(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%