Before revision: The reader may be familiar with various problems involving prisoners and lockers. A typical set-up is that there are n lockers into which a random permutation of n cards are inserted. Then n prisoners enter the locker room one at a time and are allowed to open half of the lockers in an attempt to find their own card. The team of prisoners wins if every one of them is successful. The surprising result is that there is a strategy which wins with probability about 1-\ln 2. A modified problem in which helpful Alice enters before the prisoners, inspects the whole permutationand swaps two cards improves the winning probability to 1. In our problem, there are n lockers and n cards and a helpful Alice just as before, but there is only one prisoner,Bob.If Bob may only open one locker,their chance of success is less than \frac{2.4{n}, but our main result is that, if Bob can open two lockers, their chance of success is c_n{n} where c_n\rightarrow \infty as n\rightarrow \infty.
After revision: {n}, but our main result is that, if Bob can open two lockers, their chance of success is c_n{n} where c_n\rightarrow \infty as n\rightarrow \infty.