Probability of Drawing More of One Type of Card Than Another Let's say I have a deck with three or more types of cards in it. I draw a certain number of cards at random. How do I calculate the probability that I get more of one type of card than another?
For example, let's say I have a deck of 75 cards, and it has 25 blue cards, 30 red cards, and 20 green cards. I draw 6 cards at random. What are the odds I draw more blue cards than red cards?
 A: You cannot use a binomial distribution because the card selections are without replacement.  Instead, you must use a multivariate hypergeometric distribution.
Call the total number of cards $n$, with $n_a$ of one type of interest in the comparison, $n_b$ of the other type of interest in the comparison, and $n_c$ all the others.  (It does not make a difference the makeup of the categories of the $n_c$ cards not used in the comparison.)  Of course $n = n_a + n_b + n_c$ or $n_c = n - n_a - n_b$.
The probability of getting exactly $n_a = i$ and $n_b = j$ (with $i + j \leq k$) given a total selection of $k$ cards is given by the multivariate hypergeometric distribution:
$$P(n_a = i, n_b = j; k) = {{n_a \choose i}{n_b \choose j}{n - n_a - n_b \choose k - i - j} \over {n \choose k}}$$
Noting that $0 < n_a \leq k$ and $n_b < n_a$ for "success," we have the probability of "success" being:
$$P(n_a > n_b; k) = \sum\limits_{i=1}^k \sum\limits_{j = 0}^{i-1} \left[ {{n_a \choose i}{n_b \choose j}{n - n_a - n_b \choose k - i - j} \over {n \choose k}} \right]$$
If you plug in the numbers you gave $(n = 75, n_a = 25, n_b = 30, n_c = 20, k = 6)$, the probability is:
$$P(n_a > n_b; 6) = {441153 \over 1342397} \approx 0.328631$$
A: I think you want the Binomial Distribution with some tweaking. 
You can break it down. Given this scenario, what are the chances of getting more green cards than not-green cards? When you have a binary choice like that you can use the regular Binomial distribution. Then you can re apply it to the set of not-green cards calling them say, red vs. not-red(=blue). In this way you can take the principles that apply to an either or situation and mix to cover this problem of 3 possibilities. 
I think you'll need certain weights to get the probabilities right. 
You can save yourself some work. You know if you have drawn 4 or more green card already, then you already have more green than red cards, however many blue cards you have selected. 
Let r=number of red cards, b=number blue cards, and g=number of green cards where r+b+g=6, and r,b,g are all non-negative integers. 
How many ways are there of making 6 having selected 3 green cards? You need to select 2 from the non green cards. So C(75-'greeen cards',2). What about 5 green cards? 
What are the chances of having picked 3 green cards in the first place? 4? 5? 6? 
Suppose you've selected 2 green cards. Then r+b=4. In how many ways will b>g? 
Binomial distribution gives you the chances of selecting 3 green cards from a set of green and not green cards. 
Put it all together, I think you''ll  have your answer. 
