Calculating Probability For 2 Exploding Dice Background: For those who don't know, exploding dice refers to when you roll the maximum on a die and reroll, cumulatively adding the rolled values until you don't roll the maximum anymore.
Problem: I am trying to find the probability of rolling a value $x$ when rolling 2d10, where each d10 can explode. I have already come up with an equation to calculate the probability of rolling a value on a single n-sided die:
$$P\left(x\right)=\left(\frac{\left(\frac{x}{n}-\lfloor(\frac{x}{n})\right\rfloor)}{\left\{x-\lfloor(\frac{x}{n})\rfloor\cdot n=0:1, x-\lfloor(\frac{x}{n})\rfloor\cdot n\right\}}\right)^{\left(1+\lfloor(\frac{x}{n})\rfloor\right)}$$
where $x$ is the expected value, and $n$ is the number of sides on the die.
Is there a way I can use the equation I already have that I'm unaware of to calculate the probability of value $x$ with 2 exploding dice?
 A: First, I would rewrite the probability mass function for one die as:
$$
\begin{align*}
p_1(x) &= \begin{cases}
\left(\dfrac1n\right)^{1+\left\lfloor x/n\right\rfloor}&(x\ge0) \wedge (x\bmod n > 0)\\
0&\text{otherwise}
\end{cases}\\
 &= \begin{cases}
\left(\dfrac1n\right)^{\left\lceil x/n\right\rceil}&(x\ge0) \wedge (x\bmod n > 0)\\
0&\text{otherwise}
\end{cases}
\end{align*}$$
Then for two independent dice, the sum $x$ follows the probabilities:
$$p_2(x) = \sum_{i=-\infty}^{\infty} p_1(i) p_1(x-i)
= \sum_{i=0}^{x} p_1(i) p_1(x-i)$$
Assuming $x\ge0$, splitting into cases of $x\bmod n$:

*

*if $x\bmod n = 0$, i.e. $x=kn$ for some $k$, then:
$$\begin{align*}
p_2(x) 
&= \sum_{i=1}^{n-1} \left(\frac1n\right)^{1}\left(\frac1n\right)^{k} + \sum_{i=n+1}^{2n-1} \left(\frac1n\right)^{2}\left(\frac1n\right)^{k-1}+\cdots+ \sum_{i=x-n+1}^{x-1} \left(\frac1n\right)^{k}\left(\frac1n\right)^{1} \\
&= \frac xn (n-1) \left(\frac1n\right)^{x/n+1} 
\end{align*}$$


*if $x\bmod n > 0$, then let $r=x\bmod n$, and so $x = kn+r$:
$$\begin{align*}
p_2(x) &= \sum_{i=1}^{r-1} \left(\frac1n\right)^{1}\left(\frac1n\right)^{k+1} + \sum_{i=r+1}^{n-1} \left(\frac1n\right)^{1}\left(\frac1n\right)^{k}\\
&\quad +\sum_{i=n+1}^{n+r-1} \left(\frac1n\right)^{2}\left(\frac1n\right)^{k} + \sum_{i=n+r+1}^{2n-1} \left(\frac1n\right)^{2}\left(\frac1n\right)^{k-1} + \cdots\\
&\quad +\sum_{i=kn+1}^{x-1} \left(\frac1n\right)^{k+1}\left(\frac1n\right)^{1}\\
&= \left(k+1\right)(r-1)\left(\frac1n\right)^{k+2} + k(n-r-1)\left(\frac1n\right)^{k+1}\\
&= \left\lceil\frac xn\right\rceil [(x\bmod n)-1]\left(\frac1n\right)^{\left\lceil x/n\right\rceil+1} + \left\lfloor\frac xn\right\rfloor [n-(x\bmod n)-1]\left(\frac1n\right)^{\left\lfloor x/n\right\rfloor+1}\\
\end{align*}$$
Or merging the two cases if the following looks better:
$$\begin{align*}
p_2(x)
&= \left\lceil\frac xn\right\rceil \max\left[0,\frac{(x\bmod n)-1}n\right]\left(\frac1n\right)^{\left\lceil x/n\right\rceil}\\
&\quad + \left\lfloor\frac xn\right\rfloor \left[1-\frac{(x\bmod n)+1}n\right]\left(\frac1n\right)^{\left\lfloor x/n\right\rfloor}\\
\end{align*}$$
