I enter a casino with $X_0 = 1$ dollar. I then keep playing a game: With probability $\frac{1}{3}$, my money triples. With probability $\frac{2}{3}$, my money decreases by a factor of 3. Let $X_n$ be how many dollars I have after playing this game $n$ times. Hence, $X_{n+1} = 3X_n$ with probability $\frac{1}{3}$, and $X_{n+1} = \frac{X_n}{3}$ with probability $\frac{2}{3}$. Why does $X_n$ converge to 0 as $n$ approaches $\infty$?
I got that $X_n$ converges to $\infty$ (or does not converge):
$ E(X_n) = \frac{1}{3}(3)E(X_{n-1}) + \frac{2}{3}\frac{1}{3}E(X_{n-1}) \\ E(X_n) = \frac{11}{9}E(X_{n-1}) $
And since $\frac{11}{9} > 1$, shouldn't $E(X_n)$ diverge as $n$ approaches $\infty$?