Suppose in a game, if I win in the $j$th round, I gain $+\$2^{j-1}$ and if I don't win in the $j$th, I lose $-\$2^{j-1}$. If I lose, I will keep playing until I win. Once I win, I leave the game. Otherwise, I continue to play until 30 rounds and leave the game even if I don't win. In other words, I will just stop at the $30$th round. Each round is independent and the probability of winning in each round is $\frac{9}{13}$.
I let $X$ be a random variable of my winnings. I want to find my expected winnings.
Since it works in a way that I would stop during my first win, I have a feeling that $X$ should be distributed over the Geometric Distribution. The game is either a win or a lose, so it is pretty much like a Bernoulli trial. But since I am not getting the expected number of rounds played, the Bernoulli trials cannot be just $0$ or $1$. So I thought I could modify it to become this way: $ { X }_{ j }=\left\{\begin{matrix} +2^{j-1} & if\; win\\ -2^{j-1} & if\; lose \end{matrix}\right. $ Then, $E(X)=E(X_1+X_2+\cdots +X_{30})=E(X_1)+E(X_2)+\cdots +E(X_{30})$
However, because I thought this is a Geometric Distribution, the expected value is just $\frac{1}{p}=\frac{13}{9}$. But I don't think the expected winnings is $\frac{13}{9}$ and is wrong.
Is what I have done correct?
So, instead of the usual finding of the expected number of trials of a standard geometric distribution, how can I find an expected number of another factor due to the trials (in this case, the expected number of winnings from the trials)?
Edit:
What I attempted to do was to make use of an indicator function to determine the expectation. But it doesn't seem successful.