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# The Truel Brain Teazer

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A hits his target 1/3 of the time.
B hits his target 1/2 of the time. EDIT: 1/2, not 2/3.
C hits her target every single time.

If they shoot in this order (A, then B, then C) who should A shoot to give him the best odds of winning the truel? Is there anything A can do to give himself the best odds of winning out of everyone involved?

EDIT:Everyone knows exactly how accurate everyone else is.

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His best chance of winning is to change the game to Russian Roulette.

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Been given hard homework again, eh, Hellbent?

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doesn't matter, C took his bullets out. then B misses C, C shoots B and hands A the shovel.
anyways, if he misses his first shot, the odds stay the same no matter who he aimed for. if he picks B and hits, he's dead. this isn't very hard to realize.

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nyet.

Here's a subtle hint:

Thinking about this scenario, how does it apply to the World Cup when an underdog team makes it out of its group when the other two teams are, say... Italy and England?

Correction on the original problem:
B hits his target 1/2 of the time, not 2/3.

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Do A and B both know that C never misses? In that case, A should shoot at C, because B will also shoot at C.

Edit: Either that, or A should shoot himself in the foot or something to disqualify himself because his aim is so bad.

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It's a bit fucked up here. I'm assuming the "truel" is divided in rounds, and that each gets one shot per round, and that each shooter may target each of the other (remaining) n shooters with a 1/n chance, right?

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Megalyth said:

Do A and B both know that C never misses?

duhhh. i was silly to assume hellbent's homeworks would be math tasks and not mind games. of course he should shoot his own foot and let his second take his place, cause who cares if he survives.

edit: i just realized the poor bastard has just 1/3 chance of hitting himself. what a moron.

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Keep your pants on, I was just trying to get some more details. I was thinking of it along the lines that you're an outside party that chooses who shoots at who, and who assumes that the contestants have no knowledge of the accuracy of the others.

Edit: I'm an idiot. I thought you were being sarcastic, my bad.

I'm thinking....

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I wrote a half-assed simulation of it, with the rules that the truel goes on in rounds until only one "trueler" is standing.

If the truelers CANNOT target one another, then quite predictably the odds of winning are (roughly, based on 10000 matches):

1. A=17.79%
2. B=29.37%
3. C=52.84%
Of course that doesn't say which possible intermediate outcomes are the best. Surely, if A targets and kills B then he surely loses immediately to C, while if he misses then B may target either of them etc.

However, if self-targetting is allowed, the results are surprisingly reversed:

1. A=34.03%
2. B=29.33%
3. C=25.45%

So I would not be surprised to hear that A's best choice is to try and commit suicide (with 1/3rd chance) and rely on the others NOT targetting him with prime priority (if suicide fails, then chances of being targetted and killed by B are 1/3*1/2 = 1/6 etc. and there's always the chance that C will commit suicide with either 1/3 or 1/2 chance ;-)

In any case, self-targetting (with a low chance of killing oneself) seems better than trying to kill the others (which have a better chance of doing it themselves!)

"Truel" my ass, this sounds more like the "Darwin Award Brain Teazer".

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Shoot in the air.

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Memfis said:

Shoot in the air.

he has only 1/3 chance of hitting the air. the bullet would probably slaughter a puppy.

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uhhhh

A shoots B, in case of hit and B being unable to shoot would result in C not meeting the requirements for shooting (B has not done a shot) thus making the truel last an eternity.

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j4rio said:

uhhhh

A shoots B, in case of hit and B being unable to shoot would result in C not meeting the requirements for shooting (B has not done a shot) thus making the truel last an eternity.

Nice try, but he never said that would result. I think it's supposed to be interpreted just that shooting order goes A, B, C, A, B, C...it will always fall on someone

Maes' solution sheds some light but I'm not sure what the paramaters for each experiment are. It appears in the first one...they shoot randomly at an opponent? And in the second, they always choose the opponent with the highest accuracy?

