Poll

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

12 members have voted

weaselman
weaselman
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September 29th, 2010 at 5:52:55 PM permalink
Quote: MathExtremist

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. :)


Aiming A won't help them survive, because A will shoot first anyway, and, once he has taken a shot, he is no threat.
Besides, even if they are aiming at A, he is better off aiming back at one of them because it'll improve his chances by 10% as opposed to making them 10% worse if he tried to shoot himself instead.
"When two people always agree one of them is unnecessary"
dm
dm
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September 30th, 2010 at 10:06:16 AM permalink
I say the strongest motivation is to shoot at someone who is slower than you, so as to get the first shot in against one who might choose to shoot at you. So, C fires at B. A has 2 slower draws to consider, so he should pick the best shot. A fires at B. B fires his dirty underwear, and puts on clean ones, or not.
Wizard
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October 4th, 2010 at 11:11:53 AM permalink
I've been spending a bit of time on the QI' rel='nofollow' target='_blank'>http://www.youtube.com/watch?v=qL8raEwtzVQ]QI truel puzzle. Here again are the rules.

  1. The three participants form a triangle.
  2. Each has one bullet only.
  3. A goes first, then B, and C.
  4. A's probability of hitting an intended target is 10%.
  5. B's probability of hitting an intended target is 60%.
  6. C's probability of hitting an intended target is 90%.
  7. There are no accidental shootings.
  8. Shooting in the air (deliberately missing) and shooting yourself are allowed, and are always successful.
  9. After each round, if there are two or three survivors, then each participant is given a new bullet.
  10. All three participants are perfect logicians.


Rule 9 is not stated on the show, but I think can be assumed. I also assume that if all a round results in no deaths that the same strategy is repeated the next round.

Here are my probabilities of survival according to each initial target. Anybody care to dispute this before I make an "ask the wizard" question out of it?

Air 13.887%
A 0.000%
B 12.560%
C 13.094%
It's not whether you win or lose; it's whether or not you had a good bet.
mkl654321
mkl654321
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October 4th, 2010 at 1:13:17 PM permalink
Quote: Wizard

Rule 9 is not stated on the show, but I think can be assumed. I also assume that if all a round results in no deaths that the same strategy is repeated the next round.



I'm glad that you clarified Rule 9, because if the players each had only one shot and one bullet, period, then their respective strategies would be different.

I also would like to know if enough time elapses between the three shots that B, then C, would be able to change their minds once they saw the results of prior shots. Or do they just pick a target and blaze away? The latter seems more likely, so I'll answer based on that.

A is in bad shape if either opponent remains standing after the first round. He therefore wants to give himself the best chance of eliminating both opponents. If he aims at C, B might be doing the same, and if so, C dies 64% of the time. Of the remaining 36% (when A and B both miss him), C will kill B (presuming that B is C's chosen target, which seems likely) 90% of the remainder. So 64% of the time the game is reduced to A vs. B, 32.4% of the time, to A vs. C, and 3.6% of the time, everybody misses, and we do it all over.

If A aims at B, he would actually worsen his chances, because he wouldn't WANT to hit him; that would ensure that his remaining opponent would be the deadly accurate C. If A misses B, then he will be facing B as his remaining opponent 60% of the time (ignoring the possibility of a triple miss for the moment). So this is clearly a worse outcome than the roughly 2/3 possibility that his remaining opponent will be B (by aiming at C).

The next question is, does A improve his outcome by aiming at C at all? The answer would be yes, because he gives himself the best chance that his remaining opponent will be B rather than C.

B should target C for similar reasons. He would much rather have his remaining opponent be A than C. C should target B for the same reason.

So I would expect A and B to both target C. The most likely outcome would be that A would miss C, B would kill C, and then it would be B vs. A, with a 10% chance of A winning a given round, B a 54% chance, and a 36% chance of a do-over. In a B misses, C kills B scenario, the chances would be A 10%, C 81%, do-over 9%.

