I get:
Black survives 5 times in 9
Grey survives 4 times in 9
White survives 1 time in 9
Total sums to more than one because 1 time in 9, Black and Grey both remain alive when bullets are exhausted.
… That the weaker player fares best might serve as evidence for some moral principle! 
[del] Black survives 5 times in 9
Grey survives 4 times in 9
White survives 1 time in 9
Total sums to more than one because 1 time in 9, Black and Grey both remain alive when bullets are exhausted.[/del]
Please ignore me. I detected my stupidity well within the 5-minute edit window, but Internet flakiness interfered.
If Black’s bullet is in the #1 position he’s dead no matter what he does, but he gets to choose which of White or Grey is more likely to survive. This makes the problem slightly undetermined. Either way, I think Black’s chance is (at least tied for) best at 4 out of 9.
The answer given is clever but also wrong if you add the assumption that a) the three shooters are able to communicate and b) there is a method of ensuring trustworthiness.
The reasoning goes as follows:
Assume Black’s first move is to shoot into the ground:
The only way for White to survive is:
Grey shoots at White and misses (1/3) * White shoots at Grey and hits (1) * Black shoots at White and misses (2/3) * White shoots at Black and hits (1) = 2/9
So there is a total probability of 2/9 that White survives.
Grey’s ways of surviving are:
Grey shoots at White and misses (1/3) * White shoots at Black and hits (1) * Grey shoots at White and hits (2/3) +
Grey shoots at White and hits (2/3) * Black shoots at Grey and misses (2/3) * Grey shoots at Black and hits (2/3) +
Grey shoots at White and hits (2/3) * Black shoots at Grey and misses (2/3) * Grey shoots at Black and misses (1/3) * Black shoots at Grey and misses (2/3) * Grey shoots at Black and hits (2/3) + … = 2/9+8/27+16/243+… = 2/9+4/32/9+4/3(2/9)^2+… = 38/63 ~= 60%
(By the process of elimination, Black’s chances of surviving this scenario is 11/63)
BUT, Assume White and Grey make a deal that providing Grey shoots at Black, IF Grey misses, White will shoot at Black but IF Grey hits, White will shoot at the ground.
The new probabilities are now that White has a 1/3 chance of survival and Grey has a 2/3 chance of survival which is strictly better than previously. Thus, assuming that Grey can trust White, White should propose the deal.
The biggest sticking point is that White has no incentive not to defect on the deal once Grey has made their shot. Unless White can demonstrate to Grey that defection is not possible, then both White and Grey are both worse off. As a result, it’s in White’s best interests to remove agency from his/her decision. What White should do is build a shooting robot, allow Grey to inspect and mutually agree that the robot will follow the deal struck and then move far enough out of range that White cannot get to the gun in between Grey’s shot and White’s shot. If this is possible, then this is bad news for Black and Black’s optimal move is now no longer shooting into the ground.
Ignore previous post, made a mistake in the math.
The answer given is clever but also wrong if you add the assumption that a) the three shooters are able to communicate and b) there is a method of ensuring trustworthiness.
The reasoning goes as follows:
Assume Black’s first move is to shoot into the ground:
The only way for White to survive is:
Grey shoots at White and misses (1/3) * White shoots at Grey and hits (1) * Black shoots at White and misses (2/3) * White shoots at Black and hits (1) = 2/9
So there is a total probability of 2/9 that White survives.
Grey’s ways of surviving are:
Grey shoots at White and hits (2/3) * Black shoots at Grey and misses (2/3) * Grey shoots at Black and hits (2/3) +
Grey shoots at White and hits (2/3) * Black shoots at Grey and misses (2/3) * Grey shoots at Black and misses (1/3) * Black shoots at Grey and misses (2/3) * Grey shoots at Black and hits (2/3) +
Grey shoots at White and hits (2/3) * Black shoots at Grey and misses (2/3) * Grey shoots at Black and misses (1/3) * Black shoots at Grey and misses (2/3) * Grey shoots at Black and misses (1/3) * Black shoots at Grey and misses (2/3) * Grey shoots at Black and hits (2/3) + … = 8/27+16/243+32/2189… = 8/27+8/272/9+8/27(2/9)^2+… = 8/21 ~= 38%
(By the process of elimination, Black’s chances of surviving this scenario is 25/63)
BUT, Assume White and Grey make a deal that providing Grey shoots at Black, IF Grey misses, White will shoot at Black but IF Grey hits, White will shoot at the ground.
