I am interested in the expectation of a random variable $Z$ that counts the number of independent trials needed until either mutually exclusive events $A$ or $B$ occur with probabilities $p_A$ and $p_B$ respectively. By definition, $Z \sim Geom(p_A + p_B)$, therefore $E(Z) = \frac{1}{p_A + p_B}$. Now, if I define the event, $$F \equiv \text{A occurs first}$$ using the Law of Total Expectation conditioning on $F$:
$$ E(Z) = P(F)E[Z|F] + P(F^c)E[Z|F^c] $$
with $P(F) = \frac{p_A}{p_A + p_B}$, since the trials are independent. Now, my argument goes by noting that $$ Z|F \sim Geom\left(\frac{p_A}{1 - p_B}\right) \\ Z|F^c \sim Geom\left(\frac{p_B}{1 - p_A}\right) $$
The idea being that, if for example A occurs first, then we can re-compute probabilities conditional on B not happening. However, when I plug this into the previous expression, I get: \begin{align*} E(Z) = \frac{1}{p_A + p_B} & = \frac{p_A}{p_A + p_B} \frac{1 - p_B}{p_A} + \frac{p_B}{p_A + p_B} \frac{1 - p_A}{p_B} \\ & = \frac{2 - p_A - p_B}{p_A + p_B} \end{align*}
which is obviously wrong unless $p_A + p_B = 1$, i.e. that A and B are a partition of the sample space, but then E(Z) would trivially equal one in the first place. I would greatly appreciate if someone can point me to the flaw in the argument?
I have managed to find the answer myself. My assumption that $$ Z|F \sim Geom\left(\frac{p_A}{1 - p_B}\right) \\ Z|F^c \sim Geom\left(\frac{p_B}{1 - p_A}\right) $$
was wrong. In fact, it can be shown that $P(Z = z | F) = P(Z = z)$ in the following way: $$ P(Z = z | F) = \frac{P(Z = z, F)}{P(F)} = \frac{(1 - p_A - p_B)^{z-1}p_A}{p_A/(p_A+p_B)} = (1 - p_A - p_B)^{z-1}(p_A + p_B) = P(Z = z) $$
Therefore, we simply have $$ Z|F, Z|F^c \sim Geom(p_A + p_B) $$ Plugging this into the Law of Total Expectation leads now to a consistent result.