DeletedUser
Not really very D&D but I came across this problem in a book - I could not figure out the answer myself but when I cheated and read it I was amazed. It is not a trick question, but a surprising and logical solution.
There are 50 prisoners. Their evil captor announces that the next day they will all be buried up to their necks in a line and facing the same way down the line, and either a white or black hat will be placed on each one's head. No one can see his own hat, but the first prisoner can see the hats of the 49 in front of him, the next can see the 48 if front of him, and so on down to the last prisoner, who obviously sees none of the others.
The first prisoner will be asked to guess the colour of his own hat, and if he is right he will be released - if not, he will be shot. He can only say "white" or "black" and he is within earshot of his nearest neighbours. Clearly, each prisoner has a 50:50 chance of freedom if he guesses randomly.
One of the prisoners is a mathematician however and he devises a scheme whereby, if he goes first, he can save all the other prisoners, although possibly not himself. Of course he can save the second prisoner by announcing the colour hat that that prisoner is wearing, allowing him to 'guess' right, but that will not help the next one. What was the mathematician's solution?
There are 50 prisoners. Their evil captor announces that the next day they will all be buried up to their necks in a line and facing the same way down the line, and either a white or black hat will be placed on each one's head. No one can see his own hat, but the first prisoner can see the hats of the 49 in front of him, the next can see the 48 if front of him, and so on down to the last prisoner, who obviously sees none of the others.
The first prisoner will be asked to guess the colour of his own hat, and if he is right he will be released - if not, he will be shot. He can only say "white" or "black" and he is within earshot of his nearest neighbours. Clearly, each prisoner has a 50:50 chance of freedom if he guesses randomly.
One of the prisoners is a mathematician however and he devises a scheme whereby, if he goes first, he can save all the other prisoners, although possibly not himself. Of course he can save the second prisoner by announcing the colour hat that that prisoner is wearing, allowing him to 'guess' right, but that will not help the next one. What was the mathematician's solution?
