The same sort of thing goes on here. But remember that this is a series, and so we're not looking for the total probability, but only the probability for each trial.
Let's say that r is the stopping number. In other words, you only start considering secretaries after you've interviewed at least r-1 secretaries. Let's also say that j-1 is the number of secretaries you've considered so far. That makes each secretary you interview the jth secretary.
Now that we've assigned the variables, let's ask ourselves, what is the probability that the secretary I'm interviewing--the jth secretary--is the most qualified? Well, there's only one best out of the n secretaries, so the probability of some random secretary being the best is:
That's our first condition. The other condition which must be satisfied is the probability of even considering this jth secretary. Remember, we're going to stop once we get to the best after r-1 secretaries. For us to have moved on to the jth secretary means that so far, the best secretary was one of those first r-1 secretaries, otherwise we wouldn't be considering the jth secretary. And the probability that the best so far was one of the first r-1 is:
So, since these are the only two conditions which need be met, we know that the probability of the jth secretary being the best is:
Next we add up the probabilities for each trial. This is done by converting to a series:
I know I didn't explain that last step very well. Quite frankly, I'm not sure how to put it into words. Maybe Mr. Fantastic can help out with an explanation. Hopefully you can just see how it works, though.
This can be done because r and n are both constants, and we can pull out constants. j is a variable, however, so we must keep any term which includes a j inside the sigma summation.This becomes
(r-1 / n) * ∑ (1 / j-1)
from j=r to n
I have no clue about that one. Mr. Fantastic?And then P(r) =
x * ∫ (1/t) * dt
integrating t over x and 1.