Talk:Prisoners and hats puzzle

message: At the suggestion of Ludraman, I looked over this article. Not being an expert in logic or logic puzzles, I can't speak much to that aspect. I will observe that the diagram characters have an amusing resemblance to Pacman (the yellow, probably) and that the hats might be colored in such a way as to indicate indeterminacy. Red and blue polka dots maybe?
:Ah, I was wondering why it looked familiar. Pacman, of course! Arvindn 17:19, 19 Apr 2004 (UTC)

message: Is the diagram okay? (BTW its resemblance to pacman is not intentional). When I was writing the article I thought a diagram would help but I'm not much of an artist and scribbled something in Paint that would  do for the time being. Would anyone be willing to make a better diagram? LUDRAMAN | T 02:16, 21 Apr 2004 (UTC)
replies: text: Has been fine for 8 years! Who would have thought your humble diagram would be doing the rounds in 2012 on the yet-to-be-invented Facebook? Proof enough that it's suitable, methinks. LightYear (talk) 05:22, 22 February 2012 (UTC)
thread_title: Diagram

message: I frequently fall into traps in logic puzzles, so I invite all and sundry to explain to me why I'm wrong.  That said, is it necessary for B to determine whether or not C is going to act? Wouldn't B (or whichever person saw that the other two were wearing the same color hats, as two of them must be) immediately notice the other hats' colors, and act accordingly?
:Yep, it's necessary. Remember, B can only see A's hat. C is behind him (and the prisoners aren't able to look backwards--this is an important rule), and of course the fourth man is behind the screen. The crux of this game is that even with empirical evidence that doesn't seem strong enough to guarantee an answer, B can infer from C's (assumed completely rational) action what the answer must be. Thus the prisoners can always go free. --140.103.133.75 08:02, 18 July 2005 (UTC)
thread_title:  Is waiting necessary? 

message: I was trying to solve this puzzle, and want to clarify something.  It says "The prisoners can see the hats in front of them but not on themselves or behind."  Now, based on the diagram, and the description, this means B can only see A's hat, A can see none (since he's in front), and I assume that C can only see B's hat, correct?  This is not clarified, i.e. can C see B AND A's hat?
If C can only see B's hat, and B only A's, perhaps the wording should be:  "The prisoners can see the hat of the prisoner directly in front of them, but not themselves or behind."  Before I make the change, I want to be sure I'm interpreting the description correctly.  GregTheVirus 23:25, 26 July 2007 (UTC)
replies: text: I think it's unambiguous in the current wording, particularly because each prisoner can see the hats in front of them. LightYear (talk) 05:22, 22 February 2012 (UTC)
thread_title:  Clarification 

message: The variant with three hats fails to say that the prisoners are aware which color is used three times and which only one. This information
is vital to make the case where all three (A, B and C) have the same color hat trivial. As far as I can tell, there is no solution,
if they do not have this information. Correct me if I'm wrong.. or correct the example otherwise.  —Preceding unsigned comment added by 128.84.98.99 (talk) 20:36, 16 February 2010 (UTC)
thread_title:  A different hat puzzle 

message: I felt like I was "tricked" despite it claiming not to be a trick question. Shouldn't the problem mention that the men are aware of the position of their fellow prisoners? I assumed they were completely oblivious to what was behind them. Perhaps I am just being a sore loser, but it seems like not enough information is provided, and the problem should be edited to include this information. Any thoughts? 68.59.3.58 (talk) 05:53, 29 May 2009 (UTC)
replies: text: Clarified it a little so it now sounds like the jailer explains the important rules to the prisoners. So each of the prisoners know as much as the reader. LightYear (talk) 05:22, 22 February 2012 (UTC)
thread_title: Enough Information?

