Hello all,

(BTW I misspelled Probability in the subject... apparently I cant edit it. Sorry

)

I play Magic: the Gathering, and it has been 10 years since I took probability in college.

Problem A (example): taken from wikipedia

Poker probability - Wikipedia, the free encyclopedia
What is probability of drawing 2 pair in a 5 card hand from 52 card deck?

f = 13C2 * 4 C 2 * 4 C 2 * 11C1 * 4 C 1

= (choose 2 numbers for pairs) * # combos for 4 suits pair 1 * # combos for 4 suits pair 2 * choose 11th number for last card * choose suit last card

Problem B (My problem)

I have a 60 card deck

4 card A

4 card B

10 card C

42 card D ("other")

What is the probability in 7 card, 8 card, 9 card draw that I draw at least one A, one B, and one C?

Events, 7 card draw

Choose 3 cards of 7 to be the three special cards

# ways to choose 1 card from A (4)

# ways to choose 1 card from B (4)

# ways to choose 1 card from C (10)

# ways to choose 4 cards from D (42)

f = 7C3 * 4C1 * 4C1 * 10C1 * 57 C 4

f = 2,212,056,000

N = 60 C 7

N = 386,206,920

P = 572%

**What am I doing wrong?** **The following answer seems right,** but I would appreciate validation.

If I focus on the fact that order matters for the first 3 cards, but not the last 4, I get

f = 7P3 * 4C1 * 4C1 * 10C1 * 57 C 4

N = 60P3 * 57 C 4

P = 16.36%

Thus, for these 3 cards drawn together in a hand of n, the probability is

P(3 cards in draw of n) = 160 * nP3 / 60P3

P(n+1) = (n+1)/(n-2) * P (n)

So P (3 in 7 draw) = 16%

P (3 in 8 draw) = 26%

P (3 in 9 draw) = 39%