Suppose you are dealt five cards of one suit, four cards of another. A bridge hand is found by taking $13$ cards at random and without replacement from a deck of $52$ playing cards. Suppose you are dealt five cards of one suit, four cards of another. Would the probability of having the other suits split $3$ and $1$ be greater than the probability of having them split $2$ and $2$?
My attempt
Let $S$ $=$ deck of $52$ playing cards.
Suppose that we have been dealt $5$ spades and $4$ hearts, so let $A=\{5 \text{spades}, 4 \text{hearts} \}$. Thus $A^c = S \setminus A=\{8 \text{spades},9 \text{hearts}, 13 \text{diamonds}, 13 \text{clubs} \}$
Now let $ \tilde A=\{3 \text{diamonds},1 \text{club}\}$ and let $A'=\{2 \text{diamonds},2 \text{clubs}\}$.
We can select

*

*$3$ diamonds in any one of ${13 \choose 3}=286$ ways


*$1$ club in any one of ${13 \choose 1}=13$ ways
By the multiplication principle, the number of outcomes in $\tilde A$ is $N(\tilde A)={13 \choose 3}{13 \choose 1}{17 \choose 0}=3718$, where ${17 \choose 0}$ gives the number of ways in which zero cards are selected out of the $8$ spades and $9$ hearts whereby ${17 \choose 0}=1$ (i.e. to not do anything)?
Now, the number of possible $4$ cards that can be drawn from a deck of $52-9=43$ playing cards is $ N(A^c)={43 \choose 4}=123,410$.
Thus $P(\tilde A)=\frac{N(\tilde A)}{N(A^c)}=\frac{37128}{123410} \approx 0.0301272 \Rightarrow P(\tilde A) \approx 3$ %.
Using the same reasoning, $N(A')=6084$, which means that the probability of having the other suits split $3$ and $1$ is less than the probability of having them split $2$ and $2$. However, the correct answer is yes, the probability of having the other suits split $3$ and $1$ is greater than the probability of having them split $2$ and $2$.
Would you please let me know what I'm missing here. Thanks.
 A: Your calculation appears to assume that you are drawing from $43$ cards (the full $52$ less the $9$ you know about).  But this is not correct.  We are told that the remaining $4$ cards are from the two other suits, so you are selecting four cards out of $26$ ($13$ each of suits $A,B$).
The number of ways to draw $3$ from $A$ and $1$ from $B$ is $$\binom {13}3\times \binom {13}1=3718$$  Of course, that's the same as the number of ways to draw $1$ from $A$ and $3$ from $B$.
The number of ways to draw two from each is $$\binom {13}2^2=6084$$  which is less than $2\times 3718$.
The point is that there are two ways to draw three from one and one from the other, and only one way to draw two of each.
A: Apart from the explanation provided by @lulu, if you do want to compute the whole thing, surely it is simpler to just compute, for the two alternatives,
$\dfrac{\binom{13}5\binom{13}4\binom{13}2\binom{13}2}{\binom{52}{13}}$
and compare with $\dfrac{\binom{13}5\binom{13}4\cdot 2\binom{13}3\binom{13}1}{\binom{52}{13}}$
The $2$ multiplier in the second alternative because it could either be $3$ of suit $X$ and $1$ of suit $Y$ or vice versa
