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.

  • $\begingroup$ What does "a match" means ? $\endgroup$ – user252450 Dec 16 '15 at 16:34
  • $\begingroup$ One player rolls 2,3,5; the other 1,4,5,6. Is that a match because they both rolled 5's? Or is a match 2,3,5 and 2,3,5,6 respectively? $\endgroup$ – maibaita Dec 16 '15 at 16:49
  • $\begingroup$ Thanks for the question. As long as a single die pair matches between the two players, it is considered a match. $\endgroup$ – user3203346 Dec 16 '15 at 16:54
  • $\begingroup$ (1,1,1,2) v (3,3,3,3,3,3,3,3,3,3,2) is just as valid as (1,1,1,1) vs (1,1,1,1,1,1,1,1,1) $\endgroup$ – user3203346 Dec 16 '15 at 16:56
  • $\begingroup$ When you write $x = [1,5], y = [1,10]$, do you mean that one die is thrown $5$ times, and another, $10$ times ? Pl. clarify. It is past midnight here, can try tomorrow, if unanswered. $\endgroup$ – true blue anil Dec 16 '15 at 18:57

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$=\frac{1587971}{1679616}$.


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