message: As this puzzle is stated, the prisoner could just say the color of the hat of the next prisoner, instead of all that binary - deduction stuff. That's my plan if I ever go to jail anyhow.  —Preceding unsigned comment added by 85.167.15.96 (talk) 15:16, 20 May 2010 (UTC)
replies: text: That doesn't guarantee anything. The next prisoner will know his hat color, sure. But you only have a 50% chance of having that hat color yourself. But oh you, say: the prisoner behind me said my hat color too. But that doesn't help at all, because what if the color he said is different from the color of the hat in front of you(this will happen 50% of the time). Which color should you say then? If everyone is following the plan, then you are guaranteed to die if you say the color of the hat in front of you. But if you don't say the color of the person in front of you, then he will die. Knowing this, the logical decision would be to screw the guy in front of you and say your own hat color. Now assuming everyone can figure this much out, you can't trust that the person behind you is following the plan at all. That means your strategy would be completely useless. Assuming everyone followed your plan exactly, everyone would have exactly the same chance of survival as if they had no plan at all: 50%. With a slightly modified version of your method, you can guarantee that at least 50% of the people will survive as follows: The person in the back says the color of the hat in front of him. that person says this color. then the process repeats. That guarantees that half of the prisoners will survive and that the other half have a 50% chance of survival. And that's about the best you can do with anything close to the method you described. Maybe the article description needs to be written better to be more understandable. The back prisoner is just telling the others if there is an even(blue) or odd(red) number of red hats. Then the other prisoners just have to pay attention to how many people (ignoring the first) say they have a red hat. Then they know all the information they need in order to correctly decide their own hat color, because they know then how to interpret the line in front of them. If the running total of reds(excluding the first prisoner and including the hats they can see) is odd, then they say the opposite of the first prisoner, and if it is even, then they say the same as the first prisoner.--24.24.142.225 (talk) 04:21, 30 October 2010 (UTC)

text: It is listed as a variant of the original puzzle, and that was a requirement of the original puzzle, so it has been assumed here as well. If the puzzle was being described on its own, then that would have to of course be stipulated.--24.24.142.225 (talk) 04:23, 30 October 2010 (UTC)
thread_title: Ten hat puzzle

message: Hi,
The part of the original article that says "better solution to the countably-infinite variant" is not really better at all!
Why ?  Because the "better solution" allows the prisoners to HEAR what the first person says, but the variant being discussed explicitly disallows this, so it's toe tootly different things.
I'm glad to discuss this:  eric dot osman at rcn dot com    mar 16 2011  — Preceding unsigned comment added by Ericosman (talk • contribs) 17:20, 16 March 2011 (UTC)
thread_title:  faulty advertising 

message: "When they are put into their line, each prisoner can see which equivalence class the actual sequence of hats belongs to." This is obviously false, as stated. The first prisoner can't see a goshdarned thing. I guess that this sentence is actually meant to be correct, and the problem statement is wrong -- seems to me it all works if everyone can see every hat. JoDu987 (talk) 22:20, 16 September 2017 (UTC)
replies: text: The first prisoner can see all of the other hats. They're at the back of the line. The setup is that prisoner n can see the hats on prisoner m for all m>n, not for all m<n. I think the problem description already says this (each prisoner is said to face away from, not towards, the start of the line) but maybe this could be made more clear with more description or a diagram. 2601:645:8302:AB26:F8F3:53EE:3858:5C5C (talk) 22:20, 14 October 2017 (UTC)
replies: text:  Cheers. I'm going to try editing the solution to de-confuse people like me. JoDu987 (talk) 14:21, 15 October 2017 (UTC)
thread_title:  Countably Infinite Hat Problem with Hearing 

message: I propose that Prisoners and hats puzzle be merged into Hat puzzle. The two articles are about the same type of logical problem. There is no need to have two articles on the (almost) identical topic. Vanjagenije (talk) 20:44, 6 October 2015 (UTC)
replies: text: Agreed, using the initial direction of the proposal (supported by most).  Klbrain (talk) 12:38, 28 November 2017 (UTC)
thread_title:  Merger proposal 

message: " In fact, even if we allow an uncountable number of different colors for the hats and an uncountable number of prisoners, the axiom of choice provides a solution that guarantees that only finitely many prisoners must die. The solution for the two color case is as follows, and the solution for the uncountably infinite color case is essentially the same:"
I don't think this is true. The *exact nature* of an uncountably infinite set is that it cannot be placed in a sequence. There can be no equivalence relation and no representative sequences if there are no sequences at all. I am deleting the references to uncountably infinite extensions of the problem until whoever added it can explain what they mean.  — Preceding unsigned comment added by 73.140.114.118 (talk) 22:23, 18 October 2016 (UTC)
replies: text: An uncountable set can't be "sequenced" in the sense of being indexed by the natural numbers, but it can be "sequenced" by indexing on an uncountable ordinal, and such sequencing is not necessary anyway. If we think of the hat placement as a function from prisoners to colors, we can say that any two hat placements are equivalent if they differ on only finitely many prisoners, and we get the necessary equivalence classes. No sequencing is necessary. 2601:645:8302:AB26:F8F3:53EE:3858:5C5C (talk) 22:06, 14 October 2017 (UTC)
thread_title:  Countably Infinite Hat Puzzle without Hearing