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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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Some Thoughts Very interesting. Comment 38 of 38 |

Assume:        
(1) Prisoners do not know how many players have gone before, just the number of beans in  the bag.
(2) The prisoners are neither compassionate nor psychotic, because if so in either case the first player to take beans can ensure the result.
(3) All players know that to take no beans is death, so they must take at least 1, and since everyone knows this too, they must take more than 1.

Say Player 1 chooses exactly 6 beans. Player 2's situation is then very difficult. She can be reasonably sure that she is second or third but not which.

Player 2 sees 94 beans. At least one player must have taken more than 2 beans:  (6,0), (4,2), so she is second or third. So Player 2 needs to pick either 3 or 7 beans. She won't choose 5 because this loses against (4,2) because it's larger and also loses against (6,0) because the average is greater than 5, so 5 is likely to be the low number.

          
A; Assume Player 2 chooses 3 beans: there are now 9 chosen in total.  Player 3 is either second: (9,0), or third (4,5) or 4th (4,2,3) unlikely to be 5th. Player 3's best bet is 4, and the average is moving up, so Player 2 gets left behind. But suppose that, realising this, Player 2 chooses 7 beans instead:

B: Assume Player 2 chooses 7 beans: there are now 13 chosen  in total.             
Player 3 is either second: (13,0), or third (7,6) or 4th (5,4,5) unlikely to be 5th. Player 3's best bet is at most 6.             
Player 4 sees 80 beans. With so many beans gone already, that player probably can't do much better than an averaged average; this is also around 6, and the average is falling             
Player 5 sees say 74 beans, and is in much the same boat as Player 4.             
             
So Player 1 has a good chance of survival if he plays his beans right.


  Posted by broll on 2016-06-18 06:22:29
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