I would like to calculate the conditional expectation $E(X|X>Y)$. I reckon that there are two scenarios for solving this question. But the ansewers are not the same in the end and I am confused by this.
First scenario
Firstly, we compute $E(X|X>y)$:
$E(X|X>y) = \sum x\frac{P(X \ = \ x, \ X \ > \ y)}{P(X \ > \ y)} \\ =\sum_{x\ = \ y +1} \ x\frac{(1 \ - \ p)^x \ p}{(1 \ - \ p)^y} \\ = y \ + \ \frac{1}{p}$
As $E(X|X>Y) \ = \ E(E(X|X>Y,Y)) \ = \ E(E(X|X>y))$, therefore,
$E(X|X>Y) = E(y \ + \ \frac{1}{p}) \ = \ \frac{1}{q} \ + \ \frac{1}{p}$
Second scenario(I think this is correct!)
$E(X|X>Y) = \sum x\frac{P(X \ = \ x, \ X \ > \ Y)}{P(X \ > \ Y)} \\ =\frac{1}{P(X \ > \ Y)}\sum_{x}x \ P(X \ = \ x)P(Y \ < \ x)$
Your first attempt is based on an erroneous claim that $X$ is conditionally independent of $Y$ when given $X>Y$. Surely it cannot be so. $$\mathsf P(X=x\mid X>Y, Y=y)~\neq~\mathsf P(X=x\mid X>y)$$
Rather, indeed you should indeed use the second approach. $$\begin{align}\mathsf E(X\mid X>Y)&=\sum_{x=2}^\infty x\,\mathsf P(X=x\mid X>Y)\\[2ex]&=\dfrac{\sum_{x=2}^\infty x\,\mathsf P(X=x)\,\mathsf P(Y<x)}{\mathsf P(X>Y)}\\&=\dfrac{pq\sum_{x=2}^\infty x(1-p)^{x-1}\sum_{y=1}^{x-1}(1-q)^{y-1}}{pq\sum_{x=2}^\infty (1-p)^{x-1}\sum_{y=1}^{x-1}(1-q)^{y-1}}\end{align}$$