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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

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    What do _you_ think? – 2017-01-30
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    It's good practice on StackExchange to post what you have worked through first, so we know where you are stuck. What probability distribution do you think this problem implies? What part of using that distribution is causing you trouble? – 2017-01-30
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    How many subsets of size 1000 are there and how many are monochromatic? – 2017-01-30
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    It's approximately $\frac{\sqrt{1000\pi}}{2^{1999}}$ :) – 2017-01-30
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    There are only $2$ ways to draw your monochromatic set. How many ways are there to draw any set? – 2017-01-30
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    Maybe if you wait long enough, the green ones will ripen. Otherwise, does $\binom {2000}{1000} \big / 2$ mean anything to you? – 2017-01-30
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    @TimonG. I added my work. Thanks. – 2017-01-30
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    For the denominator of your probability: How many ways to choose 2 objects from among 5 called A,B,C,D.E: That's AB, AC, AD, AE, BC, BD, BE, CD, CE, DE. That's 10 without regard to order; ${5 \choose 2} = 10.$ How many ways to choose 1000 from among 2000? (Don't try to make the list). – 2017-01-30
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    Your solution is mostly correct. What you really need is a more elegant way to present it. Can you express the numerator and the denominator in terms of factorials? – 2017-01-30
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    @Yamona Have a look at the stirling approximation. With it you´ll get the expression which is mentioned in the comment of Thomas Andrews. https://en.wikipedia.org/wiki/Stirling%27s_approximation – 2017-01-30

2 Answers 2

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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.)

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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

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    First answer is for sampling with replacement, so not applicable here, but may be good for comparison. I think the 2nd (and relevant) answer needs to be $(2000/2000)(999/1999)(998/1998)\dots (1/1001).$ Please edit if you agree. – 2017-01-30