All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Logic
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)

Comments: ( Back to comment list | You must be logged in to post comments.)
re(2):(by pcbouhid): | Comment 27 of 38 |
(In reply to re: by pcbouhid)

And I respect that you are the author of the problem so if you say they are in the same cell, they are in the same cell (and sure, let's take out the idea of seeing 21 beans missing and wondering if one took 21 or two took 10 & 11). But I would like to appeal to the fact that if deliberate ignorance of position were allowed, we could see it as a viable thread toward a much more interesting solution.

Obviously, it has been proven that if the prisoner's are all aware of their position they will all die.

I have proposed that if all the prisoner's choose to be deliberately ignorant of their position and take a random handful of beans that their chance of survival approaches 3/5.

But then comes another layer. Once prisoner's recognize that deliberate ignorance is an option, they could reason the following: "My fellow prisoner's are likely to be deliberately ignorant, so I can advance my own likelihood of survival by counting (and knowing my position)."

In this case, this would lead to this strategy: "If I check my position and am first, then I will take the mean of whatever I percieve this 'average handful' to be. If I am second and I find that the first person has taken more than what I percieve the 'average handful' is, I will take one less, otherwise I'll take one more. If I am third, fourth or fifth, I will simply take the mean of what's been taken so far (rounding toward the 'average handful')."

So I propose playing an Iterated Prisoner's Dilemma game where there is a variable probability of either choosing deliberate ignorance (cooperating) or counting (defecting) and there is also a variable for "average handful" - and while we are at it we may as well define some variable for standard deviation of the average handful. I believe we would find some equilibrium rule which all players would obseve in proportion with the respective probabilities which would maximize survival (and come very close to 3/5).

It is also possible (although this may be reaching) that we observe systems of multiple strategies symbiotically supporting one another in such an IPD game. In general, my instinct tells me that any tendency toward defecting corrupts the system rapidly, but can see that maybe there is a low sustainable probability of  defecting. In any event, I think we'd need a programmer to take a crack at this one....Charlie?

Edited on December 8, 2005, 4:06 am

Edited on December 8, 2005, 4:09 am
  Posted by Eric on 2005-12-08 04:03:27

Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (2)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (3)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2020 by Animus Pactum Consulting. All rights reserved. Privacy Information