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Re:The Logic Puzzle You Can Only Solve with Your Brightest F... by Zarel on Tuesday July 17, 2018 @05:42AM (#17548476)
Nice try, but you actually do need to work up from nothing.

> Day one: if A saw 13 or 14 they would know it was 20, so when A passes, that signals to B that A is between 10 and 12.

Your problem here is that if A saw 13, they would conclude that B saw 7 or 5. If A saw 14, they would conclude that B saw 4 or 6. In no case can A guess immediately.

So B can't conclude anything quite so strong from the fact that A passes on Day One.

Re:The Logic Puzzle You Can Only Solve with Your Brightest F... by laszlokorte on Tuesday July 17, 2018 @06:21AM (#17548611)
But by knowing that the total number is even in either case and by each one knowing that their own number is even, the odd numbers can be discarded anyway, right?
Re:The Logic Puzzle You Can Only Solve with Your Brightest F... by Zarel on Tuesday July 17, 2018 @09:57AM (#17549750)
Yes, I only brought up the odd-number cases because lucraft did.
Re:The Logic Puzzle You Can Only Solve with Your Brightest F... by lucraft on Tuesday July 17, 2018 @06:19AM (#17548603)
> Your problem here is that if A saw 13, they would conclude that B saw 7 or 5. If A saw 14, they would conclude that B saw 4 or 6. In no case can A guess immediately.

The common knowledge at this point is that B sees between 6 and 10, so A knows B doesn’t see 5 or 4.

Re:The Logic Puzzle You Can Only Solve with Your Brightest F... by yorwba on Tuesday July 17, 2018 @07:03AM (#17548769)
Your definition of common knowledge (they both know it and they both know the other knows it too) is too weak. For something to be truly common knowledge, the knowledge needs to be nested arbitrarily deep (they know, they know they know, (they know)^3, ...). Otherwise there's some remaining uncertainty about whether the other knows the common knowledge.

In particular, your assumption that if A saw 14 they'd know that B doesn't see 5 or 4 is wrong, because if A saw 14 they don't have the common knowledge that B sees between 6 and 10, because that's dependent on A seeing 12.