Suppose you have a fixed amount of money ($X$), and are playing a game against an infinitely rich adversary. At each stage you either win or lose $1$ unit with respective probabilities $p$ and $1 − p$. Show that the probability that you eventually go broke is
$$1 \space \space \space \space \space\space \space \space \space \space if \space p \leq 1/2 \\ (q/p)^X \space \space if \space p > 1/2$$ where $q = 1 − p$.
I know there is a solution by using recurrence relations, however, I am more interested in the following approach:
To lose $X$ amount of money, we can lose it in either $X$ games, $X+2$ games, $X+4$ games and so on. So, probability of going broke = $$\sum_{n = 0}^{\infty} \binom{X + 2r} {r}(1-p)^{X+r}p^r $$
I am specifically interested in the case when $p>1/2$. This suggests that,
$$\sum_{n = 0}^{\infty} \binom{X + 2r} {r}(1-p)^{X+r}p^r = \left(\frac{1-p}{p}\right)^X$$ (Using the result mentioned in the problem)
Or, $$\sum_{n = 0}^{\infty} \binom{X + 2r} {r}(p-p^2)^r = 1/p^X$$ Now,
$$\binom {X + 2r}{r} = \frac{\binom {X + 2r}{2r}}{\binom {X + r}{r}}\binom {2r}{r}$$ $$\implies \sum_{n = 0}^{\infty} \binom{X + 2r} {r}(p-p^2)^r = \sum_{n = 0}^{\infty}\frac{\binom {X + 2r}{2r}}{\binom {X + r}{r}}\binom {2r}{r} (p-p^2)^r$$
I am not sure how to proceed from here.
After a bit of manipulation on $\binom{2r}{r}$, I found that $$\binom{2r}{r} = \left(\frac{-1}{4}\right)^r \binom {-1/2}{r}$$ Combining this with the $p^r(1-p)^r$ factor in the sum gives the general term in the expansion of $(1 - p(1-p)/4)^{-1/2}$ However, I am unable to deal with the $\frac{\binom {X + 2r}{2r}}{\binom {X + r}{r}}$ part of the sum.
Is this approach correct? If so, how do I proceed?