I started trying to solve it like one of my hat puzzles — trying to divide up the permutations such that when there is going to be any misses that there are tons of misses, but that correct guesses occur together.  It is possible to make improvements over random this way.  Consider 4 books / people, 2 guesses.  (I assume throughout that we can find a way to make the boxes under the tree ordinal in some sense.  “Closest to the southwest corner at the floor with ties broken by being southernmost” or some such.  I assume that solving this isn’t the puzzle you’re aiming for and I treat it as though you can assume the boxes are numbered 1-20.)  If you have people A and B both open boxes 1 and 2 and C and D always open 3 and 4, consider the cases:

1:  A B C D

2:  A B D C

3:  A C B D

4:  A C D B

5:  A D B C

6:  A D C B

7:  B A C D

8:  B A D C

9:  B C A D

10: B C D A

11: B D A C

12: B D C A

13: C A B D

14: C A D B

15: C B A D

16: C B D A

17: C D A B

18: C D B A

19: D A B C

20: D A C B

21: D B A C

22: D B C A

23: D C A B

24: D C B A

With this strategy, you get 1, 2, 7, 8 or 1/6, which is considerably better than 1/2^4 you get for random.  This kind of idea can be extended to larger numbers of boxes and perhaps improved upon, but it doesn’t feel like a candidate for your favorite puzzle ever.  I then had one of those pull-over-to-the-side-of-the-road epiphanies that you’re throwing away a lot of information.  Strategies of this form presume that you look into a box and see either your book or “not your book.”  You actually see some particular person’s book and can alter behavior based on this event.  That’s a lot to throw away.  How can we use this to cause good guesses to collide?

Consider person A assigned to open the first box.  If he finds his own book, he’s done.  When he finds a book belonging to X, his concern is to open another box such that if if he finds his book and X happens to open that box first, that X will make a good decision.  So if he finds B’s book, and he opens box 2, he wants B, if starts at 2, to open box 1 next in the case that A’s book is there and probably not otherwise.  This same logic should apply to everyone.  So generally, that means, that if you assign each person a different starting box, if they don’t find their book, they should open the starting box corresponding to the person owning the book that was found. Assume A assigned 1, B to 2, C to 3, and D to 4.

In case 1, everyone finds their book straight off.  A happy result.

In case 2, A finds his book.  B finds his book.  C finds the book of D, and so opens 4 and finds his book.  4 finds C, opens three and finds his book.

In case 3, A finds his book.  B finds C, thus opens three and finds his book.  C finds B, opens 2, and finds his book.  D finds his book.

In case 4, A finds his book, B finds C, opens 3 and fails.

In case 5, A finds his book.  B finds D, opens 4 and fails.

In case 6, A finds his book.  B finds D, opens 4 and finds his book.  C finds his book.  D finds B, opens 2 and finds his book.

Already as many winning cases as my strategy above!  It continues:

7: Hit

8: Hit

9: Fail

10: Fail

11: Fail

12: Fail

13: Fail

14: Fail

15: Hit

16: Fail

17: Hit

18: Fail

19: Fail

20: Fail

21: Fail

22: Hit

23: Fail

24: Hit

So we get to go to Paris a staggering 10/24 times.  Since 12/24 or 1/2 seems to be a theoretical maximum, that’s just astounding!

To extend this to the case of 20, After opening your second box, you continue on by opening the box of the person’s book you find at each step.  This is the point at which I started to think I’d seen this problem before.  Since you’re following a sort of chain here, you have a finite number of states and since your box is sure to be in the chain (you started with the box to which you’re assigned) you know that if you could keep guessing and following this method that you’d find your book rather than being caught in a loop not containing your book.

The only question is whether the chain that your part of is short enough to get there in 10 hops.  If you look back at the hits and misses from the case of 4, you see that you hit exactly when the state is one that could be obtained by sorting them so each person would find their book initially and the allowing the following operations:

Leaving a book alone (generates a chain of 1).

Exchange a book with another (generating a chain of 2).  (This operation can’t be repeating on any book already used in it, of course.)

In case of a fail, something else happened.  Pick one at random, case 14: C A D B .  With this , there is a chain of positions

1->3->4->2

and only then back to 1.  A chain of 4.  There should be chains of 3 in there as well.  Looking… 4:  A C D B .  Position 1 is a chain of 1.  The other chain is:

2->3->4

and then back to two.

This chaining idea is familiar to me, but I can’t place where, and I’m pretty sure at least it wasn’t books and boxes.  But anyway… there you have it.

The next question I ask myself how often do you win with 20?  All cases where no one lands in a chain of greater than length 10 (and so in all cases where no such chain exists) is an answer of sorts, but how often is that?  I’m at a complete loss as to how to compute that.  My guess is that it stays fairly close to one half, but gets worse as you add more people and boxes.  (But it sure would be beautiful if it doesn’t change at all, except by rounding error.)  I could program it to come up with an estimate, but instead I think I’ll stop and get some work done today.  :-)

Steve, thanks for the puzzle.  This was absolutely fantastic

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