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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.

Ashish Gupta
  • 1,633

1 Answers1

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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?

Markus Scheuer
  • 108,315