What is the expected payoff of this game? 
I wager 100 Galactic Credits.  In each of two independent event, I have a $25\%$ chance of doubling my money.  If I win either the first or the second event, then I walk away with 200 Galactic Credits; if I win both events, I walk away with 400 Galactic Credits; and if I lose both events I walk away with nothing.  What is the expected payoff of this game? 

The chance to double my money the first or second time: $2 \times (25\% \times 75\%) = 37.5\%$.
So, $37.5\%$ of the time I would walk away with 200 credits, on average 75 credits. Correct?
The chance to double my money the first and the second time is $25\% \times 25\% = 6.25\%$, however I'm getting my return twice, so I would walk away with 400 credits $6.25\%$ of the time, on average, 25 credits.
So can it be that I have a $43.75\%$ chance to win at all, but because of the payout of the double win, a $50\%$ chance to double my money? Such that, on average, I walk away with an even 100 credits on average?
What am I missing, if anything?
 A: The analysis in the question appears to be correct, but it is phrased in a way which I find to be a little confused and hard to follow.  Another way of thinking of this problem is to simulate it with two four-sided dice.  If you wager 100 Galactic Credits ($\mathfrak{G}$), then the payoffs are given by:
\begin{matrix}
\text{Exactly One $4$:}\hfill & \hfill 200\ \mathfrak{G} & (1)\\
\text{Exactly Two $4$s:}\hfill & \hfill 400\ \mathfrak{G} & (2)\\
\text{Anything Else:}\hfill & \hfill 0\ \mathfrak{G} & (3)
\end{matrix}
The expected payoff is then
$$ (200\ \mathfrak{G}) \cdot P(\text{Event (1)})
+ (400\ \mathfrak{G}) \cdot P(\text{Event (2)})
+ (0\ \mathfrak{G}) \cdot P(\text{Event (3)}). \tag{$\ast$}$$
There are a total of 16 possible rolls (the first die can be any number from $1$ to $4$, and the second die can be any number from $1$ to $4$).  Event (1) (exactly one $4$) can happen in $6$ different ways:  the first die comes up $4$, and the second die can be any of the remaining three numbers; and vice versa.  Therefore
$$ P(\text{Event (1)})
 = \frac{\text{Good Outcomes}}{\text{Total Outcomes}}
 = \frac{6}{16}
 = 0.375.
$$
Event (2) can happen in only one way:  both dice must come up $4$.  Hence
$$ P(\text{Event (2)})
 = \frac{\text{Good Outcomes}}{\text{Total Outcomes}}
 = \frac{1}{16}
 = 0.0625.
$$
We could compute the probability of Event (3), but since it is being multiplied by a payoff of $0\ \mathfrak{G}$ in ($\ast$), it is not important to do so.  That is, the term doesn't matter, so we aren't going to spend any more time on it.  Substituting these results into ($\ast$) gives
$$ (200\ \mathfrak{G}) \cdot \frac{6}{16} + (400\ \mathfrak{G})\cdot \frac{1}{16}
= \frac{1200 + 400}{16}\ \mathfrak{G}
= 100\ \mathfrak{G}.$$
That is, on average, you break even (you started with 100 Galactic Credits, and walk away with 100 Galactic Credits).
