My model for the secretary problem is as in (1) in this Mathstackexchange question. More precisely, I look at the unit interval $I=[0,1]$, and $\Omega = I^n \smallsetminus D$ where $D \subset I^n$ is the set of points with at least two identical coordinates (e.g. $(\frac{1}{n},\frac{1}{n},\frac{1}{n},\frac{4}{n},\dots,1)$ would lie in $D$). On $\Omega$ there the Lebesgue probability.
Definition: Given a point $a =(a_1,\dots, a_n) \in \Omega$, an index $k \in \{1,\dots, n\}$ is said to be a maximum if $\max \{ a_i: i=1,\dots, k-1\} < a_k$. By way of definition, $1$ is always a maximum and further, we also put $n+1$ as a maximum.
Given a point $a \in \Omega$, we let $X_1(a),\dots, X_n(a)$ be its set of maxima arranged in increasing order. In particular, $X_1(a) =1$ for every $a \in \Omega$ and if we happen to have $m < n$ with $X_m(a) = k$ and $a_k = \max\{a_i: i =1,\dots, n\}$, then we set $X_{m+1}(a) = \dots = X_n(a) = n+1$. It is easy to see that the $X_i's$ are measurable functions. Indeed, the inverse image of an index $k$ is always the union of simplices of the form $\{a = (a_1, \dots, a_n)\in \Omega: a_{\sigma(1)} < \dots < a_{\sigma(n)}\}$ where $\sigma$ is some permutation.
Question: Given $i \in \{1,\dots, n-1\}$ and $i \leq k < l\leq n$, how do I prove the conditional probability formula \begin{equation}\label{1}\tag{1} P[X_{i+1} =l| X_{i}=k] = \frac{k}{l(l-1)}? \end{equation} Note, if $i=1$ I only consider the case $k=1$. I saw this formula in the book of Billingsley as well as the book of Dynkin-Yushkevich.
Attempt: For example, if I want to compute $P[X_2=k| X_1=1]$, I compute the volume of the union of simplices $$ \left\lbrace a =(a_1,\dots, a_n) \in \Omega: a_{\sigma(2)} < \dots < a_{\sigma(k-1)} < a_1 < a_k\right\rbrace $$ where $\sigma$ varies over all permutations over $k-2$ symbols. Since each simplex has volume $1/k!$, I get the formula $$ P[X_2 = k| X_1 = 1] = \frac{1}{k(k-1)}. $$ I can do something similar for $P[X_3 = l| X_2 = k]$ by counting the number for simplices occurring in the set $$ \left\lbrace (a_1, \dots, a_n) \in \Omega: \max\{a_2,\dots, a_{k-1}\} < a_1 < a_k < a_l \text{ and } \max\{a_{k+1},\dots, a_{l-1}\} < a_k. \right\rbrace$$ But if I try to go to $P[X_4=m|X_3=l]$, I become incapacitated.
Can someone please help me see the truth of formula \eqref{1}? I am weak in probability and combinatorics.
Ok, so consider $1<k<l\leq n$ and $1<i \leq k$ and the conditional probability $$ P[X_{i+1}=l | X_i =k] := \frac{P[X_{i+1}=l,\ X_i = k]}{P[X_i=k]}. $$ First note that each of the sets in the above fraction is a subset of $\Omega = I^n \smallsetminus D$ where the last coordinates $a_{l+1},\dots,a_n$ are free. Thus, in light of the product structure of Lebesgue probability, it suffices to work in the setup where $n=l$.
Then note that the set $\{a \in \Omega: X_i(a)=k\}$ is, upto a set of measure zero, of the form \begin{equation} \bigcup_{\sigma \in \mathcal{F}} (\Delta_\sigma\times I^{l-k}) = \left(\bigcup_{\sigma \in \mathcal{F}}\Delta_\sigma\right) \times I^{k-l} \end{equation} where $\mathcal{F}$ is some set of permutations of $\{1,\dots,k\}$ and $$ \Delta_\sigma = \left\lbrace (a_1,\dots,a_k) \in I^k: a_{\sigma(1)} < \dots < a_{\sigma(k)} (= a_k)\right\rbrace. $$ Let $\pi: \Omega \to I^k$ denote the projection to the first $k$ coordinates. The set $$S:= \{a \in \Omega: X_{i+1}(a) = l,\ X_i(a) = k\} \text{ projects onto } T:= \bigcup_{\sigma \in \mathcal{F}} \Delta_\sigma.$$ Thus, we have $$ S = \bigcup_{\sigma \in \mathcal{F}} \left((\pi^{-1} \Delta_\sigma) \cap S\right) $$ For a fixed $\sigma \in \mathcal{F}$, a point $(a_1,\dots,a_k,\dots,a_l)$ lies in $(\pi^{-1} \Delta_\sigma) \cap S$ only if \begin{equation}\label{2}\tag{2} \max \{a_{k+1},\dots,a_{l-1}\} < a_k < a_l. \end{equation} Since we also have that $$ \max \{a_1, \dots, a_{k-1}\} < a_k \text{ (by the defining property of } \mathcal{F}), $$ the condition \eqref{2} gives us \begin{equation}\label{3}\tag{3} k\times(k+1) \times \dots \times (l-2) \end{equation} different configurations for $\{a_{k+1},\dots, a_{l-1}\}$. In particular, this demonstrates that $ (\pi^{-1}\Delta_\sigma) \cap S$ is the union of \eqref{3} many simplices of the form $$ \Phi_\tau := \left\lbrace (a_1,\dots, a_l) \in I^l : a_{\tau(1)}< \dots < a_{\tau(l)}\right\rbrace \text{ where } \tau \text{ is some permutation of } \{1,\dots, l\}. $$ And, more importantly, this estimate \eqref{3} is independent of $\sigma \in \mathcal{F}$. Thus we can compute \begin{equation} \begin{split} \frac{P[X_{i+1}=l,\ X_i = k]}{P[X_i=k]} &= \frac{1}{P[X_i=k]}\sum_{\sigma \in \mathcal{F}} P[(\pi^{-1}\Delta_\sigma) \cap S] \\&= \frac{1}{P[X_i=k]} \left( \# \mathcal{F} \cdot \frac{k\cdots (l-2)}{l!}\right) \\ &= \frac{1}{\sum_{\sigma \in \mathcal{F}} P[\Delta_{\sigma} \times I^{l-k}]} \left( \# \mathcal{F} \cdot \frac{k\cdots (l-2)}{l!}\right) \\ &= \frac{1}{\# \mathcal{F} \frac{1}{k!}} \left( \# \mathcal{F} \cdot \frac{k\cdots (l-2)}{l!}\right) \\ &= \frac{k}{l(l-1)}. \end{split} \end{equation}