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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(4):(by pcbouhid): | Comment 30 of 38 |
(In reply to re(3):(by pcbouhid): by pcbouhid)

I am not suggesting combinatorics, I am suggesting reason alone.

It absolutely does matter if they know their position.

Before any prisoner ever even sees the bag of beans they will reason that if all five prisoners take steps to know their position, # of remaining beans, and # of beans taken, then they have no survival strategy. Therefore, if any survival strategy exists, it must follow from at least one prisoner not taking steps to know their position, # of remaining beans, or # of beans taken. As I have shown, if all five prisoners observe the strategy of total deliberate ignorance, then then they have what approaches a 3/5 probability of survival - which you have to admit is a hell of a lot better than 0. As all of them want to survive, they will necessarily give this option serious thought. Of course there will be the conflict of not wanting to be the one sucker, but when you are up against certain death, you are willing to take risks.

My reason for suggesting the Prisoner's Dilemma game theory scenario was to let a Darwinian approach to ideal strategy emerge. I first suggested that the best strategy was simply for all prisoners to grab a random handful (and this intuitively seems best to me), but I am entertaining the idea that perhaps a small % chance of 'defecting' could improve their overall chance of survival. Remember, just because the prisoners all "reason the same way as any other," doesn't mean they behave the same. They can include randomness in their strategies both in whether or not they observe the ignorance tactic as well as in how many beans to take.

No matter what, they will first all want to observe a strategy that maximizes their chance of survival, and that is the bottom line. Since the mathematical limit of that chance is 3/5, any strategy which approaches that number when universalized is appealing.

In fact, once they appreciate the deliberate ignorance strategy, they will reason that the only reason for even bothering to take steps to know their position etc. is if they suspect someone else has already observed this strategy. But because they know how their own reasoning is going, they also know the likelihood of that happening.  So, the more confident they are that deliberate ignorance is the best strategy, the more confident they are that their teammates will observe it as well.

As they become more certain that everyone else will likely observe it, they are more likely to try to improve their odds by trying to cheat the system. Of course that would mean everyone else is reasoning that way, which would again push them back toward the deliberate ignorance tactic. I believe that there is an equilibrium probability, (likely 0, but possibly something small) which would refer to the chance of defecting which corresponds with the maximum chance of survival.

Another reason this is interesting is that if this probability of defecting which maximizes survival is greater than zero, there  consequentially will be a discrete answer to the initial question: Who is most likely to survive?

Now that would be cool.

  Posted by Eric on 2005-12-08 17:24:37

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