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 The prisoners and the beans (Posted on 2005-11-27)
Five prisoners are going to take beans from a bag with 100 beans. They will do it one prisoner at a time, and only once each. No communication is allowed between them, but they can count the beans left in the bag. All prisoners who end with the largest and the smallest number of beans will die.

Who is most likely to survive?

Assume:
1. they are all smart people.
2. they will try to survive first and then try to kill more people.
3. they do not need to take out all the 100 beans.

 See The Solution Submitted by pcbouhid Rating: 3.8000 (15 votes)

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 re: | Comment 24 of 38 |
(In reply to "The Solution" is Unsatisfactory by dopey915)

I respect your point, dopey, but disagree with it.

"Who is most likely to survive?" does not imply that one or more is/are most likely t survive, and the answer to this problem is "none of them".

The problem states that first, one will try to survive, and then, try to kill more people. If the first prisoner reasons that he has nothing to do (all his possible takes result in his death, because is tacitly said that all are smart like him), then heŽll try to kill the most people he can. So, his choice of take-100 or any number 1-20, assure him that all the others will die (too).

This is not a problem to be solved by combinatorics, but by logical reasoning.

 Posted by pcbouhid on 2005-12-07 13:24:51

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