Let $X$ be a finite set and $G$ a finite group acting on $X$. By Burnside's lemma we have $$|X/G|=\frac1{|G|}\sum_{g\in G} |X^g|$$ where $X^g$ is the set of points in $X$ fixed by the group element $g$,i.e. $\{x\in X: gx=x\}$.
Let $\lambda= |X/G|$ and $x_1,\dotsc,x_\lambda$ the distinct representatives of all the orbits. That is, none of the $x_i$ are related to each other meaning $x_i \not \sim x_j$ for any $1\leq i \neq j \leq \lambda$ and for any $y\in X$, $y$ is related to some $x_i$ in the above list.
Now define $p[y\sim x_i]=|Gx_i|/|X|$ where $Gx_i$ is the orbit of $x_i$, the set $\{gx: g\in G\}$ and $y$ is a randomly, uniformly chosen element of $X$. If $G^{x_i}$ is the stabilizer of $x_i$, i.e. the set $\{g\in G: gx=x\}$, then by the orbit stabilizer theorem, we can rewrite $$p[y\sim x_i]=\frac{|G|}{|G^{x_i}|\cdot |X|}$$
If $G$ acts freely, or if even our particular $x_i$ is not fixed by any nontrivial group element, then $p[y \sim x_i]=|G|/|X|$.
Problem: I would like to prove that $\sum_{i=1}^\lambda p[y\sim x_i]=1$. We can rewrite it as $$\frac{|G|}{|X|} \sum_{i=1}^\lambda \frac1{|G^{x_i}|}.$$ If $G$ acts freely on $X$ then the result holds. For, the stabilizer is then the trivial subgroup, and the sum sums to $\lambda=|X/G|=|X|/|G|$ by Burnside's lemma and since $G$ acts freely. So the above expression simplifies to $1$.
What if $G$ does not act freely? In some examples I was doing I could verify this directly for the cases where the number of orbits was small but I could not see anything that would generalize except for getting what orders were possible for the stabilizer but I still had no idea how many multiplicities would occur in the sum (some orbits may be fixed by subgroups of the same order, but are distinct orbits). What are some references for probability and orbit spaces/group actions? Thanks!
After some thought I think its much simpler than I imagined. If I'm mistaken, please comment.
Proof: Since the orbits form an equivalence relation, they partition the set $X$ and so $|X|=\sum_{i=1}^\lambda |Gx_i|$. Now, divide by $|X|$ (assuming $|X|$ non-empty, of course), QED.