Secretary Problem - Probability of hiring n-1 people In the Hiring/Secretary Problem, what's the provability of hiring n-1 people?
At first, I calculated it as such: first, we choose the n-1 people, then we arrange them from worst to best candidate. I then noticed that this flawed, since the placement if the remaining un-hired candidate differs if he was the 1-the worst candidate or n-the best candidate.
So, I tried another way. Let Y be the indicator on hiring n-1 people.
$|Y|={n\choose1}{n-i\choose1}\cdot1\cdot{(n-i)!}{(i-1)!}$ Explanation: first choose the one you don't want to hire. Let it be i, in the subset ${a_i,a_{i+1},...,a_n}$ he mustn't come first (or he'll be hired) so choose someone from that subest to come first, arrange the rest of the subset, then arrange the others.
But I'm have doubts regarding this. What's the best approach for this?
 A: First, here's how I interpret the question based on your last comment:
Model the candidates' worthiness as $\{1, 2, \dots, n\}$ where larger number means better.  The candidates arrive in a random order (permutation) $\sigma$, where $\sigma(k)$ is the $k$-th arriving candidate, and this candidate is hired iff 
$$\sigma(k) > \sigma(j) ~\forall j < k$$
Your question is what is the probability (randomized over all permutations) that exactly $n-1$ candidates will be hired.  
First of all, is my interpretation correct?  If so, here's the answer.
Suppose $i$ is the lone unhired candidate.  Clearly we cannot have $i = n$.  Also, clearly if you consider everybody except $i$, those $n-1$ candidates are arriving in increasing order (*), although not necessarily consecutively.  So the only question is when can $i$ be arriving s.t. $i$ is not hired.
For $i=1$ (the worst candidate): this person can arrive at any time except as $\sigma(1)$.
For $i=2$: this person can arrive at any time except as $\sigma(1)$ or $\sigma(2)$.  (The latter is prohibited because by (*) we already know $\sigma(1) = 1$.)
In general, if the unhired candidate is worth $i$, that candidate cannot arrive in the first $i$ positions, i.e. there are $n-i$ possible positions.
Thus the total number of permutations where $n-1$ people are hired $= (n-1) + (n-2) + \dots + 1 = n(n-1)/2$.
And prob $n-1$ people are hired $={n(n-1)/2 \over n!} = {1 \over 2 (n-2)!}$
