  We solve the secretary problem in the case that the ranked items arrive in a
statistically biased order rather than in uniformly random order. The bias is
given by the left-to-right-minimum exponentially tilted distribution with
parameter $q\in(0,\infty)$. That is, for $\sigma\in S_n$, $P_n(\sigma)$ is
proportional to $q^{\text{LR}^{-}_n(\sigma)}$, where the left-to-right minimum
statistic $\text{LR}^-_n$ is defined by $$ \text{LR}^{-}_n(\sigma)=|\{j\in[n]:
\sigma_j=\min\{\sigma_i:1\le i\le j\}\}|,\ \sigma\in S_n. $$ For $q\in(0,1)$,
higher ranked items tend to arrive earlier than in the case of the uniform
distribution, and for $q\in(1,\infty)$, they tend to arrive later. In the
classical problem, the asymptotically optimal strategy is to reject the first
$M_n^*$ items, where $M_n^*\sim\frac ne$, and then to select the first item
ranked higher than any of the first $M_n^*$ items (if such an item exists).
This yields $e^{-1}$ as the limiting probability of success. With the above
bias on arrivals, we calculate the asymptotic behavior of the optimal strategy
$M_n^*$ and the corresponding limiting probability of success, for all regimes
of $\{q_n\}_{n=1}^\infty$. In particular, if the leading order asymptotic
behavior of $\{q_n\}_{n=1}^\infty$ is at least $\frac1{\log n}$, and if also
its order is no more than $o(n)$, then the limiting probability of success when
using an asymptotically optimal strategy is $e^{-1}$; otherwise, this limiting
probability of success is greater than $e^{-1}$. Also, the limiting fraction of
numbers, $\lim_{n\to\infty}\frac{M^*_n}n$, that are summarily rejected by an
asymptotically optimal strategy lies in $(0,1)$ if and only if
$\lim_{n\to\infty}q_n\in(0,\infty)$.
