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I 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 inte
by edanm 16y ago
I 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.