1. Card draw probability

Hallo,

I'm working on card drawing probabilities and I'm not quiet getting the results I was hoping for.
I know that the odds of drawing at least one copy of a desired card in the
first 4 cards from a deck of 12 cards containing x copies of a desired card is:

x=1 copy: 0,3333
x=2 copies: 0,5758
x=3 copies: 0,7455
x=4 copies: 0,8586
x=5 copies: 0,923
x=6 copies: 0,9697

My formula for this is:
Y=1-[(12-x)/12*(11-x)/11*(10-x)/10*(9-x)/9], which yields the above results.
The part in parantheses is the probability of not drawing on of the desired
cards for 4 draws from a deck of 12 which then gets subtracted from 1,
yielding the reverse probability (the prob. of drawing at least 1 copy in that draw).

Now i want to know, what are the odds of drawing at least 2 copies of a certain card
in the first 4 drawn cards. My formula for this would be:

y=(1-[(12-x)/12*(11-x)/11*(10-x)/10*(9-x)/9])*(1-[(12-x)/11*(12-x)/10*(12-x)/9])

I take the probability of drawing at least 1 copy, then i multiply with the probability of drawing at least 1 copy when drawing 3 cards from a deck of 11, containing x-1 copies of a desired card. I'm not sure if this is the right way of calculating these odds, but the results would be:

x=1 copy : 0
x=2 copies: 0,1570
x=3 copies: 0,3660
x=4 copies: 0,5672
x=5 copies: 0,7322
x=6 copies: 0,8522

Going forth with this line of thought, the odds of drawing at least 3 copies would then be:

y=(1-[(12-x)/12*(11-x)/11*(10-x)/10*(9-x)/9])
*(1-[(12-x)/11*(12-x)/10*(12-x)/9])*(1-[(12-x)/10*(12-x)/9])
yielding

x=1 copy : 0
x=2 copies: 0
x=3 copies: 0,0732
x=4 copies: 0,2143
x=5 copies: 0,3905
x=6 copies: 0,5681

and for drawing at least 4 copies:
y=(1-[(12-x)/12*(11-x)/11*(10-x)/10*(9-x)/9])
*(1-[(12-x)/11*(12-x/10*(12-x)/9])
*(1-[(12-x)/10*(12-x)/9])*(1-[(12-x)/9])
yielding

x=1 copy : 0
x=2 copies: 0
x=3 copies: 0
x=4 copies: 0,0238
x=5 copies: 0,0868
x=6 copies: 0,1894

Now, if x=4, drawing all 4 desired cards from a deck of 12 in the first 4 draws can also be calculated by
y=4/12*3/11*2/10*1/9=0,00202, yiedling a results that is different by a factor of >10 from the result obtained
with my formular.

Does anyone of you math wizs know how to do this properly?
card, draw, probability 