Roll a fair die until game stops. Game stops when you roll 4,5, or 6. When you roll 1,2,3, your score increases by 1 and you can roll again. When you roll 4,5, you get paid your accumulated score. When you roll a 6, you get paid nothing.
The easiest approach I can think of is using a geometric random variable approach, where we see that the game has an expected number of $2$ tosses, which means the expected accumulated score by the time the game is finished is $1$. And when the game terminates, you only get paid this accumulated score if you toss a $4,5$, which occurs with $2/3$ probability (given that the game is terminating). So the expected payoff is $2/3$.
I tried to use a law of total expectation approach, but I am not sure if I'm computing the conditional expectations correctly. I am conditioning on the disjoint events of tossing a $\{1,2,3\}$, $\{4,5\}$ and $6$, denoted as $A,B,C$. Let $P$ be the payoff.
\begin{align} E[P] = E[P | A]P(A) + E[P | B]P(B) + E[P | C]P(C) \\ E[P | A] = 1 + E[P] \\ E[P | C] = 0 \\ P(A) = 0.5 \\ P(B) = 1/3 \end{align}
I think $E[P | B]$ requires us to consider a geometric distribution, specifically the number of consecutive tosses of $1,2,3$ until we hit $4,5,6$, and that's $1$, on average. So $E[P | B] = 1$.
But something is wrong, because when I plug in these quantities, I get $$ E[P] = 5/3 $$
What am I doing wrong here?
It seems that if $E[P | A] = E[P]$, I would get the expected result, but I don't think $E[P | A] = E[P]$ since if event $A$ occurs that means our score increases by 1 and we effectively restart the game with a score of 1.
This is hard to do along the lines you propose because winning $1$ in some round does not increase your expectation by $1$, as that $1$ is not certain.
However, it is easy to compute the expected value directly.
The probability of winning exactly $n$ for $n≥1$ is $\left(\frac 12\right)^n\times \frac 13$ so the expected value of the game is $$\sum_{n=1}^{\infty} n\times \left(\frac 12\right)^n\times \frac 13=\frac 23$$
As a (very informal) sanity check: note that the game is expected to end in $2$ rounds, and of course there is a $\frac 23$ chance it ends with a payout. That reasoning also gets you to $\frac 23$. Should stress that while this sort of argument can sometimes be justified (see Wald's Lemma) the justification tends to be harder than working out the solution by other means.