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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)

Comments: ( Back to comment list | You must be logged in to post comments.)
 re(5):(by pcbouhid): | Comment 31 of 38 |
(In reply to re(4):(by pcbouhid): by Eric)

Eric, by the time this problem was submitted, I had no answer to it. And, in the original, it was not stated clearly that the prisoners knew or didnīt, the order of them to taking beans.

My initial solution, in fact, was the "take-100" strategy. As the problem was being discussed, I realize that the strategy "take 1-20" was as good as "take 100" strategy.

Discussing more, and reading your comments, I have to agree that the order (previously stipulated) does really matter. And the solution I posted assume this (but itīs a mistake not stating this clearly in the text).

If they donīt know the order, my understanding is that this problem is too complex to be analyzed.

 Posted by pcbouhid on 2005-12-09 06:29:28

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