Probability of draw 1000 balls of the same color from 2000 Without replacement. We have 2000 balls. 1000 red and 1000 green.
What is the probability of draw the 1st 1000 balls to be the same color (red or green).
 This is how I worked on it, not sure.
My work
Let's begin with a numerically simpler problem. Suppose you have 8 balls
in the urn: 4 red and 4 green. You draw 4 balls from the urn without
replacement. Then there are ${8 \choose 4} = 70$ possible outcomes. Two
of the outcomes have all balls the same color: one all 4 red and the other
all 4 green. So the probability of getting 4 balls of the same color is
$2/{8 \choose 4} = 2/70 = 0.02857.$

In the same way, the answer to the original question is $2/{2000 \choose 10000},$
which is a very small number. Maybe you are allowed to leave your answer
in this 'combinatorial' form. And maybe you are expected to use
Stirling's Approximation (as suggested by @ThomasAndrews and @callciulus).
As you can see by the Wikipedia reference, there are several forms of the
Approximation, all of which give 'about' the same answer.

The answer you linked in your Question is correct for the probability
of getting 'all red balls'. Multiplying by 2, you'd get the probability
for 'all the same color'. which matches $2/{2000 \choose 10000}.$
 (Your first factor is 2.)
Edited: I removed the second answer. Thanks for letting me know where I went wrong. ;)
This answer applies if you assume that the ball is returned to the bag every time you draw. The probability of drawing 1 ball a certain color is 1/2. Since the first 1000 balls all need to be the same color, you take probability and raise it to the power of the number of draws: so your answer is (1/2)^1000