Weighted Average?

Firstly apologies for vague title. If I knew what I was trying to find I'd probably be able to locate it on Google.

Assume x games to be played by a team using one of two different venues. Venue A charges the team 10 dollars per game and venue B charges 8 dollars per game. What I'm after is a formula where I can plug in the two values for the venues and then set a weighting so that a certain percentage of the games are played at a particular venue. Based on that weighting I'd like to get the average payment per game.

• why the down vote .. am explanation might help me improve the question??? Jan 22, 2018 at 20:55

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.

• thank you for the thorough explanation .. appreciated Jan 22, 2018 at 20:58

If $20\%$ of the game is played at venue $A$, then the average payment would be

$$\color{blue}{20}\% \cdot (10) + (100-\color{blue}{20})\% \cdot (8)$$

There is nothing special about $\color{blue}{20}\%$