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The prisoners and the beans (Posted on 2005-11-27) Difficulty: 3 of 5
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(2): Prisoner's Dilemma Deluxe | Comment 14 of 38 |
(In reply to re: Prisoner's Dilemma Deluxe by Jer)

Part of the assumption extension of all are (very) smart is that all will reason identically. To universalize their identical reasoning we can imagine the prisoners establishing a rule or a table correlating how many beansto take based on how many are missing when the bag comes to them. This table will necessarily converge in both directions from the extremes 0 and 20 for the original prisoner's choice (when the bag arrives with no beans missing) and for all prisoners afterward, albeit with more complexity. To explain:

20+ is ruled out in all cases (If the first person takes more than 20, the rest will all take less regardless of where they suspect they are in the order of choice) - But if they all reason this way, they will also realize that they can never take 20 beans, because to do so would also ensure that they held the max and were killed. So could the first take 19? Absolutely not, if no one will take 20 that person would surely die! 18? Nope. The same goes the other way. No one takes 0, therefore no one can take 1. Likewise no one can take 2 if no one would take 1, so who would consent to taking 3, or 4 even?. This logic converges to the middle, and begins to be subject to the chaos of the prisoner's deception of one another. But no matter what the first person takes, the progression is most certain to be a series of choices between two numbers. e.g. 9,8,9,8,9 - 9,10,10,10,10 ensuring everyone's death.

The prisoner's must realize that their best chances for survival is to be deliberately ignorant of their position and bean quantity. It would be best if they could not even know how heavy the bag was, that would allow them to grab larger handfuls (they can't risk the bag coming to the last person so light that they have reason to suspect they aren't the first person. Larger handfuls make the likelihood of coincidental duplication of bean counts at the extremes lower resulting in a probability of survival closer to 3/5. Any attempt to deduce their position or bean count will corrupt the system and doom them all to death.

This is a perfect extension of the classic Prisoner's Dilemma problem where prisoner choose to compete or cooperate.


  Posted by Eric on 2005-11-30 19:22:50
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