I suggest that if you do not want to know the answer, you wait until after you've taken your shot at the puzzle to scroll down, as I'm sure that answers will be discussed/revealed in the thread. So, yesterday, my philosophy professor posed a pretty good logic problem, and it took me a pretty good amount of time before i figured out the answer. Here it is. Of three prisoners in a certain jail, one had normal vision, the second had only one eye, and the third was completely blind. All were of at least average intelligence. The jailer told the prisoners that from three white hats and two red hats he would select three hats and put them on the prisoner's heads. When they were brought together, each was prevented from seeing what color hat was placed on his own head. Except for the blind man, each could see the hats of his fellow prisoners. The jailer offered freedom to the prisoner with normal vision if he could tell what color hat was on his head. In order to prevent the prisoners from guessing, the jailer told them that the wrong answer would be penalized with death. The prisoner with normal vision said that he could not tell what color hat he was wearing. After hearing the first man's response, the one-eyed prisoner was offered the same deal but he too said that he could not tell what color hat was on his head. The jailer, half-jokingly, offered the same deal to the blind man. The blind prisoner then smiled broadly and said: "I do not need my sight; from what my friends with eyes have said, I clearly see my hat is ______________ ." So, can you fill in the blank, and also figure out how it is that the blind man knew the color of his own hat? If you guess, you might get it right, but if you can't explain the logic behind the blind man's answer, then you really have not figured out the puzzle and should take no credit for having guessed correctly. Hint: There is only one combination of hats that answers this question correctly. Hint#2: There are no stupid tricks in this puzzle It took me a LONG ass time to figure out this puzzle, and then it suddenly hit me. Good luck

I'm guessing his hat was red, that the first two guys had white hats. I don't know how to logically tackle such a problem so I'm just guessing.

I think he had a white hat. 3/5 probability that the blind man got a white hat, If he was the first prisoner given a hat. 2/4 or 3/4 probability of the blind man being second and getting a white hat, and 1/3, 2/3, or 3/3 probability that he got a white hat if he was the last prisoner to get a hat. I think? I think he has a white hat but I'm still working on why exactly.

The blind prisoner knows for sure he isn't wearing a red hat, so therefore he must be wearing white. The 1st prisoner does not see 2 red hats or he'd know he was wearing white. The 2nd prisoner does not see 2 red hats or he'd know he was wearing white. The 3rd prisoner knows he isn't wearing red because that would mean the other two prisoners both would have white hats on. If it were White-White-Red, Prisoner 2 could tell his was white after Prisoner 1 says he doesn't know. If Prisoner 1 saw White-Red, then Prisoner 2 would see White-Red also. And he'd know he couldn't be wearing a red hat or Prisoner 1 would've answered white. All other combinations leave Prisoner 3 in a white hat.

The answer I came up with is that they are all wearing white hats. Here's my explanation: and for the sake of convenience... Two-eyed man=2 One-eyed man=1 Blind man=0 2 looks at 1 and 0 and sees white and white. From this, he cannot infer anything because there are still white and red hats remaining, so he gives up. Then, 1 looks at 2 and 0. He too sees white and white, once again leaving both white and red hats as possibilities. So, he too surrenders. 0 can infer from the two others' responses that he must be wearing a white hat... because if either of the other men had seen any red hats, then the competition would've ended before it even reached 0. It's very difficult to explain... here's pretty much my entire thought process explained in a good amount of detail so that you can follow where i went with this. The possible situations include: -2 looks at 1 and 0 and sees two red hats. 2 knows he is wearing a white hat. -2 looks at 1 and 0 and sees both a white and a red hat. From this he can infer nothing. Then, 1 looks at 2 and 0 and sees both a white and a red hat. If this were the case, 1 would realize that 2 must have not seen two red hats, (otherwise he would have known he was wearing white), and therefore can infer that he is wearing a white hat. -2 looks at 1 and 0 and sees both a white and a red hat. From this he can infer nothing. Then, 1 looks at 2 and 0 and sees two red hats. In this case, he would know he was wearing a white hat. -2 looks at 1 and 0 and sees both a white and a red hat. From this he can infer nothing. Then, 1 looks at 2 and 0 and sees two white hats. From this he can infer nothing because it leaves both white and red hats as possibilities. -2 looks at 1 and 0 and sees two white hats. This leaves both white and red hats as possibilities, and so he can make no inference. 1 looks at 2 and 0 and sees two white hats. He too can make no inference as to the color of his hat-- because white and red hats are still possibilities. Then, 0 is asked if he knows the color of his hat. Based on the fact that neither of the prisoners before him were able to recognize the color of their own hat, he is able to infer that they did not see any red hats. He knows that if 2 had seen one red hat, he wouldn't have been able to make an inference... but then when the question reached prisoner 1, 1 would've recognized that he must be wearing a white hat because logically... prisoner 2 would've known the color of his own hat if 1 was wearing red. Therefore, the only way for the question to reach the blind man before being answered is if both men saw only white hats. Because of the process of elimination, the blind prisoner knows that he is wearing a white hat. Pretty cool logic puzzle, and very straining on the mind, lol

