# Thread: Card draw probability

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?
Thanx in advance,
chromeboy

2. To find probability of at least two we do the following.
Find the probability of none plus the probability of exactly one.
Subtract that sum from 1.
That is your answer.