There does exists a strategy to get %50 for two people (hint below). But the case of two people does not help much to solve the more general case (imho). HINT:

There does exists a strategy to get %50 for two people (hint below). But the case of two people does not help much to solve the more general case (imho). HINT: Suppose each prisoner picks a box at random. They can (randomly) decide what box to open before they enter the room with the boxes, at a time when they can still communicate. In some cases they will then be able to determine that they have no chance of both finding their number. So there must be a better strategy.