Let $(X_n)_{n\geq 0}$ be i.i.d. with $\mathbb P(X_n = 0 ) = \mathbb P(X_n = 1) = \frac{1}{2}$. Let $\tau_a$ be the stopping times defined as $$\tau_a = \inf\{n: (X_{n-3}, ... , X_n) = (1,0,0,1)\}$$
I have already shown that $\mathbb E[\tau_a] < \infty$. Now I am supposed to choose a suitable martingale to calculate $\mathbb E [\tau_a]$.
This is what I have so far (based off an example in my notes):
Suppose that I have a gambler $k$ who bets £1 (against a fair casino) that $X_k = 1$. If she wins she has £2 which she then bets on $X_{k+1}$ being $0$. If she wins again she bets her winnings on $X_{k+2}$ being 0 and lastly on $X_{k+3} = 1$.
Let $S_n$ be the total winnings of the casino at $n$. Then $$S_n = n + \sum_{k=1}^n A_k$$ where $A_n$ is the payout from the casino at time n. So $$A_n = 2X_n+4X_{n-1}(1-X_n)+ 8X_{n-2}(1-X_{n-1})(1-X_n) + 16X_{n-3}(1-X_{n-2})(1-X_{n-1})X_n.$$
By intuition since it is a fair bet $S_n$ is a zero valued martingale but I don't think that $$\mathbb E[S_{n+1}|\cal F] = S_n.$$
Is there another martingale I should choose?
After this I intend to use Doob's optional stopping theorem to calculate $\mathbb E[\tau_a]$.
Let $F_n$ be the total fortune of gamblers $0,1,2,\ldots,n$ at time $n$. Then $G_n=F_n-n$ is a martingale with initial value $0$. And $F_{\tau_a}=2^4+2$. The $2^4$ is the fortune of gambler number $\tau_a-3$, and the $2$ is the fortune of gamble number $\tau_a$. All other gamblers are out of the picture (or have yet to start) at time $\tau_a$. By Doob's theorem, $0=\Bbb E[G_{\tau_a}]=\Bbb E[F_{\tau_a}-\tau_a]=18-\Bbb E[\tau_a]$.