[Ganchrow]

I'll take it:

All conditional probability, individual chance % listed in yes/yes/...no format. Multiplied by number of possible sequences within each outcome. All numbers rounded and results in a .2% rounding error.

0 beer card: .555 x .5 x .428 x .33 = .0395 x 1 = .040 = 4%
1 beer card: .444 x .625 x .571 x .5 = .0792 x 4 = .32 = 32%
2 beer cards: .444 x .375 x .714 x .666 = .0784 x 6 = .47 = 47%
3 beer cards: .444 x .375 x .286 x .833 = .039 x 4 = .16 = 16%
4 beer cards: .444 x .375 x .285 x .166 = .0079 x 1 = .008 = 0.8 %

Yep. Well done.

For those still a bit quizzical as to how such an answer might be quickly obtained in general:

Underlying this problem is the Hypergeometric distribution. In Excel this is represented by the =HYPGEOMDIST() function.

0 cards => =HYPGEOMDIST(0,4,4,9) = 5 126 ≈ 3.97%
1 cards => =HYPGEOMDIST(1,4,4,9) = 40 126 ≈ 31.75%
2 cards => =HYPGEOMDIST(2,4,4,9) = 60 126 ≈ 47.62%
3 cards => =HYPGEOMDIST(3,4,4,9) = 20 126 ≈ 15.87%
4 cards => =HYPGEOMDIST(4,4,4,9) = 1 126 ≈ 0.79%

In general, the probability of selecting exactly w winners out of n draws without replacement given an N sized population consisting of W winning draws (where 0≤w≤W≤N and 0≤w≤n≤N) is Excel given in Excel by:

= HYPGEOMDIST(w,n,W,N)

Using combinatorials this may be equivalently expressed as:

= combin(W, w) * combin(N-W, n-w) / combin(N,n)