I roll $n \geq 2$ ordered dice simultaneously.
Let $y$ be the number of "x-of-a-kind" sets among the ordered dice, with $x \in [2;n]$; $y \in [0; \left\lfloor\frac{n}{2}\right\rfloor]$.
For example, I roll $n=12$ ordered dice simultaneously and get the following sequence:
6 4 4 1 3 3 3 3 4 4 4 6
For this sequence, I have $y=3$ sets of "x-of-a-kind": one set of $4$s ($x=2$), then one set of $3$s ($x=4$), and finally another set of $4$s ($x=3$).
What are the probabilities for each pair $(y, n)$?
What I tried:
I defined for $r \in [2;n]$ :
$A_r$ the event for a $r$th ranked number in the sequence, which fulfills the condition that the numbers rolled for $r$ and $r-1$ are the same.
$C_r$ the event that $r$ is forming a new pair with the previous number, i.e $r-1$ is not already a part of a "x-of-a-kind".
With this, I have the following:
$P(A_r)=P(A)=\frac{1}{6}$.
$P(C_r)=P(A|\overline C_{r-1})=P(A)P(\overline C_{r-1}|A_r)$
$A_r$ and $\overline C_{r-1}$ are independant, so $P(C_r)=P(A)P(\overline C_{r-1})=P(A)(1-P(C_{r-1}))$
Wolfram Alpha gave me the explicit form: $P(C_r)=\frac{1}{7}(-\frac{1}{6})^r ((-6)^r + 6)$
Now I want to express the expected value of $y$ in function of $\sum_{r=2}^nP(C_r)$. But I think I'm missing a part here, because $P(C_r)$ and $P(C_{r-1})$ seem dependant.
I don't know how to go about this. Is my reasoning correct until now and is this the right way to go about it? I didn't manipulate probabilities since my student years and I don't remember much from them, so I may be totally wrong.