In

`HIRE-ASSISTANT`

, assuming that the candidates are presented in a random order, what is the probability that you hire exactly twice?

You hire twice when you first hire is the candidate with rank $i$ and all the candidates with rank $k > i$ come after the candidate with rank $n$. There are $n - i$ better suited candidates and the probability of the best one coming first is $1/(n-i)$ (we can ignore the other candidates and they don't affect the probability). Thus, the probability for hiring twice if your first candidate has rank $i$ is:

$$ \Pr\{T_i\} = \frac{1}{n}\frac{1}{n-i} $$

The first part reflects the probability of picking that particular candidate out of $n$.

The probability to hire twice is:

$$ \Pr\{T\} = \sum_{i=1}^{n-1}\Pr\{T_i\} = \sum_{i=1}^{n-1}\frac{1}{n}\frac{1}{n-i} = \frac{1}{n} \sum_{i=1}^{n-1}\frac{1}{i} = \frac{1}{n} \Big(\lg(n-1) + \O(1)\Big) $$