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.
The first prisoner is most likely to survive, as long as he takes 20 beans. As others have pointed out, if they all take 20 beans, no one survives. So one or more of the other prisoners would try to get an edge and take other than 20 beans. As soon as one prisoner does that, the ones following would try to take as close to the average as possible. The last prisoner would be forced to take the remaining beans, or fewer. I worked out the combinations with prisoner 1 taking 20, and prisoners 2, 3, and 4 taking 19, 20, or 21 beans. In nearly all cases, prisoner 1 survives. Other prisoners may survive also (the one(s) also taking 20 beans), but prisoner 1 survives most often. (I work in a jail in a major US city, and I can tell you that prisoner will either eat the beans or trade them for other "things" long before they use them to determine their fate!)
Posted by dopey915
on 2005-11-29 13:35:17