I roll a fair die 4 times. Let X be the number of different outcomes that I see. Find $\mathbb{E}[X]$.
My attempt:
I know that I can write X as a sum of indicator random variables and then I can use the fact that $\mathbb{E}[1_A]=\mathbb{P}(A)$. Thus,
\begin{equation}I_{A_1}=\begin{cases}1 & \text{if only one kind}\\0 & \text{otherwise}\end{cases}\end{equation}\begin{equation}I_{A_2}=\begin{cases}1 & \text{if two kinds}\\0 & \text{otherwise}\end{cases}\end{equation}\begin{equation}I_{A_3}=\begin{cases}1 & \text{if three kinds}\\0 & \text{otherwise}\end{cases}\end{equation}\begin{equation}I_{A_4}=\begin{cases}1 & \text{all different}\\0 & \text{otherwise}\end{cases}\end{equation}
Then the probabilities of event A happening for each is$$\mathbb{P}(I_{A_1})=\frac{6\cdot1\cdot1\cdot1}{6^4}$$$$\mathbb{P}(I_{A_2})=\frac{6\cdot5\cdot2\cdot2}{6^4}$$$$\mathbb{P}(I_{A_3})=\frac{6\cdot5\cdot4\cdot3}{6^4}$$$$\mathbb{P}(I_{A_4})=\frac{6\cdot5\cdot4\cdot3}{6^4}$$
The sum of these should be my desired expectation.
My question is whether or not I have found the probabilities correctly. This is my thought process for how I counted the number of choices for the numerator, using $I_{A_2}$ as an example:
There are 6 choices for choosing the first number, then we don't want to get that number again so then there are 5 choices. After that, we only want to get the first or second number again, thus there are 2 choices for the third roll and 2 choices for the fourth roll. Hence, $6\times 5\times 2\times 2=120$.
Is this the correct way of thinking about it? Thanks