Rolling a die - Conditional Probability A die is thrown repeatedly.
Let $X$ ~ First 5 is thrown and $Y$ ~ First 6 is thrown
Calculate $\mathbb{E}(X|Y=3)$
You may use the identity: $\sum_{n=k}^\infty nz^{n-k} = \frac{1}{(1-z)^2}+\frac{k-1}{1-z}$
I know from the definition of expectation, we have:
$\mathbb{E}(X|Y=3) = (1*\frac{1}{5})+(2*\frac{4}{5} * \frac{1}{5}) + (3* \frac{4}{5}* \frac{4}{5} * 0) + (5* \frac{4}{5} * \frac{4}{5} * 1 * \frac{5}{6} * \frac{1}{6}) + (6* \frac{4}{5} * \frac{4}{5} * 1 * \frac{5}{6} * \frac {5}{6} * \frac{1}{6}) + ...$, where every following term, has an extra '$*\frac{5}{6}$' term and constant increases by 1.
However I am unsure of how to apply this to the identity given to find the value of the infinite sum?

$\mathbb{P}(X=1|Y=3)=\frac{1}{5},\mathbb{P}(X=2|Y=3)=\frac{4}{25},\mathbb{P}(X>3|Y=3)=1−(\frac{1}{5}+\frac{4}{25})=\frac{16}{25}.$
Then we multiply these by the expected results, i.e. $1,2$ and $9$, giving $\mathbb{E}(X|Y=3)=(1∗\frac{1}{5})+(2∗\frac{4}{25})+(9∗\frac{16}{25})=6.28$
