Probability of disjoint dice results

Question

You have $n = n_A + n_B$ $k$-sided dice. The $n_A$ dice are thrown and a set of the resulting values, call it $S_A$, is built; likewise for the $n_B$ dice, calling the resulting set $S_B$.

What is the probability of $S_A \cap S_B = \varnothing$?

Answer 1

Over at mathoverflow there's this question: https://mathoverflow.net/q/15981, in light of it, I propose the following:

One has that $P\left(S_A \not\cap S_B \,\,|\,\,\left\vert S_A \right\vert = m \right) = \left(\frac{k-m}{k}\right)^{n_B}$ (from the comments on the original answer).

On the other hand, one has that $P\left(\left\vert S_A \right\vert = m \right) = \frac{1}{k^{n_A}}\frac{k!}{\left( k - m \right)!}\left\{ n_A \atop m \right\}$ (from the linked question).

So one could do:

$$ \begin{align} P\left( S_A \not\cap S_B \right) &= \sum_{i = 1}^{k} P\left(\left\vert S_A \right\vert = m \right) P\left(S_A \not\cap S_B \,\,|\,\,\left\vert S_A \right\vert = m \right) \\ &= \sum_{i = 1}^{k} \frac{1}{k^{n_A}}\frac{k!}{\left( k - i \right)!}\left\{ n_A \atop i \right\} \left(\frac{k - i}{k}\right)^{n_B}\\ &= \frac{k!}{k^{n}} \sum_{i = 1}^{k} \frac{(k - i)^{n_B}}{(k - i)!}\left\{ n_A \atop i \right\} \end{align} $$

Where $n = n_{A} + n_{B}$, and $\left\{ a \atop b \right\}$ is the Stirling Number of the Second Kind.

Answer 2

Setting $S:=S_{A}\cap S_{B}$ with inclusion/exclusion and symmetry we find:

$P\left(S=\varnothing\right)=1-P\left(1\in S\vee\cdots\vee k\in S\right)=\sum_{i=0}^{k}\binom{k}{i}\left(-1\right)^{i}P\left(\left\{ 1,\dots,i\right\} \subseteq S\right)$

The random set $S_A,S_B$ are independent so:

$P\left(\left\{ 1,\dots,i\right\} \subseteq S\right)=P\left(\left\{ 1,\dots,i\right\} \subseteq S_{A}\right)P\left(\left\{ 1,\dots,i\right\} \subseteq S_{B}\right)$

Again with inclusion/exclusion and symmetry we find:

$P\left(\left\{ 1,\dots,i\right\} \subseteq S_{A}\right)=1-P\left(1\in S_{A}^{c}\vee\cdots\vee i\in S_{A}^{c}\right)=\sum_{j=0}^{i}\binom{i}{j}\left(-1\right)^{j}P\left(\left\{ 1,\dots,j\right\} \subseteq S_{A}^{c}\right)=\sum_{j=0}^{i}\binom{i}{j}\left(-1\right)^{j}\left(1-\frac{j}{k}\right)^{n_{A}}$

Likewise:

$P\left(\left\{ 1,\dots,i\right\} \subseteq S_{B}\right)=\sum_{j=0}^{i}\binom{i}{j}\left(-1\right)^{j}\left(1-\frac{j}{k}\right)^{n_{B}}$

Then we end up with:

$$P\left(S_A\cap S_B=\varnothing \right)=\sum_{i=0}^{k}\sum_{j_{A}=0}^{i}\sum_{j_{B}=0}^{i}\binom{k}{i}\binom{i}{j_{A}}\binom{i}{_{j_{B}}}\left(-1\right)^{k+j_{A}+j_{B}}\left(1-\frac{j_{A}}{k}\right)^{n_{A}}\left(1-\frac{j_{B}}{k}\right)^{n_{B}}$$