OK, I am not going to give you the whole answer as this looks like homework. But let me give you massive hint as to how you can simplify your existence. If you try to compute the probability directly you will find it a very long and tedious calculation.
Call the event you are interested in $E_+$
So think about 2 more events. One being the opposite, so selecting a second card of lower face value, call this $E_-$. The other being, selecting a card of the same face value, $E_0$.
Now, it is easy to see that $\mathbb{P}(E_+) = \mathbb{P}(E_-) = p $. Also, notice that
$\mathbb{P}(E_+) + \mathbb{P}(E_-) + \mathbb{P}(E_0)=1$
since these events are disjoint and cover all posibilities.
So if you can calculate $\mathbb{P}(E_0)$ which is a lot easier than calculating $\mathbb{P}(E_+)$ directly, you are done!