Expectation and variance of dice throwing game. You roll a die, if the number on the die is higher than 2 you roll the first die again. Otherwise, you roll a second die, if the number on the second die is greater than 5 you get 1 point, if the number on the second die is less than or equal to 5 you get 2 points. What is the expected value and variance of the game if the game is played a large number of times.
What I've tried so far:
Let $X$ be the event that you roll higher than 5 on the second die and let $Y$ be the event that you roll less than or equal to 5 on the second die. Since the probability of rolling higher than 2 is $\frac{4}{6}$ and less than or equal to 2 is $\frac{2}{6}$ for the first die, I think the expected value of the game would be: $$\frac{2}{6}\cdot E(X) + \frac{4}{6}\cdot E(Y).$$
I think $E(X) = \frac{1}{6}\cdot1 = \frac{1}{6}$ and $E(Y) = \frac{5}{6}\cdot 2 = \frac{5}{3}$ which would give an expected value of the game of $\frac{7}{6}$. However, I am not confident at all that this is correct or even on the right lines and I don't know how to get the variance at all, I also don't know how to factor in the statment: "if the game is played a large number of times".
Any help would be great, thanks!
 A: As clarified in your comment, the number of points you get in a single iteration of the game can be $\ 0,1\ $ or $\ 2\ $. The first occurs if you roll $\ 3\ $ or more with the first die, which happens with probability $\ \frac{2}{3}\ $.  You get $\ 1\ $ point if you throw $\ 2\ $ or less with the first die and then $\ 6\ $ with the second, which happens with probability $\   \frac{1}{3}\cdot\frac{1}{6}=\frac{1}{18} $. You get $\ 2\ $ points if you throw $\ 2\ $ or less with the first die and then $\ 5\ $ or less with the second, which happens with probability $\  \frac{1}{3}\cdot\frac{5}{6}=\frac{5}{18} $.  Your expected score is therefore
$$
E(P)=0\cdot\frac{2}{3}+1\cdot\frac{1}{18}+2\cdot\frac{5}{18}= \frac{11}{18}\ ,
$$
where $\ P\ $ is the random variable representing the number of points you get in a single iteration of the game.  The easiest way to get the variance is to use the identity
$$
\text{var}(P)=E(P^2)-E(P)^2\ .
$$
You can work out $\ E(P^2)\ $ in essentially the same way as you work out $\ E(P)\ $, except that you replace the possible values of $\ P\ $ in the formula with their squares.
