If you look at how many possible 2nd cards satisfy the condtion for a given first card
you'll see there are 2 inital cards each that have 10-19 cards that satisfy it,
and then 30 initial cards that have 20 cards that will satisfy it.
That leads to
\(p = \dfrac{1}{25}\dfrac{1}{49}\sum \limits_{k=10}^{19}k + \dfrac{30}{50}\dfrac{20}{49}=\dfrac{89}{245}\)