We've got the following game:
You roll two dices. You get paid equal to the number rolled.
Additionally, if you roll doubles, you reroll (Same rules apply to that roll. That means there's not limit to how much you can win).
What's the expected payment of this game?
I'd managed to do the following:
$X$ is a random variable for the expected win of a single roll.
$P(reroll) = P(doubles) = {1\over 6} \\ E[X]=7$
From there, I tried to get the expected number of rolls.
$E[rolls]=\sum\limits_{n=0} P(reroll)^n={6\over 5}$
Now comes my problem, my first instinct was:
$E[payment] = E[rolls·X] = E[rolls]·E[X] = {6\over 5}·7=8.4$
Even though, that implies $rolls$ and $X$ are independent, but I'm afraid they are not.
How could I sove this problem?
Firstly I advice you to usurp the answer of @Ross. A lot of problems concerning expectation can be solved like that, and it is by far the most easy and elegant way.
Secondly your instinct did it quite well, but is not completely okay. See the comment of @Logophobic for that.
Set $P=P_1+\cdots+P_R$ where $R$ denotes the number of rolls and $P_i$ the payment at roll $i$. $$\mathbb EP=\sum_{n=1}^{\infty}\mathbb E(P\mid R=n)P(R=n)$$ We must do it with: $$\mathbb E(P\mid R=n)=\mathbb E(P_1\mid R=n)+\dots+\mathbb E(P_n\mid R=n)$$
and of essential importance is here that $P_i$ and $R$ are not independent. In principle we must calculate $\mathbb E(P_i\mid R=n)$ for each $i$.
If $i<n$ then $\mathbb E(P_i\mid R=n)=\sum_{k=1}^6\frac162k=7$.
Coincidently we also find $\mathbb E(P_n\mid R=n)=7$ and we are lucky with that.
In relief we end up with: $$\mathbb EP=7\mathbb ER=7\times\frac65=8.4$$