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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 23 of 38 |
(In reply to "The Solution" is Unsatisfactory by dopey915)

As the solution explains, if the first person takes 20 the second would be stupid to take any number but 20.  Otherwise the 3rd 4th and 5th will take the average of the preceding prisoners.

This could become sort of a prisoner's dilemma as previously mentioned.  Picking a random amount from around 10 to 20 would give each prisoner about a 40% chance of dying.  Since the second prisoner can't be sure the others wouldn't cheat, he wouldn't go along with this and would choose the same amound as the first.  This assures they all die.

 Posted by Jer on 2005-12-07 09:41:45

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