Conditional expectation of number of dice rolls I've been given the following problem and I'd like to get a better understanding of how to solve it.
A fair die is rolled successively. Let $X$ be the number of rolls needed to get a 6 Let $Y$ be the number of rolls needed to get a 5. 
Compute $E[X|Y=1]$ and $E[X|Y=5]$
Now, For the first one, I know I could do the following 
$$\sum_{x=2}^\infty x(5/6)^{x-2}(1/6)$$
and arrive at the correct answer, $E[X|Y=1]=7$, because I already know I had one failure of $X$. However, this seems like a shortcut to me and I really want to understand how to do this by definition. So what I'm trying to evaluate is the following:
$$\sum_{x=1}^\infty x P(X=x|Y=1) = \sum_{x=1}^\infty x \frac{P(X=x,Y=1)}{P(Y=1)}$$
Now since $X$ and $Y$ are both geometric random variables, I know that $$P(Y=y) = (5/6)^{y-1}(1/6)$$ $$P(Y=1) = 1/6$$
However, what I'm not sure how to do is calculate $P(X=x,Y=1)$.
 A: Note the following:


*

*For every $1\leqslant x\leqslant y$, $P(X\geqslant x\mid Y=y)=a^{x-1}$ where $a=4/5$.

*For every $x\geqslant y+1$, $P(X\geqslant x\mid Y=y)=a^{y-1}b^{x-y-1}$ where $b=5/6$.

*Finally, $E(X\mid Y=y)=\sum\limits_{x\geqslant1}P(X\geqslant x\mid Y=y)$.

A: For $k=1,2,\dots$ denote $\mu_{k}=\mathbb{E}\left(X\mid Y=k\right)$.
Here $\mathbb{E}X=\frac{1}{6}\times1+\frac{5}{6}\times\left(1+\mathbb{E}X\right)$
leading to $\mathbb{E}X=6$ and $\mu_{1}=1+\mathbb{E}X=1+6=7$. 
If $k>1$ then $\mu_{k}=\frac{1}{5}\times1+\frac{4}{5}\times\left(1+\mu_{k-1}\right)=1+\frac{4}{5}\mu_{k-1}$ leading to: $$\mu_{k}=5+2\left(\frac{4}{5}\right)^{k-1}$$
This expression for $\mu_k$ also covers the special case $k=1$. Note that under condition $Y=k>1$ the first roll cannot be a $5$ so that the probability that the first roll is a $6$ will be $\frac{1}{5}$. This corresponds with term $\frac{1}{5}\times1$. If the first roll is not a $6$ then we are back in the old situation with the following differences: we had exactly $1$ 'failure' and starting from here the number of rolls needed to get a $5$ has decreased with $1$. You could say that condition $Y=k$ has become condition $Y=k-1$ in the new situation. This corresponds with term $\frac{4}{5}\times\left(1+\mu_{k-1}\right)$

addendum focused on comments
We have $\mu_{1}=7$ and $\mu_{k}=1+\frac{4}{5}\mu_{k-1}$. In order
to find a closed form for $\mu_{k}$ it is a good custom to have a
look at $\mu_{k}$ for - let's say $k=1,2,3,4$. Quite often a pattern
can be discovered. We find:
$\mu_{1}=7$
$\mu_{2}=1+\frac{4}{5}\times7$
$\mu_{3}=1+\frac{4}{5}\left(1+\frac{4}{5}\times7\right)=1+\frac{4}{5}+\left(\frac{4}{5}\right)^{2}\times7$
$\mu_{4}=1+\frac{4}{5}\left(1+\frac{4}{5}+\left(\frac{4}{5}\right)^{2}\times7\right)=1+\frac{4}{5}+\left(\frac{4}{5}\right)^{2}+\left(\frac{4}{5}\right)^{3}\times7$
Well, doesn't this smell like: $$\mu_{k}=1+\frac{4}{5}+\cdots+\left(\frac{4}{5}\right)^{k-2}+\left(\frac{4}{5}\right)^{k-1}\times7$$?
Making use of the wellknown rule $1+r^{1}+\cdots+r^{n-1}=\frac{1-r^{n}}{1-r}$
we can find an expression for $\mu_{k}$. If you have found such an
expression then check whether it satisfies the conditions $\mu_{1}=7$
and $\mu_{k}=1+\frac{4}{5}\mu_{k-1}$. There is only one sequence
$\left(\mu_{k}\right)$ that satisfies these conditions, so you are
ready if it does. 
Concerning the calculation of $\mathbb{E}X$ denote the outcome of
the first roll by $D$. Then $\mathbb{E}X=\mathbb{E}\left(X\mid D=6\right)P\left[D=6\right]+\mathbb{E}\left(X\mid D\neq6\right)P\left[D\neq6\right]$.
Here $P\left[D=6\right]=\frac{1}{6}$ and $\mathbb{E}\left(X\mid D=6\right)=1$
(since under this condition only one roll is needed to get a $6$).
If $D\neq6$ then you land in the same situation with the only difference
that there was exactly one 'failure'. That's why $\mathbb{E}\left(X\mid D\neq6\right)=1+\mathbb{E}X$.
This leads to $\mathbb{E}X=\frac{1}{6}\times1+\frac{5}{6}\times\left(1+\mathbb{E}X\right)$.
remark
Very often a proof based on conditionals excels in simplicity and elegance. Try to get hold of that 'technique' and ask questions as: 'what has changed after one roll (or other sort of thing)?'.