I started trying to work it out and got overwhelmed:

In order to be fit for any calculation as far as I can tell, they would have to either fire at the enemy with the highest accuracy, or choose a random enemy with 50% probability.

When firing at the most accurate opponent, A should purposely miss the first round. B will then try to shoot C, and if he fails C will kill B. In this case I believe A's chance of winning by the second round is approximately 1/3:

B shoots C - 1/2
---A shoots B - 1/3 (or 1/6 overall)
---A misses B - 2/3 (or 2/6 overall)
C shoots B - 1/2
---A shoots C - 1/3 (or 1/6 overall)
---A misses C - 2/3 (or 2/6 overall)

A could miss B, giving B another shot and changing the probability, but I can't be arsed to keep going with the tree :P. As you can see, the two winning outcomes have a probability of 1/6 each, for a total of 1/3.

When firing randomly, I would think A should fire at C. This is where it gets messy:

A shoots C
---Hit - 1/3
------B hits A - 1/2 (1/6)
------B misses A - 1/2 (1/6)
---------A hits B - 1/3 (1/18)
---------A misses B - 2/3 (2/18)
---Miss - 2/3
------B shoots A - 1/2 (2/6)
---------Hit - 1/2 (1/6)
---------Miss - 1/2 (1/6)
------------C shoots A - 1/2 dead (1/12)
------------C shoots B - 1/2 (1/12)
---------------A hits C - 1/3 (1/36)
---------------B shoots C - 1/2 (2/6)
------------------Hit - 1/2 (1/6)
---------------------A hits C - 1/3 (1/18)
---------------------A misses C - 2/3 dead (2/18)
------------------Miss - 1/2 (1/6)
---------------------C shoots A - 1/2 dead (1/12)
---------------------C shoots B - 1/2 (1/12)
------------------------A hits C - 1/3 (1/36)
------------------------A misses C - 2/3 dead

A shoots B
Hit - 1/3

Miss - 2/3

BLAGH Maybe I will try to figure this out later

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Is C a hot blonde?

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Megalyth said:

Do A and B both know that C never misses? In that case, A should shoot at C, because B will also shoot at C.

Edit: Either that, or A should shoot himself in the foot or something to disqualify himself because his aim is so bad.

Yes, everyone knows exactly how accurate everyone else is.

Maes said:

It's a bit fucked up here. I'm assuming the "truel" is divided in rounds, and that each gets one shot per round, and that each shooter may target each of the other (remaining) n shooters with a 1/n chance, right?

I fail at mathematical notation, but yes, they take turns and shoot the one they think gives them the best chance of winning the truel.

You are not allowed to shoot yourself. But, I will say this... you sort of have the right idea....

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If she was, she would then have a 1/97 chance of hitting the opponent.

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[Fanfare]

Memfis said:

Shoot in the air.

[/Fanfare]

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Hellbent said:

You are not allowed to shoot yourself. But, I will say this... you sort of have the right idea....

So..are you allowed to shoot nobody???

With these new rules it appears that shooting C is the only option. If unsuccessful, B will shoot C with 50% accuracy. If THAT is unsuccessful, C will shoot B...either way A gets a free shot at B.

On the other hand, if A shoots at B first, there's a 2/3 chance of not just handing himself over to C. So if A misses, then B will shoot C with 50% accuracy, again giving A a free shot.

So, A shooting at B means 1/3 chance of immediate win by C. A shooting at C means A will not receive any shots except by B, once C is eliminated.

NINJA EDIT: Oh well, I guess you can just shoot in the air :(

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TL; DR version for everyone: poorly posed problem with ambiguous semantics (at least the way Hellbent presented it) which can be bent to suit whatever solution works the best.

Having a suicide option, even if just for A, is still the best choice though, as it actually gives the best overall chances for A to win ;-)
However it implies that everybody is as inclined to suicide themselves. It would be interesting to see what happens if each shooter is also given the option not to shoot at anything and just "passes" his turn. Will tweak the sim code to allow for that option, too, although I don't know how exactly it would differ from a miss, technically speaking. I'll set it to a 1/(n+1) chance of happening (a shooter can choose to either try to shoot one of the other n shooters AT RANDOM, or shoot none).

