Sequence waiting time, computing the mean of a stopping time using martingales This is a question about  Example 14.1.13 from "A First Look at Rigorous Probability Theory" by Jeffrey Rosenthal, 2nd Edition. The example concerns an infinite fair coin tossing experiment, $\left \{r_n \right \}_{n \ge 1}$, with the $r_n$ having values 1 for heads and 0 for tails.
$\tau = \inf \left \{ n \ge 3: r_{n-2} = 1, r_{n-1} = 0, r_n = 1 \right \}$ is the first time the sequence heads, tails, heads appears.
The purpose of the example is to show how $E[\tau]$ can be computed using martingales. This is done as follows:
At each time $n$ a new player appears and bets \$1 on heads, if they win they bet \$2 on tails and if they win again they bet \$4 on heads. They stop betting if they either lose once or win three bets in a row. Let $S_n$ be they total amount won by all betters by time $n$. Then by construction $S_n$ is a martingale with stopping time $\tau$. 
Once $S_n$ is shown to be a martingale the rest of the example is straightforward. However, I must be missing something obvious because I don't follow how $S_n$ as constructed is necessarily a martingale. I'm sure it's easy to see this and I've just missed/misunderstood something in how $S_n$ is constructed. Any pointers would be appreciated.
 A: ...and the answer is simple. With $\mathcal{F}_n$ as a series of increasing $\sigma$-fields write (as usual):
$$ S_n = S_{n-1} + W_n $$ where $W_n$ are the winnings at time $n$, then
$$E[S_n | \mathcal{F}_{n-1}] = S_{n-1} + E[W_n | \mathcal{F}_{n-1}]$$
$W_n$ only depends on the last 3 tosses of the coin so only players who joined at $n-2$, $n-1$ and $n$ contribute. For the player joining at $n-2$ only 2 possible combinations contribute to winnings at time $n$:
$$ r_{n-2} = 1, r_{n-1} = 0, r_n = 1$$ winning \$4,
and
$$ r_{n-2} = 1, r_{n-1} = 0, r_n = 0$$ winning -\$4. (The mistake I had made was thinking that $ r_{n-2} = 1, r_{n-1} = 0, r_n = 1$ gave winnings of \$7 at time $n$ and $ r_{n-2} = 1, r_{n-1} = 0, r_n = 0$ winnings of -\$1 at time $n$.) So the expected winnings for player $n-2$ at time $n$, given $\mathcal{F}_{n-1}$ (as $r_{n-2}, r_{n-1} \in \mathcal{F}_{n-1}$), are 0 (all combinations have equal probability of occurring). Similar holds for the players who joined at $n-1$ and $n$. Therefore
$$E[W_n | \mathcal{F}_{n-1}] = 0$$ and
$$E[S_n | \mathcal{F}_{n-1}] = S_{n-1}$$ Therefore $S_n$ is a martingale.
