Conditional Expectation - throwing dice 
Question: A fair die is rolled repeatedly. Let X and Y denote,
  respectively, the number of rolls necessary to obtain a 6 and a 5. For
  this question you may find it helpful tp recall how to derive a
  formula for $\sum_{k=1}^{\infty}kx^{k-1}$ from the geometric series
  forumla. 
Find
a) E(X); 
b) E[X|Y = 1];

So for a), this is a geometric distribution so $E[X] = 6$.
However I am kind of unsure how to go about doing b). So the formula for a discrete conditional expectation is 
$$E[X|y=y] = \sum_x xP\{X=x|Y=y\}$$
To find $E[X|Y=1]$ is it just:
$ (1)P\{X=1|Y=1\} + (2)P\{X=2|Y=1\} + (3)P\{X=3|Y=1\} + (4)P\{X=4|Y=1\} + (5)P\{X=5|Y=1\} + (6)P\{X=6|Y=1\} $
But then how do I compute $P\{X=1|Y=1\}$? And where does the geometric series come into play here?
 A: $\require{cancel}$The best way to get this is to notice that $X\sim \text{Geom}(p = 1/6)$ on $\{1,2,3,\dotsc\}$, and in particular $$X|Y = 1 \overset d= 1+X$$
as A.S. commented. This means
$$E[X|Y=1] = E[1+X] = 1+6 = 7.$$
There are a number of other ways to get this. To do it your way, then
$$P(X = k|Y = 1)$$
means that I tell you that you have taken one turn and got a 5, what is the probability of getting the first six on roll $k$.
So,


*

*$P(X= 1|Y = 1) = 0$

*$P(X =2|Y = 1) = 1/6$

*Consider $P(X = 3|Y = 1)$. In other words,
\begin{array}{r|c|c|c}
\text{Trial}&1&2&3\\\hline
\text{Outcome}&5&\text F&6\\\hline
\text{Chance}& 1 &\dfrac{5}{6} & \dfrac{1}{6}
\end{array}
where $\text F$ means not rolling a six. Since, it is given that $Y=1$, the chance that it happens is $1$. Hence
$$P(X = 3|Y = 1) = (5/6)(1/6).$$

*Consider $P(X = 4|Y = 1)$. In other words,
\begin{array}{r|c|c|c|c}
\text{Trial}&1&2&3&4\\\hline
\text{Outcome}&5&\text F& \text F &6\\\hline
\text{Chance}& 1 &\dfrac{5}{6}&\dfrac{5}{6} & \dfrac{1}{6}
\end{array}
Hence $P(X = 4|Y = 1) = (5/6)^2(1/6)$,


and in general
$$P(X = k|Y = 1) = (5/6)^{k-2}(1/6),$$
for $k\geq 2, 0$ otherwise.
Thus
$$E[X|Y=1] = \sum_{k = 2}^\infty kP(X = k|Y=1) = \sum_{k=2}^\infty k(5/6)^{k-2}(1/6).$$
But we can simplify our work if we play around a little.
\begin{align*}
E[X|Y = 1]&= \sum_{k=2}^\infty kP(X=k|Y=1)\\
&=\sum_{k=2}^\infty k(1-p)^{k-2}p\\
&=\frac{1}{1-p}\left[\sum_{k=2}^\infty k(1-p)^{k-1}p\right]\\
&=\frac{1}{1-p}\left[-p+\sum_{k=1}^\infty k(1-p)^{k-1}p\right]\tag 1\\
&=\frac{1}{1-p}\left[-p+E[X]\right]\\
&=\frac{1}{5/6}\left[-\frac{1}{6}+\frac{1}{1/6}\right]\\
&=7,
\end{align*}
where in $(1)$, I recognize that the series is simply the expectation of a geometric distribution with $p = 1/6$ on $\{1,2,3,\dotsc\}$. Also, at $(1)$ you can use the series they gave you if you pull out $p$ from the series. I didn't bother. I like my way better.
