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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.

 Submitted by pcbouhid Rating: 3.8000 (15 votes) Solution: (Hide) Clearly, any prisoner taking 50 or more beans dies.a) If P1 takes between 21 and 49 beans, P2 is left with 51 to 79 beans and can take 1 bean less than P1 thus ensuring his own survival as, at least, one of the remaining prisoners cannot take as many beans as he has. P2 will not take the same number of beans as P1 as this gives him the chance of dying with the equal greatest number of beans. Of course, as the others are also smart, P2 will not wish to make his action too obvious and so can take any number of beans which is greater than 1/4 of the remainder and still guarantee his own safety.As they all are smart people, P3 will have already worked out the above possibilities. He can now aim to take the number of beans matching P2īs take (thus not being the highest, nor the lowest, which must be P4 or P5 or both). However, at this point, P3 does not need to maximize the beans taken as he wishes to increase the chance of killing both P4 and P5. P3 can take any number of beans greater than 1/3 of the remainder and less than 1/2 of those already taken. He must, of course, take an odd number (to return the total to even) to try to kill both.b) If P1 takes 20 or less beans, then P2 can match this exactly; taking more than P1 will allow to P3 to get an average and survive quite possibly comdemning P2 as having the most; similarly, taking less than P1 allows the others prisoners to take the average comdemning P2 with the least. So all the prisoners take the same number of beans and die.Thus P1 having no survival strategy, will condemn all the others 4 prisoners too, taking all the 100 beans, or any number 1-20.

 Subject Author Date Very interesting. broll 2016-06-18 06:22:29 Answer K Sengupta 2008-10-17 01:24:24 Short solution Gamer 2008-10-16 20:44:55 Nice... Priya 2008-10-16 04:51:57 I think... axf 2007-02-05 05:13:23 solution paul 2006-03-17 02:34:02 answer Dean 2005-12-10 01:08:18 re(5):(by pcbouhid): pcbouhid 2005-12-09 06:29:28 re(4):(by pcbouhid): Eric 2005-12-08 17:24:37 re(4):(by pcbouhid): Cory Taylor 2005-12-08 13:27:43 re(3):(by pcbouhid): pcbouhid 2005-12-08 06:38:57 re(2):(by pcbouhid): Eric 2005-12-08 04:03:27 re(2): Cory Taylor 2005-12-07 16:01:01 unknown order Morgan 2005-12-07 14:17:37 re: pcbouhid 2005-12-07 13:24:51 re: Jer 2005-12-07 09:41:45 "The Solution" is Unsatisfactory dopey915 2005-12-07 09:03:18 re: trick question? pcbouhid 2005-12-06 08:09:01 trick question? Dean 2005-12-06 06:38:15 re: Contradiction in terms? Mindy Rodriguez 2005-12-03 00:08:23 Contradiction in terms? Gamer 2005-12-01 23:18:05 magical fruit Mindy Rodriguez 2005-12-01 00:19:24 re: Assume that... Eric 2005-11-30 20:22:03 Assume that... pcbouhid 2005-11-30 19:52:40 re(2): Prisoner's Dilemma Deluxe Eric 2005-11-30 19:22:50 re: Prisoner's Dilemma Deluxe Jer 2005-11-30 09:47:08 Prisoner's Dilemma Deluxe Eric 2005-11-30 02:00:36 ideas Morgan 2005-11-29 18:52:10 re: Minor detail = no minor detail Hugo 2005-11-29 16:44:03 Minor detail Peter 2005-11-29 16:30:27 Solution I Think dopey915 2005-11-29 13:35:17 My two cents Tristan 2005-11-28 11:23:31 middle guy marc 2005-11-28 07:59:23 Beans? Matt Van Winkle 2005-11-28 00:31:39 huh? Matt Van Winkle 2005-11-28 00:18:10 No beans left -- Easier to read version Gamer 2005-11-27 15:35:47 No beans left -- technical version Gamer 2005-11-27 15:17:44 Spoiled Beans for Bipartite Geeks Mindy Rodriguez 2005-11-27 13:02:05

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