There are hundred prisoners each standing in a column. Each one is given a hat and there are two possible colors of hat which are red and white. The 100th person can see the hats of 99 people in front of him and the 99th person can see 98 people in front of him but can't see his own and the previous hat. The number of hats aren't known and the hats are placed in random order(no sequence can be determined from the way the hats are placed). Each person can say only one word red or white. He cannot say anything more than that. If he guesses the correct hat he will be alive otherwise he'll be executed. Either ways he will be taken away from the room so we dunno if his guess was right. Come up with a stratergy such that maximum people are alive. How many people are alive? (Do not use probability constraint given)
1The question is nice and I will give answer withing 3 days or so...
1Got the answer.... Only a sudden brainwave.
The 100th prisoner counts the other 99 hats. If there are more red than white ones, he says "red", and vice versa.
The 99th prisoner says the colour that appears most as well.
Same goes for the other 98 prisoners (although the very last one will have to guess randomly).
62yes.. this is correct :)
great answer :)
so essentially there is only a chance that atmost 1 person will die.. that too with a probability 1/2.. thre is 1/2 probability that he will be alive!
1can anyone explain how this is happening....
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if the last one sees 80 red hats in front......
then the 99 th 98 th and many more will say "red" for their own hat...which may in turn be white...[7]
I could figure out only one way to satisfy what bhaiyya said..but this solution is different
1if it is known that to other people that the person is alive or executed then i think i can make a strategy that at most the 99th person will die with a probablity of 1/2 otherwise not getting a clue to the q and even the ans
1no we can use an even odd strategy to guarantee 99 lives , but as to how the posted stratergy works i have no clue