Geometric Distribution $P(X\ge Y)$ I need to show that if $X$ and $Y$ are idd and geometrically distributed that the $P(X\ge Y)$ is $1\over{2-p}$. the joint pmf is $f_{xy}(xy)=p^2(1-p)^{x+y}$, and I think the only way to do this is to use a double sum: $\sum_{y=0}^{n}\sum_{x=y}^m p^2(1-p)^{x+y}$, which leads to me getting quite stuck. Any suggestions?

It is easier to use symmetry:
$$
   1 = \mathbb{P}\left(X<Y\right) +\mathbb{P}\left(X=Y\right) + \mathbb{P}\left(X>Y\right)
$$
The first and the last probability are the same, due to the symmetry, since $X$ and $Y$ are iid. Thus:
$$
   \mathbb{P}\left(X<Y\right) = \frac{1}{2} \left(1 - \mathbb{P}\left(X=Y\right) \right)
$$
Thus:
$$
 \mathbb{P}\left(X\geqslant Y\right)  = \mathbb{P}\left(X>Y\right) + \mathbb{P}\left(X=Y\right) = \frac{1}{2} \left(1 + \mathbb{P}\left(X=Y\right) \right)
$$
The probability of $X=Y$ is easy:
$$
    \mathbb{P}\left(X=Y\right) = \sum_{n=0}^\infty \mathbb{P}\left(X=n\right) \mathbb{P}\left(Y=n\right) = \sum_{n=0}^\infty p^2 (1-p)^{2n} = \frac{p^2}{1-(1-p)^2} = \frac{p}{2-p}
$$
