Considering a random walk problem with two walls that can absorbing the particle, i.e. a particle moves randomly on a number axis and when it moves to the walls it will be absorbed.
Let p denotes the probability that in one move the particle moves right and the locations of the two walls are $x=0,\ x=a+b$ respectively. We assume that the particle won't keep still in each move and thus 1-p is the probability of moving left.
By total probability formula we can calculate the probability of being absorbed by the walls on $x=a+b$ with initial location $x=n\ (n\in\{0,1,2,\cdots,a+b\})$ $$q_{n} = \frac{1-(\frac{q}{p})^n}{1-(\frac{q}{p})^{a+b}}$$ where $q=1-p$.
I find the form of numerator and dominator pretty similar to the determinant of a special matrix $$D_n = \left[\begin{matrix} 1+(\frac{q}{p}) & \frac{q}{p} & 0 & 0 & \cdots & 0\\ 1 & 1+(\frac{q}{p}) & \frac{q}{p} & 0 & \cdots & 0\\ 0 & 1 & 1+(\frac{q}{p}) & \frac{q}{p} & \cdots & 0\\ \vdots & & \ddots & \ddots & \ddots & \vdots\\ 0 & 0 & \cdots & 1 & 1+(\frac{q}{p}) & \frac{q}{p} \\ 0 & 0 & 0 &\cdots & 1 & 1+(\frac{q}{p}) \end{matrix}\right]_{n\times n} $$ and $det(D_n) = \frac{1-(\frac{q}{p})^n}{1-(\frac{q}{p})}$. Thus $$q_n = \frac{det(D_n)}{det(D_{a+b})}$$
It seems that for every random walk process with initial location $x=n$, denoted as $r_n$, there is a corresponding matrix $D_n$ and we can use its determinant to calculate the probability.
Is it just a coincidence or is there any theory that can explain this phenomenon?