I have a problem that is very similar to gambler's ruin but with a major modification. It is similar in the sense that you have an initial capital $n$, your wealth goes up or down by 1 dollar after every turn. To generalize, we do not assume equal up/down probabilities. Instead, we have a probability of going up equal to $u$ and probability of going down equal to $d$. You will stop "gambling" once you win enough and your wealth reaches a level $M$ or once you lose too much and your wealth reaches a level $m$. Obviously, 0 < $m$ < $n$ < $M$. For even greater generality, let us assume that after every turn, your wealth can stay constant with probability $1 - u - d$. So far, I can still answer this problem. It is well-known that the probability of reaching $M$ before reaching $m$ is equal to
$$p_n = \frac{1-\left(\frac{d}{u}\right)^{n-m}}{1-\left(\frac{d}{u}\right)^{M-m}}$$
assuming $u \neq d$ and $u \neq 0$.
However, imagine that at every turn, you have a constant, non-zero probability equal to $b$ of immediately going bankrupt (i.e., your wealth immediately becomes zero). What is the probability of getting bankrupt before your wealth reaches $m$ or $M$?
I modeled this as a pseudo birth-death-process Markov chain with three absorbing states ($m$, $M$, and bankruptcy) and I was able to setup a transition probability matrix that looks like a tri-diagonal matrix with entries $d$, $1-u-d$, $u$ in every row but with an extra column filled with $b$'s. I cannot seem to find an easy, closed-form formula for the probability of bankruptcy for a given starting state in this setup.
I also tried solving this problem using a recurrence relation $$p_n = d*p_{n-1} + (1-u-d-b)*p_n + u*p_{n+1} + b$$ where $p_n$ is the probability of bankruptcy given a starting wealth of $n$. However, the roots of my characteristic equation look messy and I don't know if they will lead me to a clean answer.
Can anyone point me to the right direction?
So the probability that you will eventually go bankrupt starting from wealth level $n$ satisfies
$$(1-c)p_n=up_{n+1} + d p_{n-1} + b$$
where $c=1-u-d-b$ is the probability to not move. You can rewrite that as
$$-up_{n+1}+(1-c)p_n-dp_{n-1}=b.$$
The boundary conditions are $p_m=p_M=0$.
The homogeneous solution is given in terms of the exponents $\lambda_{1,2}=\frac{1-c \pm \sqrt{(1-c)^2-4ud}}{2u}$. A particular solution can be obtained by taking $p_n \equiv C$, in which case you find that $(-u+1-c-d)C=b$; note that $1-c=u+d+b$ so $-u+1-c-d=b$, hence $C=1$. Now you're just left to solve the linear system for the boundary conditions, which reads $c_1 \lambda_1^m + c_2 \lambda_2^m = -1,c_1 \lambda_1^M + c_2 \lambda_2^M = -1$.