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.
This is an interesting problem for a Friday afternoon.
I don't believe any of the prisoners are likely to survive, but the fifth prisoner is the most likely.
The first prisoner will probably take his share of beans: 20. He has no reason to take more, which would almost guarantee he would end up with the most, and no reason to take less, which would likely leave him with the fewest beans. This strategy is compatible with assumption #2, as it would result in the death of all 5 prisoners.
The second prisoner will also choose 20 beans for the same reason, not wanting to end up with the most or fewest.
The third will try to maximize his chances and, seeing 40 beans gone, will also take 20, trying to split the difference between whatever prisoners #1 and #2 took.
The fourth will do the same as the third.
The fifth will have 20 beans left, and will know everyone is dead. But, he will take all 20, just in case two earlier prisoneers made a mistake.
The fifth prisoner is the most likely to survive, only because there are more opportunities for other prisoners to make mistakes before it is his turn to choose.
Spoiled Beans for Bipartite Geeks
