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

 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)

Comments: ( Back to comment list | You must be logged in to post comments.)
 re(2): | Comment 26 of 38 |
(In reply to re: by pcbouhid)

I also have a problem with the solution, though it is quite different.

The solution, when it states that all prisoners will die, is correct.

However, if I were playing this game as person #1, I would take out some number less than 20, because if I were playing this game as person 2, I would take out 1 less than the person before me.  How this all works out is irrelevant, as persons 3,4&5 will ensure that nobody survives, however, person 1, being smart, will not guarantee his doom if there is ANY chance of success (his survival is higher on his particular hierarchy of needs than vidictiveness), which could (but won't) come about through many factors.  This leads to the answer that persons 3,4&5 all have a greater, but still zero, chance of survival than the first two persons.

Also, I second (or third or fourth) the suggestion that this needs more examination with the unknown order element.

 Posted by Cory Taylor on 2005-12-07 16:01:01

 Search: Search body:
Forums (2)