In either case, A's chances are greater than zero, and B's chances are better than C's (because he shoots before C), so I don't see how your percentages were derived.
The fact that a believer is happier than a skeptic is no more to the point than the fact that a drunken man is happier than a sober one. The happiness of credulity is a cheap and dangerous quality.---George Bernard Shaw
Wizard
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October 4th, 2010 at 1:47:10 PM permalink
Quote: mkl654321

I'm glad that you clarified Rule 9, because if the players each had only one shot and one bullet, period, then their respective strategies would be different.

I also would like to know if enough time elapses between the three shots that B, then C, would be able to change their minds once they saw the results of prior shots. Or do they just pick a target and blaze away? The latter seems more likely, so I'll answer based on that.



I should have wrote that after a round where there are two or more survivors that each participant is given another bullet. Additional rounds follow the same rule about going in order, skipping anybody who was already killed. Does that help? I'd like to give others a chance to get the exact probabilities before showing my math.
It's not whether you win or lose; it's whether or not you had a good bet.
thecesspit
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October 4th, 2010 at 2:12:32 PM permalink
The way i understand it is that A does their shot, then B, and B doesn't have to pre-aim before A squeezes one of...

A, B and C should all repeatedly shoot in the air. That makes all of their survival rates 100%.

If they are perfect logicians, they should come to this conclusion. ;)
"Then you can admire the real gambler, who has neither eaten, slept, thought nor lived, he has so smarted under the scourge of his martingale, so suffered on the rack of his desire for a coup at trente-et-quarante" - Honore de Balzac, 1829
Wizard
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October 4th, 2010 at 2:21:29 PM permalink
Quote: thecesspit

A, B and C should all repeatedly shoot in the air. That makes all of their survival rates 100%.



I'm not sure if I should add another rule for this, but they desire to have only one survivor. On the QI show they use the premise that they are fighting over a woman. So it would be better to die than live without her.
It's not whether you win or lose; it's whether or not you had a good bet.
mkl654321
mkl654321
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October 4th, 2010 at 2:26:33 PM permalink
Quote: Wizard

I'm not sure if I should add another rule for this, but they desire to have only one survivor. On the QI show they use the premise that they are fighting over a woman. So it would be better to die than live without her.



Which contradicts the premise that the three duellists are perfect logicans.

If they were French, they'd just all move into the same house, and A would have her on Mondays and Thursdays, B on Tuesdays and Fridays, C on Wednesdays and Saturdays, and Sunday would be her day of rest.
The fact that a believer is happier than a skeptic is no more to the point than the fact that a drunken man is happier than a sober one. The happiness of credulity is a cheap and dangerous quality.---George Bernard Shaw
Wizard
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October 4th, 2010 at 3:49:25 PM permalink
I have a draft of my next Ask the Wizard column, which shows for now how I get the probability of survival if A initially aims at B. In the writing of it, I realized I made an error in my post above, which I fixed. Later I'll add to it for the probabilities of eventual survival by aiming at C and deliberately missing.


Quote: mkl654321

Which contradicts the premise that the three duellists are perfect logicans.

If they were French, they'd just all move into the same house, and A would have her on Mondays and Thursdays, B on Tuesdays and Fridays, C on Wednesdays and Saturdays, and Sunday would be her day of rest.



Nice :-). I guess I'll have to add a rule that they are not French. How about if they were German?
It's not whether you win or lose; it's whether or not you had a good bet.
thecesspit
thecesspit
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October 4th, 2010 at 5:42:38 PM permalink
Quote: Wizard

Nice :-). I guess I'll have to add a rule that they are not French. How about if they were German?



Same time-based solution, just that they would all live in different houses.
"Then you can admire the real gambler, who has neither eaten, slept, thought nor lived, he has so smarted under the scourge of his martingale, so suffered on the rack of his desire for a coup at trente-et-quarante" - Honore de Balzac, 1829

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