First of all, you're right Crabby. I meant "quiet", not "quite". It's just that I was in such a frenzy earlier, I couldn't type straight.
OK. Your last line of reasoning, DB, gets us back to the original 1/40 (x) 4, or 10%. So we've come full circle. I'm still not sure that the reasoning that got us to 19-1 is incorrect though.
I remember from my Prob & Stat course in college, that the problems always involved colored balls in a barrel.
Our problem would be expressed this way:
"A barrel contains 356 white balls and 9 blue balls. What are the chances of picking out one of the blue balls from the barrel if you get four chances to pick?"
There's a formula for this, I just don't remember it.
It would seem to me that on the first pick you have a 9/365 chance of picking a blue ball, or 2.4657%. On the second pick (since if you got a white ball on the first pick you would discard it, not put it back in the barrel) your chances would be 9/364, or 2.4725%. Third pick 9/363, or 2.4793% Fourth pick 9/362, or 2.4861%
Total, 9.9036%
The only thing that still bothers me is an explanation for why the probability is so close to 100% after only 40 picks, when it would take 365 picks to guarantee it 100%.
And what happened to this earlier line of thought, which still seems perfectly sound to me:
"My first day there has a 1/40 chance of being one of the 9 days that she's there already.
But, if she's not there, my second day there has only a 1/120 chance of being one of her first days there of one of her three trips.
And so on, or my third and fourth days there. So, 3/120 + 1/120 + 1/120 + 1/120, or 6/120, or 19-1.
