This doesn't make much sense: $\binom{100}{1}\binom{99}{49}$ counts the number of ways of picking one out of 100 possibilities and 49 out of 99 possibilities. That is not what you want: you don't want to pick 49 other cards, you just want to pick one out of the 49 that are still "in play".
That is, I think what you meant was:
I can pick the first number arbitrarily, and there are $\binom{100}{1}$ ways of doing that; once I pick that, there are only $49$ other numbers that I can pick (those with the same parity as the one I picked), and that means $\binom{49}{1}$ of them. Multiplying we have $\binom{100}{1}\binom{49}{1}$.
(Note the $\binom{49}{1}$ rather than $\binom{99}{49}$).
That's almost right, but you are overcounting: you count the pair $(1,99)$ once when you first pick $1$ and then pick $99$; and you count it again when you first pick $99$ and then pick $1$. So, what's the easy way of taking care of that?
(To see that your answer cannot possibly be correct, note that there are only $\binom{100}{2} = \frac{100(99)}{2} = 4950$ pairs of numbers between $1$ and $100$; but $\binom{100}{1}\binom{99}{49} \approx 5\times 10^{30}$, which is way too many.)