Secretary Problem with 2 positions - Maximizing the difference
One trivial algorithm for the problem is exactly the same as the usual algorithm: observe $n/e$ of the applicants, then pick the first (if any) secretaries who are better than, or worse than, all applicants seen so far. This succeeds with probability roughly $1/e^2$. Answers to a question on math.se which includes your question claim that this algorithm (which is proposed in the answers) is obviously optimal.

To see that the success probability is roughly $1/e^2$, we follow the analysis in Wikipedia. We will analyze a more general algorithm, in which $n/e$ is replaced by $pn$, and then will show that the best choice for $p$ is roughly $1/e$. Let $q_{ij}$ be the success probability given that the $i$th secretary is the best one and the $j$th secretary is the worst one. If $i,j &gt; pn$ then $q_{ij}$ is the probability that:


$i$ is the best secretary.
$j$ is the worst secretary.
Among the first $i-1$ secretaries, the best is one of the first $pn$.
Among the first $j-1$ secretaries, the worst is one of the first $pn$.


Calculation shows that
$$
q_{ij} = \begin{cases}
\frac{pn(pn-1)}{n(n-1)(i-1)(j-2)} &amp; i &lt; j, \\
\frac{pn(pn-1)}{n(n-1)(i-2)(j-1)} &amp; i &gt; j.
\end{cases}
$$
The overall success probability is thus at least
$$
\sum_{i=pn+1}^n \sum_{\substack{j=pn+1\\j \neq i}}^n \frac{pn(pn-1)}{n(n-1)(i-1)(j-1)} \approx p^2 \left(\int_{pn}^n \frac{dx}{x}\right)^2 = \left(p\log\frac{1}{p}\right)^2.
$$
This is just the square of the usual formula, so it is maximized at $p = 1/e$, for which the success probability is roughly $1/e^2$.