gjm comments on [LINK] Scott Aaronson: Common knowledge and Aumann's agreement theorem - Less Wrong

13 Post author: gjm 17 August 2015 08:41AM

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Comment author: GMHowe 23 August 2015 09:16:41PM 0 points [-]

Maybe I'm confused, in the 'muddy children puzzle' it seems it would be common knowledge from the start that at least 98 children have muddy foreheads. Each child sees 99 muddy foreheads. Each child could reason that every other child must see at least 98 muddy foreheads. 100 minus their own forehead which they cannot see minus the other child's forehead which the other child cannot see equals 98.

What am I missing?

Comment author: gjm 04 December 2015 02:46:23PM 1 point [-]

Common knowledge means I know, and I know that you know, and I know that you know that he knows, and she knows that I know that you know that he knows, and so on -- any number of iterations.

Each child sees 99 muddy foreheads and therefore knows n >= 99. Each child can tell that each other child knows n >= 98. But, e.g., it isn't true that A knows B knows C knows that n >= 98; only that A knows B knows C knows that n>=97: each link in the chain reduces the number by 1. So for no k>0 is it common knowledge that n>=k.

Comment author: GMHowe 22 December 2015 12:20:23AM 0 points [-]

Thanks, I did end up figuring out my error.