A die is rolled $N$ times where $N\sim \operatorname{Pois}(\lambda)$. $X =$ number of different results in the rolls. For example: $(1,1,6,1)$ has 2 different results and $(1,2,3,4)$ has 4. Calculate $E(X)$
Define an indicator: $$I_i = \begin{cases} 1, & \text{$i$ was rolled at least once out of $n$ rolls} \\ 0, & \text{otherwise} \end{cases}$$ Then: $$X = \sum_{i=1}^{6} I_i$$ $$P(I_i = 1 \mid N=n) = 1 - \left(\frac{5}{6}\right)^n$$ $$E(X) = E(E(X\mid N)) = E\left(E\left(\sum_{i=1}^6 I_i\mid N\right)\right) = E\left(6\left(1-\left(\frac{5}{6}\right)^N\right)\right) \\ = 6-6E\left(\left(\frac{5}{6}\right)^N\right)$$ $$ = 6 - 6\left[\sum_{n=0}^\infty \left(\frac{5}{6}\right)^n\left(\frac{e^{-\lambda} (\lambda)^n}{n!}\right)\right] = 6 - 6\left[\sum_{n=0}^\infty \left(\frac{e^{-\frac{5}{6}\lambda}\left(\frac{5}{6}\lambda\right)^n}{n!} \right) e^{-\frac{1}{6}}\right] = 6-6e^{-\frac{1}{6}} = 0.92$$
where the last line is the sum of all probabilities of a random variable $Y\sim \operatorname{Pois}(\frac{5}{6}\lambda)$, which sums up to $1.$
This is an odd result since it didn't depend on $\lambda$ and it's too small as is.
Where is my mistake?
The problem is that $\mathrm e^{-\lambda}\ne\mathrm e^{-\frac56\lambda}\mathrm e^{-\frac16}$.