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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: Assume that... | Comment 16 of 38 |
(In reply to Assume that... by pcbouhid)

They will all die. This forced awareness rules out any advantage to cooperation through deliberate ignorance. (i.e. if the first two players choose random handfuls, the third, fourth and fifth can guarantee their own safety and the first two's death by counting. Therefore the first two will not take a random number.) In fact it doesn't matter at all what number the first person picks between 0 and 20. In fact, since the first person is doomed to death, they might as well take all the beans and enjoy their final meal.

This problem is infinitely more interesting in the case of separate prisoners unaware of their order (and more consistent with the no communication allowed concept). This is the kind of Game Theory that John Nash won a Nobel prize for.

The only other consideration is to take some number of beans and then discreetly eat some so that you "end with" a number other than the largest or smallest which you took - if this were possible, they'd all do it and again, everyone would die.

 Posted by Eric on 2005-11-30 20:22:03

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