Yes, this looks exactly like a weighted average. Suppose you play various games. If you play $N_A$ games that each cost $C_A$, and you play $N_B$ games that cost $C_B$, then altogether you pay $N_A\cdot C_A + N_B\cdot C_B$ total for all $N_A + N_B$ games.
This implies that the average amount per game is the total cost divided by the total number of games:
$$\frac{N_A\cdot C_A + N_B\cdot C_B}{N_A + N_B}$$
In your case with venue A costing \$10, and venue B costing \$8, the average cost per game is:
$$\frac{10a + 8b}{a+b}$$
where $a$ and $b$ are the number of games played, respectively, at each venue.
If you'd rather specify the percentage of games played at a certain location, we have a similar formula for the average cost per game:
$$p_A \cdot C_A + p_B\cdot C_B$$
where $p_A$ and $p_B$ are the fraction of games (or probability of a game) played at each venue, respectively. (We know that $p_A = a/(a+b)$ and $p_B = b/(a+b)$, which is how we get this new formula from the old one).
In your case with venue A costing \$10, and venue B costing \$8, the average cost per game is:
$$10\,p_A + 8p_B$$
where $p_A$ and $p_B$ are the fraction of games played, respectively, at each venue. These fractions should add up to 1.