The new probabilities are now that White has a 1/3 chance of survival and Grey has a 2/3 chance of survival which is strictly better than previously. Thus, assuming that Grey can trust White, White should propose the deal.
The biggest sticking point is that White has no incentive not to defect on the deal once Grey has made their shot. Unless White can demonstrate to Grey that defection is not possible, then both White and Grey are both worse off. As a result, it’s in White’s best interests to remove agency from his/her decision. What White should do is build a shooting robot, allow Grey to inspect and mutually agree that the robot will follow the deal struck and then move far enough out of range that White cannot get to the gun in between Grey’s shot and White’s shot. If this is possible, then this is bad news for Black and Black’s optimal move is now no longer shooting into the ground.
I’m glad to see folks have taken to trying to answer the question originally asked. ie: not bringing into it possible targets or actions not in the original problem.
Here’s what I’ve got:
When Black fires, there is a 2 in 3 chance he will miss, and thus a 2 in 3 chance that it doesn’t matter at all who he was aiming at, as play will continue from there regardless. His odds of being the last man standing are the same if he misses Gray as if he misses White.
So all we have to look at is what happens if he hits.
If his target was Grey and he hits, Grey’s turn is skipped, and then White shoots him. Game Over.
If his target was White, then Grey will fire on him. Grey only hits 2/3 of the time. While this does not mean Black would then have a 1 in 3 chance of winning, because they keep shooting, any chance at all is better than the odds he’d have if he targeted Grey.
Just playing the odds, Black should shoot at White.
Others are interpreting Biotop’s problem differently than me. Each gun is preloaded with three bullets. (Others assume a fresh bullet is provided from a probability distribution before each shot.) Assuming my interpretation is correct:
After White’s death here, Grey has two live bullets and a blank. Black has only blanks left. Unless Grey takes mercy he’s going to kill Black on his 1st or 2nd shot.
I do not see any advantage to Black shooting White (for Black) in the first round. If Black shoots anyone to death on the first round he is out of bullets and will die. What is confusing me is that if we disregard the possibility of Black shooting at the ground, does his having a REQUIRED random choice of shooting Grey or White change the odds for either Grey or White surviving * as opposed to Black simply shooting at the ground * as posted in the informative analysis by Shalmanese.
And yes, the idea is that the guns deplete as the shots are fired.
Therefore, to make it even more confusing, assuming Black has a blank on the first shot, he should have a 50 percent chance of winning on his guaranteed next shot as no one should logically shoot him on the first round, right?
Assume for simplicity that every shooter must, on every turn, attempt to shoot one living shooter, if there be any other shooter alive. Further assume that the only condition on which the target survives is if the shot was a blank.
We must also make some assumptions about the shooters’ utility functions. First assume that each shooter’s utility function is dominated by his own survival: No shooter is willing to accept any nonzero decrease in his own survival chance for the sake of choosing who wins in the event that he dies. Second, we’ll assume that, in the event that a shooter’s choice cannot affect his own chance of survival at all, he has some preference for who else dies, and will work to make the odds for that person as bad as possible.
Multiple survivors are in this case impossible: That would require that there be two people, neither of whom has a bullet in his brain nor in his gun. But there are a total of six bullets, and at most three can be in the dead man’s gun, and at most one in the dead man’s brain, leaving two bullets unaccounted for.
OK, given that, let’s find some cases where we know the outcome, and work backwards from there:
1: If White is to shoot, and he is one of two shooters currently alive, he will shoot and kill his opponent, and win.
2: If anyone else is to shoot, and is alive along with White, he must shoot at White, because doing so is his only chance to avoid scenario 1 where White kills him.
3: If anyone is out of blanks, then that person has effectively become White. Thus, if Gray misses his first shot, then White must kill Gray, since otherwise he’s left being killed by Gray.
