standard deck of cards
draw 9 cards without replacement
Probability of getting at least 2 suits with at least 3 of each suit?
I know there could be a distribution like 3,3,3,0 (this would be 3 suits with at least 3)
4,3,1,1 (this would be 2 suits with at least 3)
5,2,1,1 (this would be 1 suit with at least 3)
I know the minimum number would be 1 and the maximum would be 3.
I have no clue on how to start this other than list all combinations.
thank you
On
Consider the complementary event: we draw nine cards, and exactly three suits contain two or fewer cards.
There are $\binom{52}{9}$ hands, all of which we assume are equally likely. We want to count the number of hands in which exactly three suits contain two or fewer (i.e. zero, one or two) cards.
There are $\binom{4}{3}$ ways to pick the three suits.
Given the three suits, if they contain $i, j, k$ cards respectively, then the number of possible hands is $$\binom{13}{i} \binom{13}{j} \binom{13}{k} \binom{13}{9-i-j-k}$$
So the number of hands in which three suits contain fewer than three cards each is $$N_3 = \binom{4}{3} \sum_{i=0}^2 \sum_{j=0}^2 \sum_{k=0}^2 \binom{13}{i} \binom{13}{j} \binom{13}{k} \binom{13}{9-i-j-k}$$ The probability that there are three suits which contain fewer than three cards each is $$p = \frac{N_3}{\binom{52}{9}}$$ and the probability that there are at least two suits with three or more cards is $$\boxed{1-p = 0.570226}$$
$$ \begin{matrix} Total Cases & WaysTo Select & Total \\ 3|3|3|0 & 4 & 4*13c3*13c3*13c3 \\ 3|3|1|2 & 12 & 12*13c3*13c3*13c1*13c2 \\ 4|3|1|1 & 12 & 12*13c4*13c3*13c1*13c1 \\ 4|3|2|0 & 24 & 24*13c4*13c3*13c2 \\ 4|4|1|0 & 12 & 12*13c4*13c4*13c1 \\ 5|3|1|0 & 24 & 24*13c5*13c3*13c1 \\ 5|4|0|0 & 12 & 12*13c5*13c4 \\ 6|3|0|0 & 12 & 12*13c6*13c3 \\ \end{matrix} $$