recently I am trying to solve the following question: Let $\xi_{1}, \xi_{2}, ... $ be Bernoulli Random variable with $\mathbb{P}(\xi = 1) = \dfrac{1}{2}$ and $\mathbb{P}(\xi = 0) = \dfrac{1}{2}$. Define stopping time $\tau := \inf \{n \geq 5 \mid \xi_{n}\xi_{n-1}\xi_{n-2}\xi_{n-3}\xi_{n-4} = 10101 \}$, and our job is to find $\mathbb{E}[\tau]$.
My approach is as follows:
Consider the game where at each time $n \in \mathbb{N}$, a new player enters the game and bets 1 dollar. If $\xi_{n} = 1$, then his money doubles to 2 dollar and he stays in the game; otherwise he loses his money. If the next flip of the coin is 0, then his money doubles again and he stays in the game; otherwise he loses all his money. This continues until the player either sees 10101, in which case he leaves the game with $32.
I define martingale $X_{n} := $ (total amount of money that Casino wins from the above game), and tried to show that $X_{n}$ is indeed a martingale, which allows us to apply Optional Stopping Theorem, and conclude that $\mathbb{E}[X_{\tau}] = \mathbb{E}[\tau - 32-8-2] = 0 \rightarrow \mathbb{E}[\tau] = 42$. However, I am hard stuck at showing that $X_{n}$ is indeed a martingale. Could anyone help me to show that $X_{n}$ I defined above is a martingale? Thanks!!
Let me answer your question by first proving a more general fact. Let $(\mathcal F_n)_{n\ge 0}$ be an increasing sequence of $\sigma$-algebras, and let $(M_n)_{n\ge 0}$ be a martingale adapted to this filtration. Furthermore, let $(B_n)_{n\ge 1}$ be a sequence of random variables with the property that $B_n\in \mathcal F_{n-1}$ for all $n\ge 1$. We say that $B_n$ is predictable, meaning that you can predict $B_n$ using only the information up to time $n-1$.
Imagine that $M_n$ is the price of a certain stock or commodity at time $n$, while $B_n$ is the amount of money you have invested in that stock/commodity. Between time $n-1$ and time $n$, the stock price changes by $M_n-M_{n-1}$, which means your profit over that period is $B_n\cdot (M_n-M_{n-1})$. Therefore, letting $X_n$ be your total profit up to time $n$, we have $$ X_n=\sum_{k=1}^n B_k(M_k-M_{k-1}) $$
Proof: Clearly, $X_n\in \mathcal F_n$. Furthermore, $$ \begin{align} \mathbb E[X_n-X_{n-1}\mid \mathcal F_{n-1}] &=\mathbb E[B_n(M_{n}-M_{n-1})\mid \mathcal F_{n-1}] \\&=B_n\cdot \mathbb E[M_{n}-M_{n-1}\mid \mathcal F_{n-1}] \\&=B_n\cdot 0=0. \end{align} $$ In the second equality, we can pull the $B_n$ out of the conditional expectation because $B_n\in \mathcal F_{n-1}$, so $B_n$ is a constant with respect to $\mathcal F_{n-1}$.
To apply this to your problem, let
$\mathcal F_n=\sigma(\xi_1,\dots,\xi_n)$, representing knowledge of the first $n$ coin flips.
$M_n=\#\{k\mid \xi_n=1,1\le k\le n\}-\#\{k\mid \xi_n=0, 1\le k\le n\}.$ That is, $M_n$ is the number of ones minus the number of zeroes up to time $n$. Clearly, $M_n$ is a martingale.
$B_n$ be the total amount that all of the betters bet on the $n^\text{th}$ flip being $1$. That is, you add up the total amount which is bet on $1$, and subtract the amount which is bet on $0$ (so betting $x$ dollars on $0$ is the same as betting $-x$ dollar on $1$). It should be clear that $B_n\in \mathcal F_{n-1}$, because each better decides how much to bet based on the first $n-1$ results.
You can then verify that the total amount of money won by all betters is given by the $X_n$ expression from before, so the Lemma implies that the total amount won is a martingale.