I am looking to solve the following:
A person plays a coin tossing game where he receives $1$ point for every heads and $5$ points for every tails. The game stops when he receives $1,000$ points. Estimate within $\pm{2}$ the expectation of the length of the coin tossing game.
I know this is discrete martingale problem but I'm not sure how to approach it.
The expected number of points per game is $3$. You can then check that $M_n = \sum_{i=1}^n(X_i-3) = (\sum_{i=1}^n X_i)-3n = S_n-3n$ is a martingale, where $X_i$ is the number of points scored on round $i$ of the game, and $S_n$ is the total number of points scored by time $n$. Let $\tau$ be the first time $S_n \geq 1000$, i.e. the time the game stops. On one hand, $M$ is a martingale and $\tau$ is a bounded stopping time $(\tau \leq 1000)$, so optional stopping theorem tells us $E[M_\tau] = M_0 = 0$. On the other hand, $E[M_\tau] = E[S_\tau] - 3E[\tau]$. By definition, $1000\leq S_\tau \leq 1004 = 999+5$. Rearranging, we have $E[\tau] = E[S_\tau]/3$, so $$ \frac{1000}{3} \leq E[\tau] \leq \frac{1004}{3}. $$