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 1/2^{100} ∼ 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!
(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 20170412 02:36:48 