probability/combinatorics question with marbles An urn has 20 green out of 50 marbles. Draw all 50 marbles without replacement. Let
X = # of green marble runs of any length.
Example : GGGGBBBGGBBGBB. . .
In the above example, there are 3 runs in the first 14 trials. To get X, we would have to
examine the entire sequence of the 50 trials.
Find P(X=1)
For X=1 runs, that means that all 20 green marbles are together. There are 31 spots where that run could be. I came up with the answer
$\frac{20 \choose 20}{50 \choose 20} * 31 $
Is this correct? The biggest trouble that I have in computing my answer is that the green marbles have to be in a row. If the green marbles didn't have to be in a row then would it just be ${50 \choose 20}$?
 A: Yes.    But rather that thinking of it as "counting arrangements where green marbles are in row", think of it as "ways to split a row of green marbles into groups"; then find places for those groups.
There are $\tbinom{20}{20}$ ways to group twenty items into one group.   Though I'd rather count that as $\binom{19}{0}$ ways to put no split among twenty items (which is also $1$ way).   Then there are $\binom{31}{1}$ ways to insert that group before, between, or after thirty other items.   Finally the denominator is the total ways to arrange two types of, respectively, twenty and thirty items.
$$\require{cancel}\mathsf P(X=1) = \left.\color{silver}{\cancel{\dbinom{19}{0}}}\dbinom{31}{1}\middle/\dbinom{50}{20}\right. = \dfrac{31}{\tbinom{50}{20}}$$
Similarly, there are $\binom{19}{1}$ ways to place one split between twenty identical items, then $\binom{31}{2}$ ways to select places for these groups before, between, or after thirty items.
$$\mathsf P(X=2) = \left.\dbinom{19}{1}\dbinom{31}{2}\middle/\dbinom{50}{20}\right.$$
Oh, hey!   Can you see an obvious pattern emerging?

 $$\mathsf P(X=x) = \left.\dbinom{19}{x-1}\dbinom{31}{x}\middle/\dbinom{50}{20}\right. \quad\Big[x\in\Bbb N\cap[1;20]\Big]$$

