If any of them is aiming at A, then his chances will still be improved by shooting at the guy aiming at him.
So, despite how intriguing it sounds, I don't think it is actually in A's best interest to play Russian roulette in this situation.



Right, but they're perfect logicians so if they know A is going to aim at one of them, they might aim back (and they have a much higher chance of killing A than he has of killing himself. :)
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice." -- Girolamo Cardano, 1563

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