All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Logic
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.

  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.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
Some ThoughtsVery interesting.broll2016-06-18 06:22:29
AnswerK Sengupta2008-10-17 01:24:24
SolutionShort solutionGamer2008-10-16 20:44:55
Nice...Priya2008-10-16 04:51:57
Some ThoughtsI think...axf2007-02-05 05:13:23
solutionpaul2006-03-17 02:34:02
answerDean2005-12-10 01:08:18
re(5):(by pcbouhid):pcbouhid2005-12-09 06:29:28
re(4):(by pcbouhid):Eric2005-12-08 17:24:37
re(4):(by pcbouhid):Cory Taylor2005-12-08 13:27:43
re(3):(by pcbouhid):pcbouhid2005-12-08 06:38:57
re(2):(by pcbouhid):Eric2005-12-08 04:03:27
re(2):Cory Taylor2005-12-07 16:01:01
unknown orderMorgan2005-12-07 14:17:37
re:pcbouhid2005-12-07 13:24:51
re:Jer2005-12-07 09:41:45
"The Solution" is Unsatisfactorydopey9152005-12-07 09:03:18
re: trick question?pcbouhid2005-12-06 08:09:01
trick question?Dean2005-12-06 06:38:15
re: Contradiction in terms?Mindy Rodriguez2005-12-03 00:08:23
Contradiction in terms?Gamer2005-12-01 23:18:05
Some Thoughtsmagical fruitMindy Rodriguez2005-12-01 00:19:24
re: Assume that...Eric2005-11-30 20:22:03
Hints/TipsAssume that...pcbouhid2005-11-30 19:52:40
re(2): Prisoner's Dilemma DeluxeEric2005-11-30 19:22:50
re: Prisoner's Dilemma DeluxeJer2005-11-30 09:47:08
Prisoner's Dilemma DeluxeEric2005-11-30 02:00:36
ideasMorgan2005-11-29 18:52:10
re: Minor detail = no minor detailHugo2005-11-29 16:44:03
Minor detailPeter2005-11-29 16:30:27
SolutionSolution I Thinkdopey9152005-11-29 13:35:17
My two centsTristan2005-11-28 11:23:31
middle guymarc2005-11-28 07:59:23
Hints/TipsBeans?Matt Van Winkle2005-11-28 00:31:39
huh?Matt Van Winkle2005-11-28 00:18:10
No beans left -- Easier to read versionGamer2005-11-27 15:35:47
No beans left -- technical versionGamer2005-11-27 15:17:44
Hints/TipsSpoiled Beans for Bipartite GeeksMindy Rodriguez2005-11-27 13:02:05
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (2)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (3)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2020 by Animus Pactum Consulting. All rights reserved. Privacy Information