There are 3 black hats and 2 white hats in a box. So he calls them all 3 to his office and tells them of the situation. Three are red, and two are white. Each prisoner has to answer at the same time, but can choose to remain silent. Prisoner P1 can see P2 and P3. Here's a logic puzzle I like to give my students: _____ There were three prisoners in certain jail, one of whom had normal sight, one of whom had only one eye, and the third of whom was completely blind. each prisoner would turn around and have a hat placed upon his head so that they could see eachothers hat but not their own. The original question is here. The jailer tells them that there are two black hats and two white hats; that each prisoner is wearing one of the hats; and that each of the prisoners is only able to see the hats in front of them but not on themselves or behind. But I am not able to understand the logic. You see, on one day every year, the King would put the names of all the prisoners in a hat and draw out 3 names. There is a similar question here but with four prisoners. Lets call them Andrew, Bernard and Charlie. This would be for him to place 3 red hats and two blue hats into a bucket. He puts one hat on each of the prisoner's heads, putting out three in total. He tells them the only fair way he knew. They are all sitting facing towards Andrew. Since there are 3 Back Hats (B) and 2 White Hats (W), the various combinations possible include: BBB (Meaning all 3 prisoners wear black hats) BBW BWB WBB WWB WBW BWW So he calls them all 3 to his office and tells them of the situation. Imagine that 3 people are in one room. One such puzzle is when you have 3 prisoners, each with a random hat. The original Three Prisoners problem can be seen in this light: The warden in that problem still has these six cases, each with a 1/6 probability of occurring. He tells them the only fair way he knew. He has five hats. To these three prisoners he would present a riddle. The third prisoner has no eyes. If the two front hats were black, the third wise man would have called out the colour of his hat as white immediately. My solution is half way through. The jailer liked to play tricks on his prisoners, so one day he brought them all… Prisoner P2 can see P3 only. The prison guard plays a game with the prisoners. The jailer tells them that there are two black hats and two white hats; that each prisoner is wearing one of the hats; and that each of the prisoners is only able to see the hats in front of them but not on themselves or behind. However, the warden in that case may not reveal the fate of a pardoned prisoner. The hats are given randomly, and there's no maximum amount of hats of any color. This would be for him to place 3 red hats and two blue hats into a bucket. The hats are given randomly, and there's no maximum amount of hats of any color. The Puzzle of the 3 Hats. Prisoner P2 can see P3 only. Each scenario has a 1/6 probability. By John Tierney ... A very nice riddle. The second has one eye. Ok, here's the riddle: There are three prisoners. If the first hat were black, and the third man did not call out any colour, the second wise man could deduce that his hat was white. Riddle courtesy of William Wu’s riddle archive. Imagine that 3 people are in one room. Prisoner P1 can see P2 and P3. Lets call them Andrew, Bernard and Charlie. Each prisoner has to answer at the same time, but can choose to remain silent. One such puzzle is when you have 3 prisoners, each with a random hat. They cannot see what color hat they have chosen. Three men (we will call them A, B, & C) each reach into the box and place one of the hats on his own head. The only permutation where the 3rd prisoner could have a RED hat and the 2nd prisoner NOT have a WHITE hat is permutation 1, which is already ruled out already in (1). They are all sitting facing towards Andrew. each prisoner would turn around and have a hat placed upon his head so that they could see eachothers hat but not their own. The hats are either black or white. That is, if the 3rd prisoner has a RED hat, the 2nd prisoner would know that either permutation 2 or permutation 3 is correct, and either way, his hat must be WHITE. The first one has two eyes. The hats are either black or white.