Let $X_n$ be a (random) fortune of a gambler after the $n$th round of gambling, which evolves as follows. After the $n$th round, his fortune $X_n$ depends on the fortune $X_{n-1}$ from the prior round as follows:
$X_n$= $Z_n$ $X_{n-1}$ for $n=1,2,\ldots$
where $X_0 = 1$ is the gambler's initial investment. The variables $Z_n$ ($n\geq1$), are independent identically distributed discrete random variables with pmf given by
$$p(z) = \begin{cases} 1/3 & \text{ if } z=3 \\ 2/3 & \text{ if } z=1/3 \\ 0 & \text{ o.w.} \end{cases}$$
Q)What is the expected fortune $\mathbb{E}[X_n]$ for arbitrary $n$?
Have you tried looking computing the values of $\mathbb{E}(X_n)$ for small values of $n$? For example, $$\mathbb{E}(X_1) = \mathbb{E}(Z_1X_0),$$ by independence of $Z_1$, we have $$\mathbb{E}(Z_1X_0) = \mathbb{E}(Z_1)\mathbb{E}(X_0) = \left(3\cdot\frac{1}{3} + \frac{1}{3}\cdot \frac{2}{3}\right) \cdot 0 = \frac{11}{9}.$$
Similarly, $$\mathbb{E}(X_2) = \mathbb{E}(Z_2X_1) = \mathbb{E}(Z_2)\mathbb{E}(X_1) = \frac{11}{9} \cdot \frac{11}{9} = \left(\frac{11}{9}\right)^2.$$
In general, $$\mathbb{E}(X_n) = \left(\frac{11}{9}\right)^n.$$ As a very simple exercise you could try proving this by induction. Since I have already demonstrated the base case, you would simply need to show that $$\mathbb{E}(X_n) = \left(\frac{11}{9}\right)^n \Rightarrow \mathbb{E}(X_{n+1}) = \left(\frac{11}{9}\right)^{n+1}$$ to complete the proof. Alternatively, you could show (again by very simply induction) that $$\displaystyle X_n = \prod_{i=1}^nZ_i.$$ Then it would follow by independence that $$\mathbb{E}(X_n) = \mathbb{E}\left(\prod_{i=1}^nZ_i\right) = \prod_{i=1}^n\mathbb{E}(Z_i).$$