## Archive for **August 10th, 2009**

## Eugene Curtain and Max Washauer

If you came here because you were reading Peter Winkler’s “7 Puzzles You Think You Must Not Have Heard Correctly”, the names are supposed to be **Eugene Curtin** and **Max Warshauer**, and the paper is called “The locker puzzle”, published in *The Mathematical Intelligencer*, Volume 28, Number 1 (March 2006), pages 28ā31.

[If not, you should read the amazing “7 Puzzles You Think You Must Not Have Heard Correctly”, spend several days trying the first problem, read the brilliant solution, and then come back here if you’re interested in learning why no other solution can do better.]

The paper is available here if your institution has access. If not, here’s a sketch of the proof that the strategy cannot be improved upon. [**Update 2010-01-06**: Oliver Nash has a post about the puzzle, explaining both the original solution and the proof of optimality, here. Just the original solution is also worked out by David MacKay here.]

First, let us modify the rules slightly so that each prisoner must continue looking in boxes until he finds the box containing his name. The prisoners win if no prisoner opens more than 50 (i.e., n/2) boxes. This change obviously makes no difference to the outcome. Let’s call this (modified) game **Game 1**.

A different game involves all the prisoners being in the room at the same time, and is played as follows. The first prisoner opens boxes until he finds his name (i.e., the number “1”). Then, the lowest-numbered prisoner whose name hasn’t been revealed starts opening boxes until *he* finds his name. Then the next lowest-numbered whose name hasn’t been revealed opens boxes, and so on. The prisoners win if no one opens more than 50 boxes. Call this **Game 2**.

Let’s say we observe the prisoners as they play Game 2, and record the order in which boxes were revealed. This completely specifies what happened. For example, (with 10 prisoners instead of 100) if we record the list 2,6,1,4,9,7,10,8,3,5, we know that the first prisoner revealed boxes containing 2, 6, 1, then the third (lowest unrevealed) prisoner opened boxes with 4,9,7,10,8,3, then prisoner 5 opened 5, and they lost because the third prisoner opened 6 > 5 boxes.

**Prove:**No matter what strategy the prisoners follow, each permutation has the same probability (1/n!) of being the list recorded.**Prove:**“The classical records-to-cycles bijection”. It sends 2,6,1,4,9,7,10,8,3,5 to (2 6 1)(4 9 7 10 8 3)(5), for example.- So the probability of the prisoners winning Game 2 (no matter what strategy they follow) is exactly the probability that a random permutation has no cycle of length greater than n/2.
**Prove:**Any strategy for Game 1 corresponds to a strategy for Game 2 with the same probability. (Trivial: the only change is that you don’t have to open boxes you’ve already seen opened.)

This proves that the pointer-chasing strategy is optimal for Game 1.

Here’s the puzzle as it was originally considered, still open: suppose there are n prisoners and 2n boxes, half of them empty. The prisoners can each examine n lockers. The pointer-chasing strategy doesn’t work as empty boxes point nowhere. Does the probability of winning go to 0 as nāā?