Consider a fair 6-sided die. If you roll a 1,2,3 you increment your score by 1 and roll again. If you roll a 4,5, the game terminates and your payoff for the game is your accumulated score. If you roll a 6, the game terminates and your payoff is nothing.
If we define $N$ to be the number of rolls to get a success (termination state), then we know $E[N] = 2$. The accumulated score is 1 less of this, so it's just 1. The expected payoff is thus $\frac{2}{3}$ since when terminating, 2 out of the 3 states result in profit and the other does not. I know this is the correct answer.
However, I tried a different approach for this problem, and I don't know why it's not working. Let $P$ be the payoff.
Then we have the following
$$ E[P] = \sum_{i=1}^6 E[P|E_1]P(E_1) \\ P(E_1) = \frac{1}{6} $$ where $E_i$ is the event of rolling $i$.
We know that
$$ E[P|E_1] = E[P|E_2] = E[P|E_3] = 1 + E[P] \\ E[P|E_4] = E[P|E_5] = E[P] \\ E[P|E_6] = 0 $$
If you plug this in and solve for $E[P]$, you get $E[P] = 3$. What is wrong with this latter approach? I suspect it's something to do with my conditional expectations, but those seem right to me.