4: If anyone is out of bullets when anyone else dies, that person is screwed, since he cannot kill the other opponent, and the other opponent is therefore guaranteed to eventually get his shot and kill the person out of bullets. Only Black can ever get into this situation, though, because Gray and White both have enough bullets to kill their opponents.
5: Thus, if Black hits with his first shot, he’s dead. And if he misses with his first shot, it doesn’t matter who he shot at. Let’s look at the case where he misses, and the cases where he hits.
If Black shoots Gray and hits, White is guaranteed to win. Likewise, if Black shoots White and hits, Gray is guaranteed to win. If Black shoots himself and hits, then Gray has a 2/3 chance of winning, and White has a 1/3 chance of winning. No matter what his preference, shooting at himself is thus dominated by shooting at his hated target.
If Black misses, then it’s now Gray’s turn. If Gray also misses (no matter whom he shoots at), then he has become inerrant, and so becomes White’s priority target, and dies. If Gray hits Black, then he becomes White’s only target, and so dies. If Gray hits himself, of course, he also dies. Therefore, Gray’s only chance of survival is to target White and hope he hits, and so that’s what he’ll do. This gives him a 2/3 chance of hitting White, then a 1/2 chance of being missed by Black, then a 1/2 chance of hitting Black, for a chance of survival (conditioned on Black’s first shot missing) of 1/6.
If both Black and Gray miss, then White kills Gray, and then Black has a 50% chance of killing White. So White’s chance of survival (conditioned on Black’s first shot missing) is 1/3 (for Gray missing him) times 1/2 (for Black missing him), or 1/6.
Black then, by subtraction, has a 2/3 chance of winning (conditioned on missing his first shot). Since Black missing his first shot was also a 2/3 chance, Black’s overall chance of victory is 4/9.
Meanwhile, if Black hits, then whoever Black didn’t aim at has a 100% chance of victory.
So, if Black’s more-hated enemy is Gray, then Gray’s chance of victory is (2/3)(1/6), or 1/9, and White’s chance of victory is 4/9. And if Black’s more-hated enemy is White, then White’s chance of victory is (2/3)(1/6), or 1/9, and Gray’s chance of victory is 4/9.
To put it all together, then: Black has a 4/9 chance of victory, whichever of the other two Black hates less also has a 4/9 chance of victory, and whichever one Black hates more has a mere 1/9 chance of victory.
Or to state this another way: If we assume that all players make their choices deterministically, and that the only randomness is in where the live bullets are as opposed to the blanks, then there are nine possible states for the initial setup (three possible positions for Black’s live bullet, and three possible positions for Gray’s blank), and each one will lead deterministically to one particular victor. In states 11, 12, and 13 (that is, where Black’s bullet is in the first position, and Gray’s blank is in the first, second, or third position), Black’s hated enemy and Black himself die, leaving the less-hated enemy the victor. In state 21 (that is, where Black’s bullet is in the second position, and Gray’s blank is in the 1st), White kills Gray and Black kills White, leaving Black the victor. In state 22, 23, or 32, Gray kills White and then Black kills Gray, also leaving Black the victor. In state 33, Gray kills White and then Black, leaving Gray the victor, and in state 31, White kills Gray and then Black, leaving White the victor.
This is what actually happened:
Mr Black, who was good at logic but a terrible shot, fired at the ground. Unfortunately, he shot himself in the foot, which caused him to scream in agony, and start hopping around.
Mr Gray, who was a fair shot, but very nervous, inadvertently pulled the trigger of his gun, and killed an innocent logician who was busy taking notes. He then, under the rules, took a shot at White, but missed.
Mr White coolly ignored the mayhem unfolding around him and aimed at Grey, but just as he was about to fire, Black blundered into him. As he tried to fend the injured man off, White fell onto his own gun and died instantly when it went off.
Gray, on hearing the approaching sirens, slipped away into the crowd, and Black, after treatment, was arrested for 2nd degree murder. Police are yet to find Mr Grey, and with half a dozen different descriptions from the assembled logicians, they are not optimistic.
