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 The prisoners and the beans (Posted on 2005-11-27)
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.

 See The Solution Submitted by pcbouhid Rating: 3.8000 (15 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 "The Solution" is Unsatisfactory | Comment 22 of 38 |

The posted solution fails to answer the question "Who is most likely to survive?"  The question implies that there is one (or more) prisoners who is/are most likely to survive, rather than the scenario that none survive (so all having an equal chance).

I stand by my earlier "Solution I think" posting, that the first prisoner is the most likely to survive if he takes 20.  If one works out all the combinations after prisoner 1 takes 20, it turns out that prisoner 1 survives in more of those combinations than any other prisoner.

 Posted by dopey915 on 2005-12-07 09:03:18

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