Suppose you've got a game wheel that can be divided into m slices where each slice contains a symbol and an associated multiplier w. The game is played by placing bets on one or more symbols where the total number of bets is non 0. The wheel is spun and a slice on the wheel is selected. Bets corresponding with the symbol on the selected slice are multiplied by the multiplier on the selected slice.
If given n symbols with an associated set of multipliers and an overall expected value $\overline{x_{total}}$, how can we calculate the relative probabilities $p_i$ (number of slices in m) for each multiplier in each symbol so that the game has the same expected value no matter what is bet on?
So far this is what I've got:
$$ \overline{x_{total}} = \frac{\sum_{i=1}^{n} \overline{x_{i}}}{n} $$
where $\overline{x_{i}} = \overline{x_{total}} $ always. and $\overline{x_{i}}$ is defined as:
$$ \overline{x_{i}} = \sum_{i \in n}w_i p_i $$
where $\sum_{} p_i = m = 1$, $0 \leq p_i \leq 1$, and $w_i \geq 0$.
Example of information that can be given:
$\overline{x_{total}} = 0.8$ ($\$0.80$ is expected to be returned to the player for every dollar spent in the game on average.)
$Symbol_A = \{5,2, 0\}$
$Symbol_B = \{10, 15,20\}$
This would result in:
$$ \overline{x_{A}} = \sum_{i \in A} w_i p_i = 5p_1 +2p_2+0p_3 = 0.8$$
$$ \overline{x_{B}} = \sum_{i \in B} w_i p_i = 10p_4 +15p_5+20p_6 = 0.8$$
$$ \overline{x_{total}} = \frac{\overline{x_{A}}+\overline{x_{B}}}{2} = 0.8$$
By trial and error on a calculator I found these values for the above example: $$p_1 = 0.04, p_2 = 0.3, p_3 = 0.6, p_4 = 0.03, p_5 = 0.02, p_6 = 0.01, $$
My gut tells me that variance should play a role I'm just not sure how.
Besides the equations for $\overline{x_A}$ and $\overline{x_B}$ that you have given, we have $$p_1+p_2+\cdots+p_6=1$$ so three linear equations in $6$ unknowns. If there is one solution, there are infinitely many. You can find them as usual by Gaussian elimination. Of course, we also need $p_i\geq0$, for $i=1,\dots,6$, but you can deal with that after finding the general solution.
Variance plays no role in the problem as you have stated it.