Solved by James Pryor

easy solution for screwy pirates

Here's a less elegant solution. How many different combinations of 6 pirates are there? 13 choose 6 is: 13!/6!7! That's a hell of a lot of locks. But you could say, put that many locks on the chest. And for each lock, you distribute keys to all the pirates EXCEPT the selection of 6 corresponding to that lock. Then any selection of 7 pirates is guaranteed to be open all the locks. For consider the first 6 of them. There will be exactly ONE lock that those six are unable, between the 6 of them, to open. But the seventh pirate will be able to open that lock. Because keys to that lock went out to everybody except those 6.

Also, any selection of 6 pirates is guaranteed to encounter exactly one lock that they are unable to open. (So any selection of fewer than 6 is guaranteed to encounter 1 or more locks they are unable to open...)

So that will do the trick, and the explanation of why it would do the trick is elegant. But I don't feel that it's an elegant solution, because of there being so many locks... Anyone have something better?