Context: In calculating the optimal policy for MDP's an algorithm called Value Iteration is used. I am using this algorithm to calculate the optimal policy for a small game and test my knowledge in the field.
In the game, $d$ normal dices (1-6) are rolled simultaneously, and you can either pick all dices with the largest value, or all dices with the smallest value. To not have to compute all possible $6^d$ dice rolls, I limit it to $x$ dices getting the smallest values, and $y$ dices getting the highest values, where $1 \leq x \leq d$ and $y \leq d - x$.
Now my question is: With $d$ dices, what is the probability that $x$ dices fall on a minimum value $v_x$, $y$ dices fall on a maximum value $v_y > v_x$, and $z = d - (x+y)$ dices are between (not including) $(v_x, v_y)$?
I have the feeling that the $z$ in-between dices can be modeled with a binomial distribution with $binom(z, d, \frac{v_y - v_x - 1}{6})$, but I am not sure how to reconcile this with the probabilities of $x$ and $y$.
Assuming that $v_x<v_y$, $0<x,0<y,x+y\le d$ the probability reads: $$ p(x,y,v_x,v_y)=\frac{d!}{x!y!(d-x-y)!}\frac{(v_y-v_x-1)^{d-x-y}}{6^d},\tag1 $$ with the convention $0^0=1$. The probabilities do not sum to $1$ because the allowed events do not include the cases with $v_x=v_y$, i.e. the events with all dice showing up the same value, the probability of such events being $\frac1{6^{d-1}}$. One can notice that the formula $(1)$ is applicable in the case $v_x=v_y$ as well provided the convention that one of $x,y$ is $d$ and the other is $0$.