<s>[INST] How do you find the probability of A winning if the probability of getting a favourable outcome in the $r^{th}$ turn is a function of $r$? Problem:
Two players A and B are playing snake and ladder. A is at 99 and he needs 1 to win in rolling of a dice. However, he is always allowed to re-throw the dice if 6 appears.
What is the probability that A will win eventually before B gets a chance, if the probability of getting 1 in the $r^{th}$ throw is $\frac{1}{3^r}$ and that of getting 6 in the $r^{th}$ throw is $\frac{2r-1}{4r}$?
My attempt:
We know that A can win before B gets a chance only if he rolls {$1$},{$6$,$1$},{$6$,$6$,$1$} and so on.
In the $r^{th}$ turn, we have the probability: $$\frac{1}{3^r}\cdot\frac{1}{4}\cdot\frac{3}{8}\cdot\cdot\cdot\frac{2r-3}{4(r-1)}$$
$$=\frac{1}{3^r}\cdot\frac{1}{4^{r-1}}\left(\frac{1\cdot3\cdot5\cdot7\cdot\cdot\cdot(2r-3)}{1\cdot2\cdot3\cdot4\cdot\cdot\cdot(r-1)}\right)$$
$$=\frac{1}{3^r}\cdot\frac{1}{4^{r-1}}\left(\frac{(2r-2)!}{(r-1)!\cdot(r-1)!\cdot2^{r-1}}\right)$$
$$=\frac{1}{3}\cdot\frac{1}{24^{r-1}}\left(\frac{(2r-2)!}{(r-1)!\cdot(r-1)!}\right)$$
Therefore, we have the probability as
$$\sum_{r=1}^{\infty} \frac{1}{3}\cdot\binom{2r-2}{r-1} \frac{1}{24^{r-1}}$$
Taking $r-1$=$n$
$$\frac{1}{3}\sum_{n=0}^{\infty} \binom{2n}{n} \frac{1}{24^{n}}$$

I got stuck at the last step because I do not know how to evaluate that summation. Any help with the summation/providing an alternate way to solve this question will be appreciated.
 [/INST] 
The generating function for the central binomial coefficients is
  \begin{align*}
\sum_{n=0}^{\infty}\binom{2n}{n}z^n=\frac{1}{\sqrt{1-4z}}\qquad |z|<\frac{1}{4}
\end{align*}
This is an application of the binomial series 
  \begin{align*}
(1+z)^{\alpha}=\sum_{n=0}^{\infty}\binom{\alpha}{n}z^n\qquad |z|<1, \alpha\in\mathbb{C}
\end{align*}
  and the relation
  \begin{align*}
\binom{-\frac{1}{2}}{n}=\frac{(-1)^n}{4^n}\binom{2n}{n}
\end{align*}

Can you proceed?
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