Summary: 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 \(\sigma \in S_n\), the probability of \(\sigma\) is proportional to \(q^{\mathrm{LR}_{n}^{-}(\sigma)}\), where the left-to-right minimum statistic \(\mathrm{LR}^-_n\) is defined by \[ \mathrm{LR}^-_n (\sigma) =|\{j \in [n] : \sigma_j = \min\{\sigma_i: 1 \leq i \leq j\}\}|, \,\,\sigma \in S_n.\] 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, where the highest ranked item is denoted by 1 and the lowest ranked item is denoted by \(n\). In the classical problem, the asymptotically optimal strategy is to reject the first \(M_n^\ast\) items, where \(M_n^\ast \sim \frac{n}{e}\), and then to select the first item ranked higher than any of the first \(M_n^\ast\) items (if such an item exists). This yields \(e^{-1}\) as the limiting probability of success. With the above bias on arrivals, and for the parameter \(q=q_n\) depending on \(n\), we calculate the asymptotic behavior of the optimal strategy \(M_n^\ast\) and the corresponding limiting probability of success, for all regimes of \(\{q_n\}^\infty_{n=1}\). In particular, if the leading order asymptotic behavior of \(\{q_n\}^\infty_{n=1}\) is at least \(\frac{1}{\log n}\), and if also its order is no more than \(o(n)\), then the limiting probability of success logn 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, \(\lim_{n \rightarrow \infty} \frac{M_{n}^{\ast}}{n}\), that are summarily rejected by an asymptotically optimal strategy lies in \((0,1)\) if and only if \(\lim_{n \rightarrow \infty} q_n \in (0, \infty)\).