Can it be solved using probabilities? I mean, if, for example, the Guru says ‘I see nobody on this island who has blue eyes’, then the whole 100 brown-eyed people can pack and leave at once. If the Guru says ‘I see at least one blue-eyed person’ AND thereare two BE present, neither of them leaves that night and so next morning they both know they are BE and can leave the following night, and so on?
I mean, the ‘perfect logicians’ part put me into thinking like ‘...and if there are 3 BlE and 100 BrE, and the Guru says ‘I see at least 1 BlE’, then at that moment each one of the three BlE thinks there’s 2⁄3 chances she means either of the other two, so next time just before noon, when they converge again, one of the three BlE finds the other two and goes away without saying anything. Then if next morning the two others are found to have left, he knows he’s chance of being the only BlE left has gone up and presents himself for inspection. When he is confirmed as BlE, next time it automatically releases all BrE. Now let’s consider the case of four BlE present...′ etc. I doubt it can be done easily for large groups of people, though, unless they cooperate. The easiest way to do it is to appear before the Guru in pairs:)
Can it be solved using probabilities? I mean, if, for example, the Guru says ‘I see nobody on this island who has blue eyes’, then the whole 100 brown-eyed people can pack and leave at once. If the Guru says ‘I see at least one blue-eyed person’ AND thereare two BE present, neither of them leaves that night and so next morning they both know they are BE and can leave the following night, and so on?
The approach you describe is sensible, but I don’t see what it has to do with probabilities; all the probabilities involved are either 0 or 1.
I mean, the ‘perfect logicians’ part put me into thinking like ‘...and if there are 3 BlE and 100 BrE, and the Guru says ‘I see at least 1 BlE’, then at that moment each one of the three BlE thinks there’s 2⁄3 chances she means either of the other two, so next time just before noon, when they converge again, one of the three BlE finds the other two and goes away without saying anything. Then if next morning the two others are found to have left, he knows he’s chance of being the only BlE left has gone up and presents himself for inspection. When he is confirmed as BlE, next time it automatically releases all BrE. Now let’s consider the case of four BlE present...′ etc. I doubt it can be done easily for large groups of people, though, unless they cooperate. The easiest way to do it is to appear before the Guru in pairs:)