Originally posted by plawrence: 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.
Plaw, I think the flaw with this reasoning (which I didn't catch earlier) was that in the second, third and 4th days, there is not a 1/120 shot because in each day, there is an equal chance of it being the first day in any of her 3 trips. You got 1/120 through 1/40 (her chance of being there that day) times 1/3 (the probability of it being the first day of one of her trips - let's say her first trip). However, there is an equal chance that it is the first day of her second trip or her third trip. Therefore, the true probability of the second day is:
1/3*1/40 + 1/3*1/40 + 1/3*1/40 = 1/40.
In any case, I believe it's a coincidence that the original 4/40 you got was close to the 10% I calculated. If we carry out my reasoning, (converse reasoning - which actually uses the formula that you are looking for), we can see what the probability is after 40 days.
