Re: Riddle Minute
Posted: Sat Dec 19, 2009 3:26 am
Prisoner #10 counts the number of black hats. If it is even, he says white; if it is odd, he says black. He has a 50/50 chance of being right, but either way, everyone will know if he was right or wrong based on if he was killed or not. #9 can use this information to determine his own hat color: count the number of black hats he can see, and keep in mind that the total number of remaining black hats must be even if #10 said white and must be odd if #10 said black. Thus:
#9 sees an even number of black hats and #10 said white: #9's hat is white
#9 sees an even number of black hats and #10 said black: #9's hat is black
#9 sees an odd number of black hats and #10 said white: #9's hat is black
#9 sees an odd number of black hats and #10 said black: #9's hat is white
Thus, #9 can get his color right 100% of the time.
Going down the line, each prisoner will know the colors of all the hats behind him, as well as the ones in front of him that he can see, and whether the total number of black hats is odd or even, which is all the information anyone needs to figure out his own hat color.
Prisoners #1 through #9 live 100% of the time; prisoner #10 lives 50% of the time.
#9 sees an even number of black hats and #10 said white: #9's hat is white
#9 sees an even number of black hats and #10 said black: #9's hat is black
#9 sees an odd number of black hats and #10 said white: #9's hat is black
#9 sees an odd number of black hats and #10 said black: #9's hat is white
Thus, #9 can get his color right 100% of the time.
Going down the line, each prisoner will know the colors of all the hats behind him, as well as the ones in front of him that he can see, and whether the total number of black hats is odd or even, which is all the information anyone needs to figure out his own hat color.
Prisoners #1 through #9 live 100% of the time; prisoner #10 lives 50% of the time.