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The prisoners and the beans (Posted on 2005-11-27) Difficulty: 3 of 5
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)

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ideas | Comment 11 of 38 |
I think Peter is correct in that the prisoners do not know in which order they are drawing beans except for the first person because there would still be 100 beans in the bag (except if pesron one took zero beans but he wouldn't do that because then he would die for having the smallest number if beans.)  If they knew their draw order then everyone would take the average of the previous people and all would either tie for highest or lowest and all would die.

So, to make it more interesting assume they don't know the draw order.  Each prisoner other than #1 is going to know how many beans are missing and are going to have to guess how many people have already drawn from the bag.  If say 20 beans are missing, does the prisoner assume that one person drew 20, two prisoners drew 10 each, or 3 prisoners drew 6,7,7, or 4 prisoners drew 5 each?  or some other combination?

If the first prisoner picks a number that is too high, the next prisoner is more likely to guess that at least 2 people must have already picked and will pick a number either at or between 1/2 to 1/3 of the number of beans missing.  Then prisoner #1 will be highest and die.  If the first prisoner picks to low then the #2 prisoner may guess that only one person has picked before him and will then take one more than the number missing in the hope that prisoner #3 will guess that he is actually #2 and take 1 less than the total of #1 and #2 prisoner.

Anyway, there are a lot of combinations for prisoners #2 - #5, but all involve the prisoners guessing.

I believe the 1st prisoner has the best chance of living as he is the only one who knows his own order when picking from the bag.  The others are all guessing.  Since he is smart (assumption#1) then I'm assuming that he can use this knowledge to his advantage.

  Posted by Morgan on 2005-11-29 18:52:10
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