Let $N_n$ be the number of occurrences of 5 or 6 in $n$ throws of a fair die. Show that as $n\to\infty$, $$ \frac{1}{n}N_n\to \frac{1}{3}, \mbox{ in mean square.} $$
My thoughts so far:
I order to use the Mean square law of large numbers we need to have that the $N_i$s have the same mean and the same variance, as they are obviously independent.
Proving that they have the same mean, namely $\frac{1}{3}$ is not hard, however I do not know how to go about finding the variance. I will present the way I found the mean, maybe there is a better way of finding that as well, and to extrapolate that formula to the variance.
If we throw a die $n$ times, then there will be $6^n$ possible outcomes, each with the same probability. Hence, there will be $n\cdot 6^n$ numbers, each with the same probability. By symmetry, the number of $5$s and $6$s will be the same, hence we only need to calculate the number of $5$s.
If we fix a die as $5$ then we will have $6^{n-1}$ numbers of 5 thrown with that die fixed. Given that we have $n$ dice, the total number of $5$s will be $n6^{n-1}$
$$ E[N_n] = 2\cdot\left ( \frac{1}{6^n}\cdot (n\cdot 6^{n-1})\right ) = \frac{1}{3} $$
How would I be able to apply this logic (or any other) in order to calculate the variance?
Hint: Try to use the Central Limit Theorem applied to $N_n=\sum_{i=1}^nX_i$
where $X_i=1$ if the throw is $5$ or $6$ and $X_i=0$ otherwise.
This immediately gives $\mathbb{E}[X_i]=\frac{1}{3}$, $var(X_i)=\frac{2}{9}$