Tue Aug 30, 2016 4:48 am by tartle 


A warden places a hat on the head of each of 100 prisoners, each with a random number from 1 to 100. There may be duplicates. Each prisoner can see everybody's hat but their own. Each prisoner then guesses their own number; if any guess correctly, they all go free. The group may not communicate in any way during the trial, but may strategize beforehand. What strategy has the best chance of success? 




Tue Sep 06, 2016 9:08 pm by jatekchhateja@ 


What if a prisoner says his number wrongly?
And just to clarify again, only one of them needs to say it correctly? 




Mon Sep 12, 2016 4:36 am by Dzallen 


Do the prisoners hear each other's guesses?, because if so, the first prisoner can just say the number to his right as his guess, and the person to his right would then know his number.
Alternatively, if each prisoner had to guess correctly to save his own life, then the first guesser should just say the sum of all the numbers he sees, so the others can figure their numbers out. 




Tue Nov 15, 2016 4:30 pm by schizomidget 


I think the question is a bit short on hints. I'll assume that the trial is either 1. extended to an infinite amount of time (at least 5050 days) or 2. the prisoners have access to a clock.
So each prisoner will add (assuming that they are fairly good at mental math) all the numbers they see. This way, if two prisoners have the same hatnumbers, the two will have the same result number (sum). Let's say that this number is X. Before the trial, the prisoners decide to come to (gather in) a specific place after X days or X minutes. If two prisoners have the same number, they will meet each other, realize that they have the same number, and will be able to view the other person's number. If no prisoners are able to meet, that means that all of them have different numbers (1100 each). If they realize that no one has got the answer until the 5050th day or 5050th minute, they will realize that their number is the number that is missing from 1100. 






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