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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?

Ricky_Nelson
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    $X$ and $Y$ aren't independent though -- e.g. $$\mathbb P(X=2\mid Y=2)=0\neq\mathbb P(X=2).$$ – jlammy Jan 09 '21 at 00:33
  • @jlammy Oh, any ideas on how else can the joint distribution might be obtained? – Ricky_Nelson Jan 09 '21 at 00:36
  • Sadly I think this method is doomed to fail -- the way we actually compute $\mathbb E[\mathbb E(X\mid Y)]$ is via the formula $\sum\mathbb E[X\mid Y=y]\cdot\mathbb P(Y=y)$, which requires knowing $\mathbb E[X\mid Y=k]$... – jlammy Jan 09 '21 at 00:40
  • And in terms of being more efficient, I think you'll be hard pressed to do better than Christian Blatter's lovely solution – jlammy Jan 09 '21 at 00:44
  • @jlammy I found this question in a book and the book specifically asked to use the double expectation formula. so I am sure the formula can be used to solve the problem. It might be that this formula might be cumbersome to use in this case. – Ricky_Nelson Jan 09 '21 at 00:47

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$f(x,y)=f(x|y)f(y)$

$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.

Arash
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