How to conceptualize conditional expectation inductively? Attempting the solve the following problem 

A fair die is successively rolled. Let $X$ and $Y$ denote,
  respectively, the number of rolls necessary to obtain a    6 and a 5.
  Find 
$E[ X|Y = 5]$

I am confused about how to conceptualize such a problem?
I know $E[X|Y= 1] = 1 + \frac{ 1}{ p} = 1 + \frac{ 1}{ 1/6} = 7$ building off the that we have a geometric random variable. But, I am unsure of how to extend this to a higher given expectation for $Y$.  
I would like to solve this in an inductive mannor but


*

*I am unable to conceptualize an inductive solution?

*I am not sure that an inductive solution would even valid?

 A: I am not quite sure what kind of induction you are looking for, so I will just try giving a more general solution to this problem, i.e. $E[X|Y=y]$.
The point is to notice that random variable X distributed differently before and after the y-th roll, for tosses before the first one which yield 5 cannot have 5 as result, hence there are 1 out of 5 chances to get a 6.
The conditional probability is 
$$P_{X|Y}(x,y)=\left\{ \begin{array}{1}(4/5)^{x-1}\times 1/5 ~ ~(X<Y)\\0 ~ ~(X=Y)\\(4/5)^{y-1}\times(5/6)^{x-y-1}\times1/6~~(X>Y)\end{array} \right.$$
Then follow the definition of conditional expectation we have:
$E[X|Y=y]\\=\sum\limits_{x}xP\{X=x|Y=y\}\\=\sum\limits_{x=1}^{y-1}x(\frac{4}{5})^{(x-1)}\frac{1}{5}+0+\sum\limits_{x=y+1}^\infty x(\frac{4}{5})^{y-1}(\frac{5}{6})^{(x-y-1)}\frac{1}{6}\\=\frac{1}{5}\sum\limits_{x=1}^{y-1}x(\frac{4}{5})^{(x-1)}+\frac{1}{6}(\frac{4}{5})^{y-1}\sum\limits_{x=y+1}^\infty x(\frac{5}{6})^{(x-y-1)}\\=\frac{1}{5}[5^{1-y} (-5(4^y)+5^{1+y}-4^yy)-y(\frac{4}{5})^{y-1}]+\frac{1}{6}(\frac{4}{5})^{y-1}6(y+6)\\=\frac{5}{4}(4-(\frac{4}{5})^y(y+4))+(\frac{4}{5})^{y-1}(y+6)$
To quickly verify this, plug in y=1 we will get 7 and for y=5 we have 5.8192.
Note: 
The part of the equation to obtain $\sum_{x=y+1}^\infty x(\frac{5}{6})^{x-1}$ may not be obvious but the trick here is to use the series of $\sum_{k=0}^\infty(k+1)x^k=\frac{1}{(1-x)^2}$, which can be obtained by geometric series in the following way:
$$
x\sum (k+1)x^k-\sum (k+1)x^k=(x-1)\sum (k+1)x^k =1+x+x^2+...=\frac{1}{1-x}\Rightarrow\sum (k+1)x^k=\frac{1}{(1-x)^2}
$$
A: For any $i$ and $j$ with $i < j$, what is $P(X=i, Y=j)$?  This means the first $i-1$ rolls are neither 5 nor 6, the $i$'th is a 6, the $i+1$'th to $j-1$'th (if any) are not 5's, and the $j$'th is a 5.  So $P(X=i,Y=j) = (4/6)^{i-1} (5/6)^{j-i-1} (1/6)^2$.  Similarly for $i>j$, interchanging $i$ and $j$.  Of course for $i=j$, $P(X=i,Y=i)=0$.  Divide by 
$P(Y=j)$ to get the conditional probability $P(X=i | Y = j)$. 
