4 ms·
with 2 prisoners the probability of success is 0.5, so you are correct. with 4 prisoners the probability of success is about 0.37, which is better than chance.
by dododo 16y ago
with 2 prisoners the probability of success is 0.5, so you are correct.
with 4 prisoners the probability of success is about 0.37, which is better than chance. (the probability of success turns out to be one minus the difference of two harmonic numbers.)
it's hard to give a hint without giving the answer. the best i can do is that it has to do with the structure of cycles in permutations.
- edanm 16y agoI don't know the answer, but I think I know the general direction. And unless I'm mistaken, you're wrong about the "2 prisoners". Semi-Spoiler alert As I interepreted the parent post, he hasn't found a way, with 2 prisoners, to do better than the strategy of every prisoner picking a random box. Under that strategy, your chance of success is 0.25 - 0.5 * 0.5 (each has a 0.5 chance of being right). So if, as you say, you can get the probability to 0.5 for 2 prisoners, you are doing better than chance. (And this I already know how to do, I'm assuming for more prisoners it's the same idea, but generalized). Am I making sense at all?
- sesqu 16y agoYou're right, a 0.5 strategy is better than random chance. Unfortunately, the simplest generalization of that strategy yields only 0.5^(n-1), which is not good enough - you need to start with a bigger base case.
- alphaBetaGamma 16y agoThere 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.