you know, I've been thinking more about this... maybe the problem does contain more than one solution for the combination of hats being worn. I'm gonna map out a little chart using the numbers I used before to refer to the prisoners and W and R to refer to white and red. This shows all the possible combinations of hats that can be distributed 2 | 1 | 0 -------------------------- 1) W | W | W 2) W | W | R 3) W | R | W 4) R | W | W 5) R | R | W 6) R | W | R 7) W | R | R Corresponding to above numbers 1) 2 sees two white hats. He can't infer anything because both white and red hats are still possibilities, and therefore passes to 1. 1 also sees two white hats, and cannot infer anything because white and red hats are still both possible for the color of his own hat. 0 then knows that he must not be wearing a red hat, otherwise 1 would recognize that 2 did not see two red hats, and was therefore himself wearing a white hat. original solution 2) here, 2 sees a white hat and a red hat. He can't infer anything so he passes to 1, who sees a red and a white hat. Since 1 knows that 2 must not have seen two red hats (otherwise he would have known the color of his own hat), he knows the color of his hat is white. 3) 2 sees a red and a white hat. He can't infer anything and passes to 1, who sees two white hats. 1 recognizes many things about the situation, but ultimately would be forced to guess between red and white if he answered... so he passes to the 0. 0 realizes that given the situation, he cannot feasibly be wearing a red hat because if he was, then 1 would have answered the question correctly. sweet, i knew there was another possible combination. 4) 2 sees two white hats. He can't infer anything so he passes to 1, who sees a red and white hat. 1 cannot make an inference either because he could potentially be wearing a red or white hat (remember, 1 is under the assumption that two red hats are potentially in the field, because 2 couldn't see that he himself was also wearing a red hat). Therefore, 1 passes to 0. 0 recognizes that if he was wearing a red hat, the question would have been answered by prisoner 1. Therefore he knows he's wearing a white hat. damn, another one. i thought i had this whole problem figured out til I did this. 5) 2 sees a red and a white hat. He cannot infer anything and passes to 1, who also sees a red and a white hat. Based on what he knows, 1 cannot know 100% the color of his own hat. He cannot be sure if 2 saw red and white, or white and white-- so he passes to 0, who knows that he must be wearing a white hat, otherwise 1 would have been able to figure out the color of his own hat. 6) 2 sees a white and a red hat. He can't infer anything and passes to 1, who sees two red hats and knows he is wearing a white hat. 7) 2 sees two red hats. Game over, son. So apparently there are three possible combinations that will solve the puzzle. Care to prove me wrong or agree with me? good chance there's a flaw somewhere in all that thinking. by the way, this was an assignment for philosophy that i did blazed while on grasscity (this post in particular was the assignment). assignment directions were to list all possible combinations of hats and the outcomes of their situations. edit: anyone besides me keep thinking white rhino and white widow seeing all those Ws and Rs next to eachother? WR, WW... damn... I probably should be thinking more like World War...

i just rediscovered this old ass thread... i want more logic problems like these. i need to take another logic course. this shit was fun as hell. maybe some other people can be entertained by this. lol i'm finding tons of errors in my solution though. i don't think there are actually multiple solutions-- i was probably just baked and forgot that 0 is blind. i'm gonna proof my solution tonight or something.

1. If P2 and P3 both have red hats, then P1 knows what hat is on his own head. 2. It isn't the case that P1 knows what hat is on his own head. 3. It isn't the case that P2 and P3 both have red hats. 4. If P3 has a red hat then P2 knows what hat is on his own head. 5. It isn't the case that P2 knows what hat is on his own head. 6. Therefore it isn't the case that P3 has a red hat on. Thus we conclude P3 has a white hat on.