A fair die is rolled repeatedly. Let $X$ be the number of rolls needed to obtain a $5$ and $Y$ be the number of rolls needed to obtain as $6$. A fair die is rolled repeatedly. Let $X$ be the number of rolls needed to obtain a $5$ and $Y$ be the number of rolls needed to obtain as $6$.
Calculate $E(X|Y=2)$.
My attempt
As $X$ is the first time a 5 is rolled among independent trials, $X$ follows a geometric distribution with $p=\frac{1}{6}$.
$Y=2 \Rightarrow$ the first roll was not a $6$ and the second roll was a $6$.
This means a $5$ was obtained for the first time either on the first roll with $p=\frac{1}{5}$ or on at least the third roll with $p=1-\frac{1}{5}=\frac{4}{5}$.
$P(X=1|Y=2)=\frac{1}{5}, P(X=2|Y=2)=0, P(X=3|Y=2)=\frac{4}{5} \cdot 1 \cdot \frac{1}{6}, P(X=4|Y=2)=\frac{4}{5} \cdot 1 \cdot \frac{5}{6} \cdot \frac{1}{6}$
$E(X|Y=2)=(1/5)+(3)(4/5)(1/6)+4(4/5)(5/6)(1/6)+...= (1/5)+(4/5)\sum_{x=3}^{\infty}x \cdot (\frac{5}{6})^{x-3} \cdot (1/6)$
By letting $k=x-3$, I see that this sum equals $0.2(1)+0.8(8)=6.6$
One solution to this problem has the following: $E(X|X \ge 3)= \frac{1}{p}+2=6+2=8$. Thus $E(X|Y=2)=0.2(1)+0.8(8)=6.6$. But why is $E(X|X \ge 3)=8$?

$\mathbb E(X\mid X\geq3)$ is the expectation of the number of rolls needed to obtain a $5$ under the condition that the rolls $1$ and $2$ do not provide a $5$.
We might say that the sequence of relevant rolls actually starts after $2$ rolls that are in vain, so that:$$\mathbb E(X\mid X\geq3)=\mathbb E(2+X)=2+\mathbb EX$$
Here $\mathbb EX=6$ since $X$ has geometric distribution (number of trials needed) with parameter $p=\frac16$.
This together explains that: $$\mathbb E(X\mid X\geq3)=2+6=8$$