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Hellbent said:

[Fanfare]
[/Fanfare]

I was there first!!!!! Sure, I had Stephen Fry provide the answer for me, but I still got that answer in first! xP

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Damn, I wish I could help with this but it's too much for my little brain to process.

Oh yeah, check your PMs. I've sent you a little 'treat' ;)

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He should actually throw away (or delope) his shot by firing into the ground, not the air.

Especially so if he's known to be a terrible shot, since a stray shot into the air could be interpreted as an attempt to hit B or C, while a clear shot into the ground could potentially signal that he is willing to end the duel honorably, without killing anyone. A lot of speculation about the famous Hamilton-Burr duel says that Burr might have misinterpreted Hamilton's intentions because Hamilton wasted his shot into the trees behind Burr, rather than into the ground, leading to his death. Of course, just as many people say that Hamilton was suicidal, or wanted to spite his opponent through death, so it's tough to say.

I guess the whole thing would also depend on how many rounds there are (i.e. does A fire again after C has fired?), as well as whether or not it's possible for the participants to end the dispute with more than one of them still standing.

(It's also probably safer to the public at large, though marginally. :P)

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Marnetmar said:

Damn, I wish I could help with this but it's too much for my little brain to process.

Oh yeah, check your PMs. I've sent you a little 'treat' ;)

Didn't get nothin.
D=

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Perhaps to Hellbent he was talking too?

Anyways, since C never misses, A would most likely have to shoot at C. With 1/3 chance of hitting, he may most likely miss. However, B, would also most likely shoot at C, since if B shot at A, and hit, C would win by shooting B. For A to have a chance at winning, both him and B must shoot at C. C will then shoot at one of them, most likely B, as he has a better chance at hitting him than A does. This will give A another shot. However, this would be the only chance A has to score a hit.

Now of course, if A hit C on shot one, B would still have that 50/50 chance of taking out A. If B misses, A has a shot, albeit possibly only one.

Either way though, of course, two of these guys are fucked. Shooting into the ground will only be a wasted turn, and then B will most likely shoot C, and if C hits B, and if A tries his ground 'Lets have Peace' shot or whatever, since that doesn't seem part of the rules, C will shoot and kill A.

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'A' waits for 'B' to aim at 'C' then aims at 'B' and says "bang" really loud, being the trickster that he is.
'B' misses his shot at 'C'
'C' shoots and kills 'B'
'A' grabs the quad and spams 'C' to death for the win.

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Here are some more simulation results: I added a variable shooting strategy for each shooter, and if using a "Shoot the most accurate of your foes" strategy for EVERY shooter, then yeah, C gets it badly and A gets just about his best chances.

• A: 36.7 %
• B: 41.09 %
• C: 22.21 %
However, this implies that ALL shooters use the same strategy, and not the one that's best for them. C is fucked by having to shoot last, even if he does hit his target everytime, regardless of strategy.

However, for what regards shooter A, the strategy of shooting C first is just marginally more efficient that everyone using a random shooting strategy with suicide included.

For A to waste his shot is not a good long-term option, because then he won't be able to kill the last man standing, beside himself, and will thus always lose (0 % chance of winning). Even if A changes his strategy after the first shot from "Waste a shot" to "Shoot randomly" or "shoot accurately", the odds are better but not optimal:
• A: 20.24 %
• B: 25.29 %
• C: 54.47 %
In any case, I hate problems where certain details and parameters are left ambiguous such as this one, because they often are posed in such a way that you'll never "get" the right answer unless you also happen to guess the particular -implicit- problem parameters the proposer had in mind.

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The best chance A has is to waste the shot...then B is forced to shoot at C (or he gets killed by C himself), if B hits then A gets a free shot at B and if B misses then C has to kill B (he has a higher chance of being killed by him) and A gets a free shot at C.

If A fired at B he risks being killed by C, if A fired at C he risks leaving himself open to a free shot from B.

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