Dice: Product of Counts

A fair six‐sided die is rolled 10 times. For each face $i\in\{1,2,\dots,6\}$ , let $$N_i$$ denote the number of times face $$i$$ appears. Compute the expected value $\displaystyle E[N_1N_2N_3N_4N_5N_6].$

Answer

The product

N_1N_2\cdots N_6

counts the number of ways to get a $$1$$ , times number of ways to get a $$2$$ , …, times number of ways to get a $$6$$ . It only cares about the total number of ways all those six positions are taken.

Now, the main idea is to express this product as the sum of indicator random variables which represent all the possible ways of choosing faces $1,2,\dots,6$ (i.e. $10\cdot9\cdot8\cdot7\cdot6\cdot5$ ways), which then take a value of $$1$$ exactly when the selected positions coincide with the actual positions of the six faces.

Given this, then we clearly have:

\boxed{ E[N_1\cdots N_6] =(10\cdot9\cdot8\cdot7\cdot6\cdot5)\,(1/6)^6 =\frac{10\cdot9\cdot8\cdot7\cdot6\cdot5}{6^6}. }

Notes and comments

Comment 1: Note that if face 1 appears in three slots $\{2,5,9\}$ , then every indicator term with $t_1\in\{2,5,9\}$ (and fixed $t_2,\dots,t_6$ ) equals 1, yielding exactly 3 matches. This correctly reflects $$N_1=3$$ .