In the case of a generalized Wheel of Fortune, the Expected Value is simply the dot-product of a Probabilities vector with a corresponding Values vector:
$$E(W)=\sum p_i v_i$$
By way of example, a three-slice wheel with two slices having a $25\%$ chance of "hitting" and the third slice having a $50\%$ chance of hitting, using values for the three slices respectively of 1, 4, 10 would yield : $$E(W) = (.25\cdot 1) + (.25\cdot 4) + (.50\cdot 10) = 6.25$$
If the goal were to have the same $ E(W) $ but make the probabilities for each slice as close in value as possible, one could define a simple error function acknowledging that perfectly equal slices would each have $\frac{1}{3}$ probability of hitting.
$$Err(p) = \sum ( p_i - \overline{p} )^2$$
This error could never be zero unless the $E(W)$ value were changed to 5, which is not desirable.
$$E(W) = \left(\frac{1}{3}\cdot 1\right ) + \left(\frac{1}{3}\cdot 4 \right) + \left(\frac{1}{3}\cdot 10\right) = 5$$
An alternate error function may be considered without regard to the mean probability:
$$Err(p) = \sum p_i^2$$
These $p_i$ values are probabilities:
$$\sum p_i = 1$$
$$0 < p_i < 1$$
Question: Given $E(W)$ and each value on the Wheel of Fortune $(v_1 \ldots v_n)$ as a priori constant, is there a well known way to establish $(p_1 \ldots p_n)$ such that either error function $Err(p)$ is minimized? Its safe to assume the error function is continuous.
I don't think simplex can work because of the constraints about probabilities summing to 1.
The error functions appear to be the same if one expands out the $(p_i - \bar p)^2$ term. But this seems to be a standard constrained optimization problem: $\min \|p\|^2$ subject to $\langle \vec{1},p\rangle = 1$ and $\langle v,p\rangle = \mathbb{E}[W]$, so you can just solve it with Lagrange multipliers. Let $w := \mathbb{E}[W]$, so the Lagrangian is
\begin{align*} \max_{\lambda_1, \lambda_2 \in \mathbb{R}} \min_{p \in \mathbb{R}^n} \big( \langle p,p \rangle - \lambda_1 (\langle \vec 1,p\rangle -1) - \lambda_2 (\langle v,p\rangle -w) \big). \end{align*} The first order condition for $p$ implies $p = \frac{\lambda_1}{2} \vec1 + \frac{\lambda_2}{2} v$, so after plugging in, our goal is now \begin{align*} &\max_{\lambda_1, \lambda_2 \in \mathbb{R}} \frac 14 \big( -\lambda_1^2 \langle \vec1,\vec1 \rangle - 2 \lambda_1 \lambda_2 \langle \vec1,v \rangle - \lambda_2^2 \|v\|^2 + 2 \lambda_1 + 2 \lambda_2 w \big) \\ =& \max_{\lambda_1, \lambda_2 \in \mathbb{R}} \frac 14 \big( -\lambda_1^2 n - 2 \lambda_1 \lambda_2 \langle \vec1,v \rangle - \lambda_2^2 \|v\|^2 + 2 \lambda_1 + 2 \lambda_2 w \big). \end{align*} The first order conditions for $\lambda_1$ and $\lambda_2$ then give the system \begin{align*} -\lambda_1 n - \lambda_2 \langle \vec1,v \rangle + 1 &= 0 \\ -\lambda_2 \|v\|^2 - \lambda_1 \langle \vec1,v \rangle + w &= 0, \end{align*} which can then be solved to find $\lambda_1$ and $\lambda_2$.