4 today, but DH got it in 3!
Are any others here "math people" who are trying to figure out a way to calculate more stats than what it shows? I'd like to know average number of tries, but I'm not sure what to do with the ones you don't solve at all - do you just set an arbitrary number (7?, 10?) for the ones you don't solve, or is it better to reverse it to a sort of "points" system - like 6 points for solving it on 1, 5, for getting it in 2...1 for taking all 6 attempts, and 0 if you don't get it at all?