Probability Puzzle - I disagree with their solution

These answers all presume that each shooter knows the reliability of the others’ shots. So Mr. Black has gotten into a three-way duel (I don’t like “truel”) with someone who is twice as good as him, and someone else who is thrice as good as him, and he knows it.

Clearly, Mr. Black is already a gambler with a death wish.

I say he shoots Mr. Gray, knowing that he’ll either miss or wound him, really pissing him off, but that Gray will strategially have to shoot White. If he doesn’t kill Gray, White will, and Black has the next shot. If he does kill Gray, White will kill him but it will fulfill the same weird little cowboy fantasy that got him into the situation. Win-win.

After talking more and more about this with one of my very smart friends, I am more convinced than ever that I am right in my initial analysis.

Intending to shoot Mr. Gray would result in a 1/3 probability of hitting him, and a 2/3 probability of missing. Intending NOT to shoot Mr. Gray would result in a 2/3 probability of hitting him, and a 1/3 probability of missing him.

Therefore, the best course of action is for Black not to shoot at all (or call his intended shot as "neither Mr. Gray nor Mr. Black which would result in the same probability of intending to shoot either of them), hope that he misses, hope that Gray shoots at White and kills him, and then his next shot should be “I intend not to hit Mr. Gray” which would result in a 2/3 probability of hitting Mr. Gray. If Gray does not kill White, then White will kill Gray on his next hit, thinking he’s more of a threat, but forgetting that Mr. Black can just then say “I intend for my next shot to not hit Mr. White” and thus have a 2/3 chance of hitting him. Either way, Black has pretty good odds of living.

you cannot be serious. the best way to hit something is to miss?

So even though all these bizarre probabilities are in play (e.g. black choosing to deliberately miss means he has a 2/3 chance of hitting), the players are unaware of this? This is the solution with the least amount of assumptions on our part?

I think this is really stretching the logic. If I’m at target practice, and I can only nail my mark 1 out of 10 times, aiming at not the mark doesn’t magically make me hit 9 out of 10 times.

This is not “real life” this is a logic puzzle and you follow the rules of logic. The wording is “makes 1/3 of his shots” which any reasonable interpretation would mean “his shots land where he intends them to land 1/3 of the time.”

If he intends for his shot land at Mr. Gray, it will have a 1/3 chance of happening. If he intends for his shot to land at Not Mr. Gray (IE, anywhere else, IE calling out “I intend for my next shot to miss Mr. Gray”) then by a logical conclusion of the rules of the riddle, he has a 2/3 chance of missing his shot and hitting Mr. Gray.

Ok. formal logic time then.

Or the way you’re phrasing it, possibly denying the antecedent

I aim at Grey -> then I miss him 2/3ds of the time.
(not) I aim at Grey -> then I hit him 2/3ds of the time.

No, even within the framework of the puzzle, this isn’t true. You’re excluding the middle.

You can’t say “I’m aiming at the ground and have a 2/3 chance of missing it, therefore I have a 1/3 chance of hitting Mr Gray.” The real odds are you have a 1/3 chance of hitting something that isn’t the ground - which could be Mr Gray or Mr White or yourself or you own gun or Mr Green who lives down the street or the sky or the Moon.

As for saying “I’ll just aim at everything that isn’t Mr Gray. Being as I miss what I aim at 2/3 of the time, I’ll have a 2/3 chance of hitting Mr Gray.” that’s another false premise. You have a 2/3 chance of missing a target; “everything that isn’t Mr Gray” is not a single target.

Put it this way. We’ve established Mr White always hits what he’s aiming at. Can he end the duel on his first shot by deciding “I’m aiming at everyone I’m dueling with.” and then hitting both Mr Black and Mr Gray with a single shot?

Thank you. Otherwise, suppose Mr. Purple shows up who hits 0% of his shots. Therefore he could hit 100% of his shots simply by trying to hit something else other than the target.

By the way, although the puzzle does in some sense involve probability, it is actually a game theory problem.

I make 90% of my free throws. That means whenever I drop a ball, one out of ten flies off into space.

Does he need to say it out loud, or can he just think it? Which way does he point the gun? :rolleyes:
If you choose to use this kind of nonsensical “logic” on a logic puzzle, why bother?

Would a literal reading of “at each other” mean that Black (a) can’t shoot at the ground or whatever, but (b) should fire his first shot at himself?

Black probably misses; Gray then shoots at White, and probably kills him; if so, Black then shoots at Gray and possibly gets the win.

This isn’t really a logic puzzle nor does it have a trick answer. It’s really just a simple application of probabilities.

As another example of how to overthink a problem:

If a chicken and a half lays an egg and a half in a day and a half, how many eggs will three chickens lay in three days?

  1. Three
  2. Six
  3. It depends. Maybe the farmer got hungry on the second day and ate one of the chickens.
  1. Half-chickens can’t lay eggs.

Half-chickens can too lay eggs. It just needs to be attached to another half-chicken at the time.

I don’t think real world assumptions like that are allowed in this thread.

I revive this old thread for two reasons.

First, I would like to thank drewtwo99 for starting it. My sister is recovering in the Baptist hospital in Winston-Salem, North Carolina from serious back surgery. She is a computer science professor and loves lateral thinking and math puzzles. My idea was to find a puzzle my sister had never encountered before for her to ponder while recovering. I was searching through the many great threads on this theme that have showed up on the SDMB over the years when I stumbled upon this one. Many hours of debating this puzzle with her and my brother followed. It was especially nice as none of us had encountered this puzzle before even though we all are fans of the genre and know most of the favorites. I believe that this eight month old thread provided as great of a distraction from my sister’s pain as did the morphine. I love this place.

Secondly, I wanted to post the variation that we never resolved. Suppose Black, Grey, and White are all excellent shooters who never miss. Each has a gun with three bullets. However, Black 's gun has only one bullet and two blanks. Grey 's gun has two bullets and one blank. White 's gun has three bullets and no blanks. The bullets and blanks are loaded randomly in the guns, and each of the shooters knows the number of bullets and blanks in all the guns. Black gets the opportunity to shoot first because he has the fewest real bullets. Grey follows and then White and back then again to Black. Being dead, as in the original puzzle, means the duel skips to the next shooter. The duel continues until only one man is left alive. Our question : Assuming each man is logical, who has the best chance of winning this duel? Note: each shooter’s gun must either shoot a bullet or a blank per turn.

Must each shooter shoot at a person on each turn, or may they “shoot at the ground”?

In the latter case, there is, at least a priori, a possibility of ending in a state where all the bullets have been exhausted but more than one shooter remains alive. Does a shooter have any preference between remaining alive alone and remaining alive along with one, the other, or both opponents, and if so, how strong is this preference?