Secretary problem and two almost the same consecutive applicants

We present a new variant of the secretary problem. Let $A$ be totally ordered set $n$ \emph{applicants}. Given $P\subseteq A$ and $x\in A$, let $rr(P,x)=\vert\{z\in P \mid z\leq x\}\vert\mbox{ }$ \emph{relative rank of} $x$ \emph{with regard to} $P$, $rr_n(x)=rr(A,x)$. $x_1,x_2,\dots,x_n\in random sequence distinct applicants. The aim is to select $1<j\leq n$ such that $rr_n(x_{j-1})-rr_n(x_j)\in\{-1,1\}$. $\alpha$ real constant with $0<\alpha<1$. Suppose following stopping rule $\tau_n(\alpha)$: reject first $\alpha applicants then $x_j$ $rr(P_j,x_{j-1})-rr(P_j,x_j)\in\{-1,1\}$, where $P_j=\{x_i\mid 1\leq i\leq j\}$. $p_{n,\tau}(\alpha)$ probability $\tau_n(\alpha)$ selects show \[\lim_{n\rightarrow\infty}p_{n,\tau}(\alpha)\leq \lim_{n\rightarrow\infty}p_{n,\tau}\left(\frac{1}{2}\right)=\frac{1}{2}\mbox{.}\]