Four dice are tossed let $N$ be the number of distinct outcomes. Find $E[N]$
If we let $X$ be the number of distinct outcomes in a given toss.
E.g. if we get the toss $1,2,2,4$ then $X=3$
Calculating some probabilities:
$P(X=1) = 6 \cdot 6^{-4}$ since we have six ways we can end up with all four dice having the same value face-up.
Now according to a solution I have for this:
$\displaystyle P(X = 2) = 6^{-4}\cdot 6 \cdot 5 \cdot \binom{4}{2}$
This is where I am not so sure; why $4$ choose $2$ as opposed to $6 $ choose $2$ ?
$P(X= 4) = 6^{-4}\cdot 6 \cdot 5 \cdot 4 \cdot 3$
and finally $P(X = 3) $ I haven't calculated as I am not clear on $P(X=2)$.
The issue for me seems to be the combinatorial part of the argument. i.e. The number of ways each event occurs.
Best Answer
Alternatively, split into indicator variables and use linearity of expectation.
For $1\leq i\leq 6$, let $X_i$ be $1$ if we roll $i$ at least once, and $0$ otherwise. Then $$\mathbb E[X]=\sum_{i=1}^6\mathbb E[X_i]=6\mathbb E[X_1].$$ But $\mathbb E[X_1]=\mathbb P(\text{roll a }1)=1-\left(\frac56\right)^4$, so we get $\mathbb E[X]=\frac{671}{216}$.
This method has the benefit of generalising easily. For a $d$ sided die rolled $n$ times, the expected number of distinct outcomes is $$\mathbb E[X]=d\left[1-\left(\frac{d-1}{d}\right)^n\right].$$