The dice A has 2 red faces and 4 green, and the dice B conversely: 4 red and 2 green. We toss a symetric coin; if come up heads, we choose the dice A, otherwise - the dice B. Next we execute a series of throws by the chosen dice. Let: $X_n= 0 $, if a red face in a n-th throw and $X_{n}=1$, otherwise. Find the limit of $\frac{X_1+X_2+\dots+X_n}{n}$ in a.s. sense.
Because $(X_n)$ are iid and $\text{E}X_1=\frac{1}{2}$, we can apply SLLN. Hence:
$$\frac{X_1+X_2+\dots+X_n}{n} \to \frac{1}{2}$$ almost sure.
I'm not sure if $X_n$ are independent. Are they?
Try to calculate what $Pr(X_i = 1)$ is. You would get $Pr(X_i=1) = Pr(Coin=heads)\cdot Pr(dice A = green) + Pr(Coin=tails)\cdot Pr(dice B = green) = \frac{1}{2}\cdot \frac{4}{6} + \frac{1}{2}\cdot \frac{2}{6} = \frac{1}{2}$.
Similarly $Pr(X_i=0)$ will also be $\frac{1}{2}$.
Now, for purpose of argument, calculate $Pr(X_i=1 | X_j=1)$ (with $i\neq j$). Note that the way the problem is worded, a single coinflip will determine the choice of which die to use for all throws of the dice, and that we do not reflip the coin each time. (I had not noticed this when I initially posted my answer.). As such, as explained by another user in a comment to the question, we calculate $Pr(X_i=1 | X_j=1) = \frac{Pr(X_i=X_j=1)}{Pr(X_j=1)} = \frac{\frac{1}{2}(\frac{1}{3})^2 + \frac{1}{2}(\frac{2}{3})^2}{\frac{1}{2}} = \frac{5}{9}\neq \frac{1}{2} = Pr(X_i=1)$
Noting that $Pr(A|B)=Pr(A)\Leftrightarrow A$ and $B$ are independent events, the above shows that $X_i=1$ and $X_j=1$ are not independent events. You will get then that the SLLN does not apply in this case.
Re limits, given that the coinflip was heads, the limit is $\frac{1}{3}$, and, given that the coinflip was tails, the limit is $\frac{2}{3}$.