OK, since no-one even replied to that prisoner's dilemma, I guess it's time to produce some clues. We'll name the prisoners P1 through Pn and boxes B1 through Bn. We'll also, to avoid complications, assume the warden is going to arrange the boxes purely at random (so we don't need to cope with the warden just putting P1 in B2 and P2 in B1 for the 2 prisoner case, for example).¹

How do two prisoners play our game? If they both pick randomly, their joint chance of success is 1/4. However, if P1 picks B1 and P2 picks B2, they're either both correct or both incorrect, each with a 1/2 chance. This must be optimal or there'd be some one-prisoner strategy which is better than just picking one box.

What does this tell us?

1. The prisoners each need to pick different strategies.
2. The prisoners could do with knowing each other's strategies.

¹ You can get around any strategy the warden produces by simply numbering the prisoners randomly after the box order is chosen, so that the warden doesn't know the prisoners' chosen ordering. (It doesn't have to be the same numbering as the warden used to call them in.)