Multivariate Hypergeometric Distribution With Wildcard In the wikipedia link http://en.wikipedia.org/wiki/Hypergeometric_distribution it is obvious you can do things like "I draw 5 cards from a deck of 50 cards where there are 10 cards that equate to success, 40 that fail, and I want to know the probability of drawing 2 success cards", but in the multivariate version at
http://en.wikipedia.org/wiki/Hypergeometric_distribution#Multivariate_hypergeometric_distribution
it seems you have to pick the exact amount of things you want to draw, and draw exactly that many. i.e.
I draw 10 cards from a deck with 10 red cards, 15 blue cards, and 5 blank cards, and want to draw exactly 3 blank cards, 6 red cards, and 1 blue card.
How do I use the multivariate distribution method, to allow for wildcards or unsure draws, i.e. I want to draw at least 3 blank cards given the same example above, but don't care about what else I draw?
I tried using the binomial coefficient "X choose Y" given that X is the total number of marbles/cards/whatever in the pile, and Y is the number of wildcards, but it didn't work. It gave me wildly incorrect results.
Help! D:
EDIT: Probability without replacement question The top answer seems good, and it solves one of my problems (for example now I can put a deck of cards that has three types of unique cards in it, 3 of each kind of card, pick one card and figure out the probability of drawing it if I draw 1, 2, 3, etc. cards from the deck) but now I can't pick more than one type of card that I want to draw from the deck. So... Is there a way to use the technique in section "A" in the above question's answer, to do this if I want to grab multiple different types of cards?
 A: Hypergeometric Probability Distribution.
1.) What is the probability of making a flush in Poker given you are holding four cards of the same suit ?
(hypergeometric 1 47 1 9) = 9/47 ~ 19.148...%
2.) what is the probability of making a flush in Poker given you are holding three cards of the same suit ?
(hypergeometric 2 47 2 10) = 45/1081 ~ 4.1628... %
Similiar to your problem but not exactly so.
A Sceme program that can be modified to extend the hypergeometric probability distribution. 
(define (factorial-iter-aux product counter max-count)
(if (> counter max-count)
product
(factorial-iter-aux (* counter product)
(+ counter 1)
max-count)))
(define (factorial-iter n)
(factorial-iter-aux 1 1 n))
(define (choose j k)
(/ (factorial-iter j) (* (factorial-iter k) (factorial-iter (- j k)))))
(define (hypergeometric x N n k)
(/ (* (choose k x) (choose (- N k) (- n x))) (choose N n))) 
A: Codefun, I'm honestly not sure if this is a useful idea, but you could try summing over all the possible outcomes you don't care about. 
Say you only want to know "What is the probability that with 10 draws I draw at least 3 blank cards." Well, there are many ways you could do that: $3+B$ blank, $R$ red, and $U$ blue, where $B+R+U = 7$. To figure out the total probability of drawing at least 3 blank cards, not caring about what the others are, we just have to add up the probabilities for each configuration where we happen to draw 3 blank cards. I will use the notation $P(D,B,R,U)$ to denote the probability of drawing $B$ blank cards, $R$ red cards, and $U$ blue cards in $D$ draws.
I claim that the probability to draw at least $B_1$ blank cards is:
$$\sum_{B=B_1}^{D}\sum_{R=0}^{D-B} P(D,\ B,\ R,\ D-B-R)$$
The nested sum can be difficult to unpack, but just remember what we are doing : summing the probabilities of all configurations that match our condition. Notice also that once we have specified the number of blank cards and the number of red cards, the number of blue cards is fixed (hence why $U = D-B-R$ in my expression).
