Gambler's ruin

We will use the Optional stopping theorem on two associated Martingales to solve the problem.

Let us denote the gain of the gambler on the $i$ th game by the random variable $X_i, i\geq 1$ , where $X_i$ 's are i.i.d. Each of the $X_i$ s take values from the set $\{1,-1\}$ with equal probability, thus $\mathbb{E}X_i=0, \mathbb{E}X_i^2=1, \forall i\geq 1$ . Total winnings of the gambler upto $n$ th game is given by the r.v. $S_n$ where $S_n=\sum_{i=1}^{n}X_i$ Let the threshold for the gambler for quitting be $a$ and $-b$ . Here $a=5000, b=2000$ . The gambler quits when $S_n$ reaches either $a$ or $-b$ . Let $T$ denote the random time when the gambler quits. Note that $T$ is a stopping time , w.r.t. the $\sigma$ field generated by $\{S_n\}_{n\geq 1}$ . It is also clear that the sequence $|S_{n\wedge T}|$ is uniformly bounded. Hence using the optional stopping theorem on the zero-mean martingale Sequence $\{S_n\}$ and the stopping time $T$ we conclude that, $\mathbb{E}S_{T}=0$ i.e., $a\mathbb{P}(S_T=a)-b\mathbb{P}(S_T=-b)=0$ Hence, $\mathbb{P}(S_T=a)=\frac{b}{a+b}, \mathbb{P}(S_T=-b)=\frac{a}{a+b}$ Finally, we check that the sequence of random variables $\{Z_n\}, n\geq 1$ , where $Z_n=S_n^2-n$ is also a zero-mean martingale. Hence using the Optional Stopping Theorem on $\{Z_n\}$ with the stopping time $T$ , we conclude that $\mathbb{E}Z_T=0, \text{i.e. }\mathbb{E}\big(S_T^2-T\big)=0$ . Hence, $\mathbb{E}(T)=\mathbb{E}(S_T^2)=a^2\frac{b}{a+b}+(-b)^2\frac{a}{a+b}=ab$ Hence the expected number of games that the gambler plays is $5000\times 2000=10,000,000$ .