The original proposal isn’t worded well for two reasons. First, it assumes everyone knows everyone else’s shooting ability, but doesn’t say it. Second, it should say “take turns firing” rather than “shooting at each other”.
But if we resolve those two nits, it’s a nice little game theory experiment.
That’s an interesting one too, and the strategies would be different, since after each shot, the odds change.
FTW
zugzwang
If Mr. Black is compelled to shoot, and he kills Mr. Grey, Mr. Black dies next turn. If he kills Mr. White, then he only enjoys a 1/3 chance of getting a second shot. If Mr. Black skips his turn, then he’s guaranteed a second shot.
zugzwang
All choices are bad.
Chronos, that was an excellent detailed explanation in Post #71. I plan to credit you, copy that post and send it both to my sister and brother. I think your idea of working backward from the outcome makes the problem easier to understand. Kudos! We also never really considered the possibility that Black would have a “more hated” enemy, and thus be able to alter the odds against that greater enemy.
I think the variation problem is also very interesting if Black doesn’t despise one enemy more than the other. Therefore, when he has a chance to shoot first, Black is indifferent. The choice of going after Grey or White would be purely random… perhaps determined by a coin flip. Black doesn’t care. It doesn’t help or hurt his chances who he fires at on the first round.
In this scenario, if I’ve got it figured out right, Black still has the best chance of winning the duel despite having the gun with the fewest real bullets. Grey and White have equal chances of winning, though both less than Black:
Black’s odds: 4/9
Grey’s odds: 5/18
White’s odds: 5/18
The problem is especially bizarre if in the first round Black decides to shoot at the ground rather than at Grey or White-- (Say Black doesn’t want to be the one to shoot the fatal shot unless he has to).
If I’ve got this figured out correctly, and please tell me if I am wrong, then:
Despite the fact that, in a coin-flip scenario, Black is equally as likely to shoot at Grey as White in the first round and is also equally as likely on that first shot to kill Grey as White…if Black instead passes on the coin flip and instead shoots at the ground, this immediately increases Grey’s chances of survival by over 5.5% and decreases White’s chances by the same amount. It has no effect on Black’s chances of survival because if the first bullet is real Black must die regardless of whether he shoots at a foe or at the ground. So Black’s shooting at the ground only costs White— the guy with the most real bullets.
Weird!
It’s also interesting to note that in the case where the only inaccuracy comes from bullets vs. blanks, all of the counterintuitive decisions disappear: Nobody ever wants to shoot at themselves, Black shoots at whomever he likes less, and Gray and White both shoot at whomever is most dangerous.
My brother suggests re-writing the same bullets/blanks riddle this way, letting you determine whether you want to be “Black,” “Grey,” or “White” instead of asking for the odds on who wins:
There is to be a three-way duel to the death between you and two others. You all hate each other and want to win the duel. You are all also logical and honorable and will follow the rules. There will be three guns. One gun will be loaded with three bullets. One gun will be randomly loaded with two bullets and a blank. The third gun will be randomly loaded with only one bullet and two blanks. The duel will proceed with each person taking a single shot at another foe of choice until only one combatant remains. You will all be close enough that a miss will be impossible. The person with the single real bullet in his gun will shoot first. Next will come the two bullet gun holder if he is still alive, then next the person with the three bullets in his gun gets his turn to shoot (that is, of course, presuming he is still alive). If the duel is not finalized after one round, the rounds continue in the same order with the same rules until the duel is decided.
You have all drawn lots and you have won. You therefore have been rewarded with the choice of which gun you want to possess for the duel. Which should you pick?
While it exactly the same riddle as the variation we have been discussing, the counterintuitive correct selection of the gun with only one bullet is a fun surprise answer.
Eh, I don’t know that it is necessarily counterintuitive. That guy has the disadvantage of only having one bullet, but also the advantage of going first. A sensible person should at least recognize the possibility that one of those might outweigh the other.
True enough I suppose. Still, the first person also obviously dies if he actually hits anyone in the first round, so it seems easy to dismiss that as the best option when you also consider that the well-armed forces are coming up next.