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(3):(by pcbouhid): by pcbouhid)
But knowing the order is incredbly important to the "deny all"
attitude. If the first person who picked (and he knows he is
first as any pickers before him would not have picked zero bean as they
are smart) takes, say 9 beans, how will the future of the game work
out. I would contest it incredibly unlikely
that all prisoners die, which would then give the first picker some
small chance of survival - negating the take 100 beans strategy as his
survival is a higher need than killing his competitors (this is the
result of the wording in point #2).
If you were handed a bag, and there were 12 beans missing, how many
would you take? (If 1 person has gone, I should take 11, if two people
have gone, I should take 6, if three people have gone I should take 4
and if I am last I should take 3)
More interestingly, if you were handed a bag with 100 beans, how many would you take?
re(4):(by pcbouhid):
