The probability for the experiment of rolling two different colored dice is 
$$\frac{162601421574468954588}{2^{2\times36}}\approx0.0344322.$$

---

Indeed, for the set $[36]_0:=\{0,\dots,36\}$, let $T$ denote the set of all triples $(m,n,k)\in[36]_0^3$ such that 
$\dfrac m{36}\dfrac n{36}=\dfrac k{36}$ or, equivalently, $mn=36k$. 

Let $I$ denote the set of all pairs $(a,b)$ of subsets of the set $[36]_0$ such that $a$ and $b$ are independent. Then $(a,b)\in I$ iff $(|a|,|b|,|a\cap b|)\in T$, where $|\cdot|$ denotes the cardinality. 

For any given $(m,n,k)\in T$, 
$$N_{m,n,k}:=\big|\{(a,b)\in I\colon (|a|,|b|,|a\cap b|)=(m,n,k)\}\big|=\binom{36}m\binom mk \binom{36-m}{n-k}.$$

So, the probability for the experiment of rolling two different colored dice is 
$$\frac1{2^{2\times36}}\sum_{(m,n,k)\in T}N_{m,n,k}
=\frac{162601421574468954588}{2^{2\times36}}.$$

---

Here are details of the calculations, with Mathematica: 

[![enter image description here][1]][1]


  [1]: https://i.sstatic.net/5rGH5.png