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.
(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.
re(2):
