Hi Moxen,

This is a fairly complicated problem if I understand it correctly. To recap my understanding, you have a non-standard 60-card deck with

1 One

21 Twos

4 Fours

3 Fives

31 Others

and you would like to know the probability of drawing a 7-card hand with at least 1 One, 2 Twos, 1 Four, and 1 Five.

There are $\displaystyle \binom{60}{7} = 382,606,920$ possible 7-card hands, all of which are equally likely.

We need to count the number of acceptable hands. As a first step, let's list all the possibilities in terms of the numbers of Ones, Twos, etc. Let's say there are a Ones, b Twos, c Fours, d Fives, and e Others. Then the possibilities for an acceptable 7-card hand are

Code:

a b c d e
1 2 1 1 2
1 4 1 1 0
1 2 3 1 0
1 2 1 3 0
1 3 1 1 1
1 2 2 1 1
1 2 1 2 1
1 3 2 1 0
1 3 1 2 0
1 2 2 2 0

As you can see, there are 10 combinations of (a,b,c,d,e) possible. For each of these possibilities, the number of hands is

$\displaystyle \binom{1}{a} \binom{21}{b} \binom{4}{c} \binom{3}{d} \binom{31}{e} $

Sum up these products for the 10 (a,b,c,d,e) possibilities. (I used a spreadsheet.) If you do it correctly, I think you will find the sum is 1,980,720. So the probability of drawing an acceptable hand is

$\displaystyle \frac{1,980,720}{382,606,920}$,

which is approximately 0.00512.

If you are not familiar with the notation above,

$\displaystyle \binom{n}{m} = \frac{n!}{m! (n-m)!}$

is the number of combinations of n objects taken m at a time.