A fair die is rolled repeatedly. Let $X$ be the number of rolls needed to obtain a 5 and $Y$ the number of rolls needed to obtain a 6. Calculate $E[X \mid Y=2]$.
This problem has been discussed here and here, but none of the solutions in those posts use the double expectation formula. I want to solve the problem above using the Double Expectation formula because I think that will be more efficient. If nothing else, I'll gt more practice with the formula. The formula is $E[Y] = E_X[E_Y[Y\mid X]]$ for jointly distributed random variables $X$ and $Y$. I think both $X$ and $Y$ follow the geometric distribution with mean $1/6$. So, we can write
$P(X=x) = (5/6)^{x-1} \cdot (1/6)$
$P(Y=y) = (5/6)^{y-1} \cdot (1/6)$
Assuming $X$ and $Y$ to be independent, we have $f(x, y) = (5/6)^{x-1} \cdot (1/6)^2 \cdot (5/6)^{y-1}$. This is where I get stuck. Can someone please explain how this solution can be completed from here?
$f(x,y)=f(x|y)f(y)$
$f(x|y)= \begin{cases} (4/5)^{x-1}(1/5), & x<y \\ 0, & x = y \\ (4/5)^{y-1}(5/6)^{x-y-1}(1/6), & x>y \end{cases} = \begin{cases} (4/5)^{x-1}(1/5), & x<y \\ 0, & x = y \\ (24/25)^{y}(5/6)^{x}(1/4), & x>y \end{cases} $
It doesn't seem to give an elegant solution.