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?
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.
Many people have commented that the last person has the biggest
advantage. I agree. But it is not so obvious as it might
seem. He can count the number of beans left, but he does not know
how many each took. (I assume he knows that he is last.) What is
this "strategy" that will give the last person an advantage?
The last person will take the average of the number of beans taken by
the previous four people. If this does not help him survive,
nothing will. Even so, this strategy prevents him from
accidentally saving anyone's life, since at worst, he will tie for
highest or lowest. Will he round up or down? Flip a
coin. It makes it harder for previous prisoners to predict his
strategy (though perhaps a 50-50 chance isn't always best). If
there are not enough beans, his best bet is to take all of them.
Similarly, the fourth and third person share this strategy. They
know that the prisoners after them will not save them, so they must
save themselves. (Exception: if they know there are not enough
beans to support the later prisoners' strategy.) Both will choose
the average of the number of beans taken by the previous prisoners.
The first and second prisoners are different. They have no way of
saving themselves--they must rely on the later prisoners. I'm not
sure yet what their strategy will be, but it will depend on their
knowledge of later prisoners' strategies.
Hey, this is pretty interesting. Maybe I'll write more on this later.
Posted by Tristan
on 2005-11-28 11:23:31