Consider a re-combining binomial tree with probability of up = $p$ and probability of down = $(1-p)$. Let $n$ be the number of time steps in the binomial tree (the $x$-axis is time, and each column of nodes corresponds to "a time step". Also, each column of nodes is evenly spaced along to $x$-axis).
A re-combining binomial tree is in this picture. "Re-combining" just means that we end up in the same place given (i) Up movement --> Down movement, (ii) Down movement --> Up movement.
The set $\{0,1,...,n\}$ corresponds to the location of each node(s) on the $x$-axis. So, in the above picture, "1" would correspond to the nodes $S_{\Delta t}$, and "2" would correspond to the nodes $S_{2\Delta t}$. Consider the sets of nodes that correspond to to the following locations on the $x$-axis:
(i) $m \in \{1,...,n-2\}$,
(ii) $m+1$ and
(iii) $n$.
Let the $j$-th node at $m$ be the node that has had $j$ up-movements in all the sample paths leading to it. Let the $i$-th node at $n$ be similarly defined.
I want to find the probability that the sample path goes through node $j$ at $m$, node $j$ at $m+1$, then through node $i$ at $n$. Denote this $P\{(j,m)\cap(j,m+1)\cap(i,n)\}$.
We already know that $P\{(i,n)\cap(j,m)\} = P\{(j,m)\}P\{(i,n)|(j,m)\} = \frac{m!}{j!(m-j)!}p^j(1-p)^{m-j}*\frac{(n-m)!}{(i-j)!(n-m-i+j)!}p^{i-j}(1-p)^{n-m-i+j} = \frac{m!(n-m)!}{j!(m-j)!(i-j)!(n-m-i+j)!}p^i (1-p)^{n-i}.$
I'm looking for a way to arrive at $P\{(j,m)\cap(j,m+1)\cap(i,n)\}$.
I already know the answer is obviously $P\{(j,m)\cap(j,m+1)\cap(i,n)\} = P\{(i,n)|(j,m+1)\}P\{(j,m)\}*(1-p)$ from observation ... But I would like a "proper" way to derive it.