Gambler's Ruin (Infinite Capital) here's a question that I've been stuck for very long:
Two gamblers $A$ and $B$ initially have capitals $\$k$ and infinite capital respectively, where $k$ is a positive integer. At each round of the game, $A$ wins $\$1$ from $B$ with probability $p$, or loses $\$1$ to $B$ with probability $1-p$, where $0.5 < p < 1$. If, however, A has exactly $\$1$ just before a certain round and he then loses $\$1$ at this round then, with probability $s$ (where $0 \leqslant s<1$), he can receive $\$1$ from someone. Different rounds of the game are independent. $A$ and $B$ play until one of them is ruined.
(a) Find the probability that the capital of gambler $A$ has ever been $\$1$ given that $k = 2$.
(b) Find the ruin probability of A given that $k = 1$.
(c) Find the ruin probability of A given that $k$ is a positive integer.
I've tried this:
Given that $p > q$, I have used the formula $A + B\cdot[(1-p)/k]^k$ where $k = 2$, giving me 
$$
1 - \left(\frac{q}{p}\right)^2 > 0
$$
(since both gamblers $A$ and $B$ won't be ruined)
Not sure if I'm correct, please help!
 A: HINTS:
Presumably you're already familiar with the more traditional gamblers ruin problems. What makes this tricky is the possibility of someone swooping in and saving $A$ before he's completely ruined. The three parts of this question will guide you past this quirk.
Part a is strangely worded. It seems to be asking you for the probability that $A$ had $\$1$ at some point in the past given that you're currently observing him with $\$k$. This would also depend on the time at which we're observing him, which is not actually mentioned in the question. I think this is a mistake, and I'll instead talk about the probability that $A$ eventually has $\$1$ given that he starts with $\$2$.
For part a, consider a slightly different game. Rather than saying that $A$ has been ruined when he reaches $\$0$, we'll say that $A$ is ruined when he reaches $\$1$. What is the probability that $A$ becomes ruined in this new game? How does this relate back to the question?
For part b, note that if we start from $\$1$, there are two possibilities. Either $A$ loses and is instantly ruined, or $A$ wins, and has $\$2$. Before $A$ is ruined, though, he'll need to go back to having $\$1$ at some point first. How does this relate to part a?
Part c should drop out easily by generalising your answer to part a.
SPOILERS:
Consider a standard Gambler's Ruin game in which $A$ starts with $\$1$, $B$ has infinite capital, and $A$ wins w.p. $p$. Let $q = 1-p$ From standard results listed on Wikipedia, the probability that $A$ is ever ruined is
$$
\lim_{n_2\rightarrow \infty} \frac{\frac{q}{p}-\left(\frac{q}{p}\right)^{1+n_2}}{1-\left(\frac{q}{p}\right)^{1+n_2}} = \frac{q}{p}
$$
This all works out fine, since we're given that $p \in (0.5,1)$, so $q<p$. This game is identical to the one in which $A$ starts with $\$2$, $B$ has infinite capital, and $A$ is ruined at $\$1$, which is also the probability that $A$ reaches $\$1$ at some point in the future given that he starts at $\$2$.
For part b, suppose $A$ has $\$1$, and let $R$ be the probability of eventual ruin given that $A$ starts out with $\$1$. There are three possibilities.


*

*$A$ loses and is not saved with probability $(1-s)(1-p)$, leading to ruin with probability 1.

*$A$ loses but is saved with probability $s(1-p)$, leading to ruin with probability $R$

*$A$ wins with probability $p$. He then returns to $\$1$ with probability $\frac{q}{p}$, leading to ruin with probability $R$
This then tells us that $R = (1-p)(1-s) + (1-p)sR + p(\frac{q}{p})R$, which solves to
$$
R = \frac{(1-p)(1-s)}{1-(1-p)(s+1)}
$$
Check your sanity: whenever you get a statement like this make sure it'a actually possible for this to be a probability. You should be able to do this yourself, and there's no excuse beyond hubris not to.
For part c, we just throw in the probability that we eventually reach 1 if we start from $k$. If you could follow my reasoning up there, you should also be able to do this. If you can't follow my reasoning, ask for clarification.
