Solution:
- The probability the best person falls in the first \(n\) is \(\frac{n}{N}\)
- The probability the best two people do not fall in the first \(n\) candidates is
\begin{align*}
&& 1-P &= \frac{\binom{N-2}{n}}{\binom{N}{n}} \\
&&&= \frac{(N-2)(N-3)\cdots(N-2-n+1)}{n!} \frac{n!}{N(N-1)(N-2) \cdots (N-n+1)} \\
&&&= \frac{(N-n)(N-n-1)}{N(N-1)} \\
\Rightarrow && P &= 1- \frac{(N-n)(N-n-1)}{N(N-1)} \\
&&&= \frac{N(N-1) - N(N-1)+n(N-n-1)+Nn}{N(N-1)} \\
&&&= \frac{n(2N-n-1)}{N(N-1)}
\end{align*}
If \(N = 4, n = 2\) the possibilities are, the best candidate can be first \(3!\) ways, or second \(3!\) ways, which is \(\frac{12}{24} = \frac{1}{2} = \frac{2}{4} = \frac{n}{N}\) so our formula works.
In the case neither of the best two candidates are in the first half, the possibilities are \(3412, 3421, 4312, 4321\), ie \(\frac{4}{24} = \frac16\) chance, so the probability they are selected in the first \(n\) is \(\frac56\). our formula says it should be \(\frac{2 \cdot (2 \cdot 4 - 2 - 1)}{4 \cdot 3} = \frac{2 \cdot 5}{4 \cdot 3} = \frac56\) as desired.