3 ms·

I thought about that too, if he passes 6 times (and I've passed 6 times) I know that he has at least 6 (and he knows I have 10 or 12), so I pass a 7th time, and

• owenversteeg 8y ago
  Hmm. I was pretty convinced that my method would work, then I got convinced it couldn't work, now I'm confused again. A few notes: You can see 12, your friend can see 8. So your friend knows you have either 12 or 10. You know your friend has either 8 or 6. Step 1: you pass, friend knows you have >=1. Friend passes, you know they have >=1. Step 2: you pass, friend knows you have >=2. Friend passes, you know they have >=2. [...] Step 7: you pass, friend knows you have >=7. Friend passes, you know they have >=7; you have 12, so there must be 20, you win. I'm actually pretty convinced that it would work now... anyone want to poke a hole in it?
  • AstralStorm 8y ago
    Only if you use + 1 as a rule. What if your friend uses more efficient + 2, as you know the difference is by two? (Then unless you know who has the lower number, you're out of luck.)
  • nomynous 8y ago
    I thought of something along those lines at first as well. However, the problem is that the answer is not logically deducible using this method as it requires both of you to agree on a counting system and an end condition. And since the puzzle states that the two of you are not able to communicate, this method would not be viable. The given solution allows you to deduce the result with only the information and actions you have readily available to you.