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Topic: three-way pistol duel (Read 63196 times) |
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srn437
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the dark lord rises again....
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Re: three-way pistol duel
« Reply #100 on: Sep 23rd, 2007, 9:14pm » |
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doesn't that count as you getting hit?
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JiNbOtAk
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Hana Hana No Mi
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Re: three-way pistol duel
« Reply #101 on: Sep 23rd, 2007, 10:09pm » |
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Not if you miss.
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Quis custodiet ipsos custodes?
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srn437
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the dark lord rises again....
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Re: three-way pistol duel
« Reply #102 on: Sep 23rd, 2007, 10:29pm » |
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still, if b and c are infinitely smart, one of them might shoot you.
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shasta
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Re: three-way pistol duel
« Reply #103 on: Jan 19th, 2008, 3:29pm » |
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According to a strict interpretation of the wording of the puzzle; you should aim at the entire portion of the universe that doesn't contain the 100% accurate robot. 2/3rds of the time you will miss what you are aiming at, in this case resulting in killing your most deadly opponent on your first shot. Should you survive the other cyborg's shot at you, you then repeat the process, aiming at the entire universe that doesn't contain it until you hopefully win.
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vinylsiding
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Re: three-way pistol duel
« Reply #104 on: Feb 12th, 2008, 2:12pm » |
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In order to maintain the spirit of the puzzle, let's eliminate the loophole that each could survive by simply refusing to shoot. We state that after a period of time, if no one has attempted a shot, a survivor will be chosen at random.
Note: even with that natural condition, there are still instances where each is best off shooting straight at the ground. Consider the trivial case of p1=p2=p3=1.00.
NOW, onto the puzzle and arguably the most natural followup question.
For what values of p1, p2 and p3 will an optimal strategy be to shoot straight at the ground?
More generally, how do we partition the unit cube into the three regions that correspond to the three strategies available for A: shoot at the ground, shoot at B, shoot at C?
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towr
wu::riddles Moderator
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Some people are average, some are just mean.
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Re: three-way pistol duel
« Reply #105 on: Feb 12th, 2008, 3:23pm » |
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on Feb 12th, 2008, 3:20pm, vinylsiding wrote:In order to maintain the spirit of the puzzle, let's eliminate the loophole that each could survive by simply refusing to shoot. We state that after a period of time, if no one has attempted a shot, a survivor will be chosen at random.
Note: even with that natural condition, there are still instances where each is best off shooting straight at the ground. Consider the trivial case of p1=p2=p3=1.00.
NOW, onto the puzzle and arguably the most natural followup question.
For what values of p1, p2 and p3 will an optimal strategy be to shoot straight at the ground?
More generally, how do we partition the unit cube into the three regions that correspond to the three strategies available for A: shoot at the ground, shoot at B, shoot at C? |
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Is there really any need to repeat yourself four times? Let alone making that fourth one an hour after the first?
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| « Last Edit: Feb 12th, 2008, 3:24pm by towr » |
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Wikipedia, Google, Mathworld, Integer sequence DB
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Talabeh
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Re: three-way pistol duel
« Reply #106 on: Apr 5th, 2012, 6:13pm » |
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I propose the following working conjuncture (solution): Since the problem asks one to determine "whome/what we should shoot at to maximize one's chances of survival, it would be prudent to approach the problem from the neutral observer perspective (non-dueler).
If i am C, i know that since i can take out both A and B with a single shot, my best option would be to take out B - it has 50 percent accuracy- in the first around.
If i am A i know that the probability of hitting B and C is the same. Therefore, it makes no difference to whome A shoots at because his chances of survival is depended on the success and failure of both B and C.
Let also look at how B would reason. B is in the worst possible situation. No matter what option B takes, the odds are always stacked against it.
If B shoots at either A or C, its success of survival are always lower than both A and C. Thus, B's object of target is irrelevant to his survival.
From the above analysis, it follows that if both A and B shoot at C (combining their probability of hitting the target: 30% +50%), the probability is higher that they will hit C. Hence, maximizing their chances of survival. Since C is always in a win win situation. it would be pragmatic that A (and B) shoot at C!
It seems that most think that A shouldn't shoot at anything. Hence, maximizing his/or her chances of survival. However, given the nature of the puzzle, this solution is irrelevant (Re-Read the puzzle to see what i mean!!!!).
Just a thought!!!
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| « Last Edit: Apr 8th, 2012, 7:52am by Talabeh » |
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v_blade
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Re: three-way pistol duel
« Reply #107 on: Jun 1st, 2012, 8:04am » |
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You should shoot at neither of the two...
Case 1:
if you shoot at Cyborg100 and hit, there is a 50% chance that you will die in the next turn
if you shoot at Cyborg100 and miss, Cyborg50 will obviously shoot at Cyborg100. Even if he misses(then Cyborg100 obviously kills him) or hits, you have 33% chance of killing the one who is left.
Case 2:
if you shoot at Cyborg50 and hit... you're dead.
if you miss, then again you have 33% chance of killing the one who is left as above.
Cyborg100 and Cyborg50 will only shoot at each other.
So, you have best chance of survival if you miss, or rather not shoot either of them at all
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roady
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Re: three-way pistol duel
« Reply #108 on: Mar 27th, 2014, 6:56pm » |
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Running with the assumption that the cyborgs will always shoot when there are only 2 left.
But in otherwise still allowing a cyborg to pass their turn:
I get either, everyone passes, or you should not shoot at C giving a chance of survival of 5/12 (1/4 and 1/3 for A and B respectively) compared to a chance of 13/36 (5/12 and 2/9 for A and B respectively) if you were to shoot C.
How exactly the case where C died should be considered may have some effects on the result.
(it changes things somewhat if B is very likely to not shoot in the AB pairing, making it better to shoot C on the first turn in that case)
...
I basically drew out some kind of expectimax tree.
allowing some parts to loop back on themselves.
Setting values the case ABC survive and C passes turn, to X, and comparing it to the value obtained at the root. (which should be equal since they are basically the same state)
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| « Last Edit: Mar 27th, 2014, 7:35pm by roady » |
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