Expectation of sum of RVs: what if there's conditional RV inside the sum?

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I want to calculate an expected payoff from playing the following two games:

GAME 1: by some rules of the game which I'm not going to bother you with, you earn $z_1$ after playing the first game.

GAME 2: Here you earn $z_2$ after playing, but the distribution i.e. probability mass function of the random variable $Z_2$ will depend on $z_1$.

$$ f_{Z_2}(z_2) = \begin{cases} \text{(foo)} & \text{if } z_1 < 0.5 \\ \text{(bar)} & \text{if } z_1 > 1. \end{cases} $$

Hence the expected payoff from playing the two games would look like $$ E(Z_1 + (Z_2|Z_1)) = E(Z_1) + E(Z_2|Z_1). $$

And here's my dilemma: I know that $ E(Z_1)= 2/3 $. So it's tempting to calculate the expected payoff by

$$ E(Z_1) + E(Z_2 | Z_1=2/3) = (2/3) + E(Z_2 | Z_1=2/3) $$

However, I think this ignores the fact that $ E(Z_2|Z_1) $ is also a random variable. So perhaps what I should calculate is

$$ E(Z_1) + E ( E(Z_2|Z_1) ) = E(Z_1) + E(Z_2)? $$

(This equality is a consequence of the law of total expectations.)

Which one of the two possible `solutions' sounds more plausible? Or perhaps I should use a totally different method?

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The second one is correct and you have justified it using the law of total expectations.

The first one can't be correct. Note that $E[Z_2|Z_1]$ is a function of $Z_1$ and also, it is possible that $Z_1$ never take value of $E[Z_1]$.