# Probability Puzzle - I disagree with their solution

I don’t think the puzzle assumes that a “miss” will kill an unintended target. But consider:

Black shoots at White and misses, or shoots the ground. Grey and White go at it, and Black gets a shot at the survivor before they fire back.

Black shoots at White and hits. Grey then gets a chance to fire at Black, with a 2/3 chance of killing him.

The difference is who gets to go first in the second round (Black vs Grey/White survivor). If Black forfeits or misses in the first round, he lives to get a shot at the last man standing and if he hits, he wins. If he kills White, he loses that shot and the odds are against him living to get another one.

On preview: or what Folly just said more succinctly.

I think people are reading too much into the puzzle.

The point is for the reader to realize when they do the math - that shooting at either person actually harms you. If this occurs to you, but you felt like it wasn’t a legit answer - you still passed!

By wording it differently - it limits the reader to come to this conclusion on their own.

There was a “similar” problem recently posted about Jeopardy - where it turned out - strategically - it was better for someone NOT to answer a question they knew - in order to have a chance of coming in second (I think this was the point). It is counterintuitive to think you shouldn’t answer a question you know the answer to - just as it is not to shoot at someone in a duel.

It does seem after review that the best course of action for all of the men is to shoot at nothing (assuming shooting at nothing is guaranteed to not hit either of the others, which I think is an assumption that FLIES IN THE FACE of the wording of the riddle).

I still contend that by the reading of the riddle, if your target is NOT PERSON A, and you have a 1/3 chance of hitting your target, then you have a 2/3 chance of hitting person A when making that shot.

The issue is if Black aims for white and actually hits him. Now Grey will aim for Black and with 2/3 probability, kill him.

If Black doesn’t shoot at anyone, then Grey will shoot at white and kill him with a 2/3 probability. Then Black will kill Grey with a 1/3 probability, etc.

If Grey misses White, White will kill Grey with 100% probability and Black will kill White with a 1/3 probability.

If you run through all the math, it makes sense for Black to miss everyone.

"If you run through all the math, it makes sense for Black to miss everyone."

I do see that now. Thanks all for elucidating that.

You can’t just mix pure logic and a real universe like that. For instance, if black shoots at gray and he has a 1/3 chance of hitting him and therefore a 2/3 chance of hitting white…
Does that mean he has a (1/2)*(2/3)=1/6 chance of hitting both of them? Is hitting both of them a possibility in this universe?

There are so many things wrong with this post that I don’t actually know what you’re saying. For starters, (1/2)*(2/3) does not equal 1/6.

You’re quite right. 2/6 or 1/3.

But anyway, I’m saying he’s inventing an empty universe where the only possible targets are grey and white. As someone else said, it’s not cheating to assume you are on earth with earth-like conditions.

Setting aside the ambiguity OP draws attention to, and assuming the firing order is by increasing probabilities which are
0 < p < q < r < 1
For which (p,q,r) is it best for weakest shooter NOT to aim at the ground? (I don’t know the answer but am afraid the solution is some very complicated inequality.)

I think this is the problem. You’ve phrased the puzzle in such a way that it prohibits the answer that you’re supposed to come up with.

First, you say it doesn’t list the ground as a possible target. But I think you’re wrong. The puzzle you wrote says “If you are Mr. Black, where should you shoot first for the highest chance of survival?” There’s nothing here that says Mr. Black has to shoot at another person. “At the ground” is a perfectly acceptable answer to “Where do you shoot?”

If the puzzle had said “If you are Mr. Black, who do should you shoot at first for the highest chance of survival?” then you’d have an argument that the puzzle requires you to shoot at a person.

Second, there’s the issue of what “hits his shot” means. That is imprecisely defined in your puzzle and does leave open the possibility that a person could aim at the ground and miss and perhaps hit himself or another person. So you’d have been better off writing “Mr. Black, who can hit a person 1/3 of the time, gets to shoot first.” That way, it’s clear that the difficulty of hitting a person does not translate into an equivalent difficulty in hitting any target.

There are a lot of assumptions in a question like this that are not spelled out. For example, it apparently is an assumption that killing or missing someone aimed at are the only possibilities, though in real life most shots wound. We are also assuming that the participants are motivated only to maximize their survival, although in practice other considerations might come into play, such as getting revenge on whoever just shot at them. The scenario is also quite artificial: A thiree-way gunfight, with people taking turns to give the weaker shots a chance, has never occurred. In such an artificial situation, the failure to inform us that shooting elsewhere is an option should be seen as a significant oversight. I do agree, however, that missing the ground just means that nonground parts of the environment will be struck, and the possibility of aiming at the ground and hitting another participant may be disregarded, just as the possibiity of aiming at one participant and hitting another is disregarded.

It’s helpful to get a sense of just what the chances are that Mr. Black will survive a given situation. Suppose, for example, that the gunfight ends up as a two-person fight with Mr. Gray, in which Mr. Black shoots first. There is a one-third chance that Mr. Black will kill Mr. Gray with his first shot. There is a four-ninths chance that Mr. Gray will kill Mr. Black with his first shot (two-third chance that Mr. Black will miss, times a two-thirds chance that Mr. Gray will then hit). There is a two-ninths chance that neither will hit, in which case the round will be repeated. Thus, if Mr. Black goes up against Mr. Gray and gets to shoot first, he has a three-sevenths chance of survival - not great, but better than some alternatives.

