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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(4):(by pcbouhid): | Comment 29 of 38 |
(In reply to re(3):(by pcbouhid): by pcbouhid)

But knowing the order is incredbly important to the "deny all" attitude.  If the first person who picked (and he knows he is first as any pickers before him would not have picked zero bean as they are smart) takes, say 9 beans, how will the future of the game work out.  I would contest it incredibly unlikely that all prisoners die, which would then give the first picker some small chance of survival - negating the take 100 beans strategy as his survival is a higher need than killing his competitors (this is the result of the wording in point #2).

If you were handed a bag, and there were 12 beans missing, how many would you take? (If 1 person has gone, I should take 11, if two people have gone, I should take 6, if three people have gone I should take 4 and if I am last I should take 3)

More interestingly, if you were handed a bag with 100 beans, how many would you take?

 Posted by Cory Taylor on 2005-12-08 13:27:43

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