Given 2 players, one rolling $x$ d6 dice and the other rolling $y$ d6 dice, what is the probability of a match between the two players? I'm getting stuck on the sub-set comparisons - I can calculate for $(x+y)$, but get stuck when I break it into two sets. I can brute force for $x=1,2~~~~ y=1,2$ but I think a formula would be more practical when expanding out to $x=[1,5]$ and $y=[1,10]$. I'm sure I'm making this harder than it needs to be...
Clarification update: A match is any single die match between two players. i.e. as long as each player rolls a 1, it doesn't matter what any of the other dice rolled are.
They are using standard d6 dice.
First, we want the probability $\alpha_{L,n}$ that a given set of $L$ faces appears in a roll of $n$ dice. Call the multiset of the rolled faces $X$.
Let the faces be $i_1,\dots,i_L$.
Then, using the inclusion-exclusion principle, we have $$ \begin{align*} \alpha_{L,n} &= 1 - \sum_{J \subseteq \{1,\dots,L\}, J \neq \varnothing} (-1)^{|J|-1} P( \cap_{j \in J} i_j \not \in X) \\ &= 1-\sum_{J \subseteq {1,\dots,L\}, J \neq \varnothing }} (-1)^{|J|-1} \left( 1 - \frac{|J|}{6} \right)^n \\ &= 1 - \sum_{r=1}^L (-1)^{r-1} \left( 1-\frac{r}{6} \right)^n \binom{L}{r} \end{align*} $$ Suppose player $1$ rolls $n$ dice and player $2$ rolls $m$ dice. Let $X$ be the multiset of player $1$'s rolls, and $Y$ be the multiset of player $2$'s rolls.
Let $A_i$ be the event that $i \in X$ and $i \in Y$.
We want $P(\cup_{i=1}^6 A_i)$.
By another application of the inclusion-exclusion principle, we have $$ P(\cup_{i=1}^6 A_i) = \sum_{i=1}^6 (-1)^{i-1} a_{i,n} a_{i,m} \binom{6}{i} $$
For example, for $n=m=1$, probability$=\frac{1}{6}$.
For $n=m=2$, probability$=\frac{37}{72}$.
For $n=m=3$, probability$=\frac{6151}{7776}$.
For $n=3, m=6$, probability$=\frac{1587971}{1679616}$.