Now, suppose that Mr. Black shoots at one of the other antagonists for his first shot. If he shoots at Mr. Gray and misses, it’s the same as if he shot at the ground. If he shoots at Mr. Gray and hits him, Mr. White will then shoot him dead. Not a lot of upside there.

Suppose instead Mr. Black shoots at Mr. White. Again, if he misses, he might as well have shot at the ground. If he hits Mr. White, then, then Mr. Gray will try to shoot Mr. Black. Mr. Gray has a two-thirds chance of killing Mr. Black with his first shot. If he misses, then the situation is like that already described in which Mr. Black shoots first. Thus, the chance of surviving a shootout with Mr. Gray, in which Mr. Gray shoots first, is 3/21 or one-seventh (one-third chance of surviving the first shot, times the three-sevenths chance of surviving when Mr. Black shoots first).

And the sum total possible outcomes if Mr. Black aims at Mr. White? There is a two-ninths chance that Mr. Black and Mr. Gray will both miss, Mr. White kills Mr. Gray, and Mr. Black gets to shoot again at Mr. White, whom he must hit (with a one-third chance) or he will die. There is a one-third chance that Mr. Black will kill Mr. White, in which case Mr. Gray will then shoot at Mr. Black. There is a four-ninths chance that Mr. Black will miss, Mr. Gray will kill Mr. White, and Mr. Black will get the first shot at Mr. Gray. The sum chances of survival are (2/9)(1/3) + (1/3)(1/7) + (4/9)(3/7) = .3122, if I haven’t made a mistake with my calculations.

What about if Mr. Black fires at the ground (and hits or misses, but does not hit Mr. Gray or Mr. White)? Then there is a two-thirds chance that Mr. Gray will kill Mr. White, in which case Mr. Black will have the first chance to shoot Mr. Gray, and a one-third chance that Mr. White will kill Mr. Gray, in which case Mr. Black will have a chance (there’s only one) to shoot Mr. White. The probability sum is (2/3)(3/7) + (1/3)(1/3) = .3968.

So Mr. Black should aim for the ground, if this is an option.

Actually, I like drew’s alternate answer, rephrased thusly: “Mr. Black should aim at the birthday balloons at his 200th birthday party.” As long as the length of the turns of the duel isn’t specified, he might piss off the other duelers so much that they challenge him to another duel, but at least he won’t die in this one.

This isn’t true. Black has a unique characteristic that Grey and White do not have; Black is not a primary target for anyone. Grey is the primary target for White and White is the primary target for Grey and Black. So Grey and White have a different strategy than Black does.

The thing is, the entire point of these types of puzzles is a form of nitpickery. Since they didn’t say “Which person should he shoot?” you’re supposed to have another option, e.g., the ground. But, in doing this, they also open up other forms of nitpickery, including the fact that the puzzle never said that someone was better at hitting the ground than other things.

You can’t have your cake and eat it, too. Either you need to take the most natural reading of the puzzle, which limits your shooting choices to the two other people, or you can take a rather technical interpretation that both allows you to shoot the ground, but also means that shooting the ground has a 2/3 failure rate.

To put it more simply: this is a gotcha question, so the answer is also open to a gotcha.

If I was going to take any issue with the wording of the original puzzle, it’s the definition “take turns shooting at each other”

This opens up the interpretation that Black’s shot at the ground isn’t a turn because he didn’t shoot at another person. We’re left to infer from usual duel rules that a missed shot (of any sort) constitutes a turn even if it doesn’t match the literal description.

If you remove the words “at each other” and simply state “take turns shooting” then Black’s shot at the ground clearly meets the definition of a turn.

Maybe I’m missing something, but in the original answer they gave, why do they take it as a given that Mr. White will shoot back at Mr. Grey and not shoot Mr. Black? I guess they assume White would want revenge since Grey shot at him?

But if White picks Black, then Black’ll die in the first round and he didn’t even attempt to shoot anybody else.

… and yes I agree that the question specifically says they shoot at each other, so shooting at the ground shouldn’t be allowed.

I worked this through with numbers before reading the given solution or the OP’s nitpick. I didn’t consider the possibility of shooting the ground, so obviously shooting White dominates shooting Green, since killing Green guarantees Black’s death.

In terms of numbers, I ran them with two different assumptions. One is that Black has no idea who the others will aim at (maybe they don’t know the accuracies of each dueler) so it’s 50-50. In this case, Black has an 11% chance of surviving if he shoots at Green, a 16% chance of surviving if he shoots at White, and a 17% chance of surviving if he shoots the ground. So it hardly matters between White and the ground.

However, assuming that Green and White know the accuracies and so target each other first, the odds of Black surviving are 26% if he shoots at Green, 31% if he shoots at White, and 39% if he shoots the ground.

Long story short, I agree with the published solution, but I think it’s interesting that Black has a much better chance if he makes sure his opponents know he’s a lousy shot.

They assume this because Grey is twice as likely to hit as Black (2/3 compared to 1/3). So if you calculate the probabilities on all of White’s options, he’ll have a better chance of living by killing Grey first.

Yeah but that’s an assumption. Maybe Black’s a huge jackass and forced them all into this, and once he’s dead White and Grey will go off and have a nice meal together. EDIT: and as Bad Astronaut brings up, the problem doesn’t state whether the shooters know each other’s accuracies.

The problem doesn’t state that all the players make the “best” decision. So as phrased, White could shoot Grey and screw up their “official” answer.

Maybe:

p > (qr - rr + 2qrr - qqrr + qr) / (qqr - qq - rr + 2qrr - qqrr)