3 ms·

It’s strange because at the beginning you know for a fact that your friend sees either 6 or 8, and your friend knows you see either 10 or 12, but the given proc

It’s strange because at the beginning you know for a fact that your friend sees either 6 or 8, and your friend knows you see either 10 or 12, but the given process throws that information away. I think it’s possible to escape on Day 2 if you keep that knowledge: A is 12, B is 8 A knows B is either 6 or 8. If 6, B knows A is either 12 or 14. If 8, B knows A is either 10 or 12. B knows A is either 10 or 12. If 10, A knows B is either 8 or 10. If 12, A knows B is either 6 or 8. So the common knowledge (they both know it and they both know the other knows it too) at the beginning is: A is between 10 and 14 and B is between 6 and 10. 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. Now, if B saw 9 or 10, they would know it was 20, so B must be between 6 and 8 Day two common knowledge: A between 10 and 12. B between 6 and 8. If A saw 10 or 11 they would know it was 18, so when they pass B knows it must be 12 B knows it’s 20, and wins. ... Right?

Zarel 8y ago
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.
- lucraft 8y ago
  > 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.
  - yorwba 8y ago
    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.
- laszlokorte 8y ago
  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?
  - Zarel 8y ago
    Yes, I only brought up the odd-number cases because lucraft did.

Zarel 8y ago

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.

lucraft 8y ago
> 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.
- yorwba 8y ago
  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.
laszlokorte 8y ago
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?
- Zarel 8y ago
  Yes, I only brought up the odd-number cases because lucraft did.