This question builds off of the previous MO question Number of collinear ways to fill a grid.
Let $A(m,n)$ denote the set of binary words $\alpha=(\alpha_1,\alpha_2,\ldots,\alpha_{m+n-2})$ consisting of $m-1$ $0's$ and $n-1$ $1's$. Evidently $\#A(m,n) = \binom{m+n-2}{m-1}$.
For $\alpha \in A(m,n)$ and $1\leq i \leq m+n-2$, set $$ b^\alpha_i := \#\{1\leq j < i\colon \alpha_i\neq\alpha_j\} +1.$$ $$ c^\alpha_i := \#\{1\leq j \leq i\colon \alpha_i=\alpha_j\}=(i+1)-b^\alpha_i.$$
The resolution of the above-linked question implies that $$ \sum_{\alpha \in A(m,n)} \frac{b^\alpha_1b^\alpha_2 \cdots b^\alpha_{m+n-2}}{b^\alpha_{m+n-2} (b^\alpha_{m+n-2}+b^\alpha_{m+n-3})\cdots (b^\alpha_{m+n-2}+b^\alpha_{m+n-3}+\cdots+b^\alpha_1) } = \frac{mn}{(m+n-1)!}$$ Meanwhile, in this answer, it is explained that $$ \sum_{\alpha \in A(m,n)} \frac{1}{c^\alpha_{m+n-2} (c^\alpha_{m+n-2}+c^\alpha_{m+n-3})\cdots (c^\alpha_{m+n-2}+c^\alpha_{m+n-3}+\cdots+c^\alpha_1) } = \frac{2^{m-1}2^{n-1}}{(2m-2)! (2n-2)!}$$
Considering the similarities of these two remarkable weighted sums over binary words, we ask:
Question: Is there a more general formula which specializes to the above two formulas?
EDIT:
Since for any $\alpha \in A(m,n)$ we have $$ \{c_1^{\alpha},c_2^{\alpha},\ldots,c_{m+n-2}^{\alpha}\} = \{1,2,\ldots,m-1,1,2,\ldots,n-1\},$$ we can rewrite that second sum to be $$ \sum_{\alpha \in A(m,n)} \frac{c^\alpha_1 c^\alpha_2 \cdots c^\alpha_{m+n-2}}{c^\alpha_{m+n-2} (c^\alpha_{m+n-2}+c^\alpha_{m+n-3})\cdots (c^\alpha_{m+n-2}+c^\alpha_{m+n-3}+\cdots+c^\alpha_1) } = \frac{2^{m-1}(m-1)!2^{n-1}(n-1)!}{(2m-2)! (2n-2)!},$$ to be even more similar to the first sum.
If we set $$ d^{\alpha}_i = xb^{\alpha}_i+yc^{\alpha}_i,$$ then the above commentary explains that $$ \sum_{\alpha \in A(m,n)} \frac{d^\alpha_1 d^\alpha_2 \cdots d^\alpha_{m+n-2}}{d^\alpha_{m+n-2} (d^\alpha_{m+n-2}+d^\alpha_{m+n-3})\cdots (d^\alpha_{m+n-2}+d^\alpha_{m+n-3}+\cdots+d^\alpha_1)}$$ has a product formula for $x,y\in \{0,1\}$. Maybe it has a product formula for general $x,y$.
