Consider a 1-dimensional random walk,we have a absorbing barrier at -a.The probability of moving to the left or right on every step is p and 1-p.
My question is how to calculate the probability of being absorbed eventually if we start at the origin.
Attempt I
I try to imitate the solution to solve the similar question with two absorbing barrier(eg.a and -a).
In detail,let $$p_k=P(\{\text{start at k and be absorbed at a eventually\}})$$ Obviewsly,we have $p_{-a}=0$ and $p_{a}=1$
Besides,we can draw the equation: $$p_k=p_{k+1}*p+p_{k-1}*(1-p)$$
By this equation and initial condition above,we can attain the expression of $p_k$ and $p_0$ is what we need.
But this method is noneffective for the problem at hand because there is only one barrier,thus we only have one equation in the initial condition.
Attempt II
I try to simplify the question.Consider there is no barrier,we only need to start at the orign and end at -a with n steps.
It's simple to calculate such $p_n$.But some moves are invalid for our real question because they have reached -a before n steps.If we can find the number of invalid moves,which I can't resolve,we can attain $p_n'$,probability of being absorbed at -a right on the n_th step.Finally what we need is $\sum\limits_{n=a}^{+\infty}p_n'$
Hint: Attempt 1 is close. You can indeed leverage the two-barrier solution; instead of having the other barrier be $a$, have it be some independent quantity (say, $b$). Compute the probability of reaching $b$ before reaching $-a$; then, take a limit as $b \to \infty$.