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Names in Boxes (Posted on 2017-04-09) Difficulty: 3 of 5
The names of 100 prisoners are placed in 100 wooden boxes, one name to a box, and the boxes are lined up on a table in a room. One by one, the prisoners are led into the room; each may look in at most 50 boxes, but must leave the room exactly as he found it and is permitted no further communication with the others.
The prisoners have a chance to plot their strategy in advance, and they are going to need it, because unless every single prisoner finds his own name all will subsequently be executed.
Find a strategy for them which has probability of success exceeding 30%.

Comment: If each prisoner examines a random set of 50 boxes, their probability of survival is an unenviable 1/2100 ∼ 0.0000000000000000000000000000008. They could do worse—if they all look in the same 50 boxes, their chances drop to zero. 30% seems ridiculously out of reach—but yes, you heard the problem correctly!

No Solution Yet Submitted by Ady TZIDON    
Rating: 3.7500 (4 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Some Thoughts re: This is not possible. | Comment 3 of 10 |
(In reply to This is not possible. by Math Man)

I'm not sure this is quite correct. Assume a simpler case with 4 boxes and 4 prisoners. Each prisoner selects his own box first (i.e prisoner 1 chooses box 1, prisoner 2 selects box 2, etc.) and then selects the number of the box the first box contained.

Prisoner 1 has a 50/50 chance of finding his own number. These permutations work: {1, 2, 3, 4} | {1, 2, 4, 3} | {1, 3, 2, 4} | {1, 3, 4, 2} | {1, 4, 2, 3} | {1, 4, 3, 2} | {2, 1, 3, 4} | {2, 1, 4, 3} |{3, 2, 1, 4} | {3, 4, 1, 2} | {4, 2, 3, 1} | {4, 3, 2, 1}.

However, once prisoner 1 has succeeded, the order of the boxes is fixed. Since Prisoner 2 is committed to the same strategy, only those cases where prisoner 1 already succeeded need to be considered. These are: {1, 2, 3, 4} | {1, 2, 4, 3} | {1, 3, 2, 4} |{1, 4, 3, 2} | {2, 1, 3, 4} | {2, 1, 4, 3} | {4, 2, 3, 1} |{4, 3, 2, 1} with probability 1/3, for a total of 1/6.

But note that that probability doesn't change at all if we add two more prisoners, 3 and 4. In every case where prisoners 1 and 2 succeeded, prisoners 3 and 4 using the same strategy are bound to succeed. So the overall probability is 1/6, not around 1/16.

I am still thinking about this however.



  Posted by broll on 2017-04-12 02:36:48
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