Consider that if X and Y are identically independent and geometrically(p) distributed, show the following:
P( X ≥ Y ) = 1 / ( 2 - p )
$\displaystyle \Pr(X \geq Y) + \Pr(X \leq Y) - \Pr(X = Y) = 1$.
By symmetry, $\displaystyle \Pr(X \geq Y) = \Pr(X \leq Y)$.
Therefore $\displaystyle 2 \Pr(X \geq Y) = 1 + \Pr(X = Y)$.
$\displaystyle \Pr(X = Y) = \Pr(X = 0, Y = 0) + \Pr(X = 1, Y = 1) + \Pr(X = 2, Y = 2) + \, ....$
$\displaystyle = p^2 + (1 - p)^2 p^2 + (1 - p)^4 p^2 + \, .... $
(since X and Y are independent)
$\displaystyle = p^2 (1 + (1 - p)^2 + (1 - p)^4 + \, .... ) = p^2 \cdot \frac{1}{1 - (1 - p)^2} = \frac{p}{2 - p}$.
Therefore $\displaystyle 2 \Pr(X \geq Y) = 1 + \frac{p}{2 - p}$
etc.