# 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.

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.

• I don't understand why we're comparing $2 \times 3718$ to $6084$. Commented Sep 2, 2021 at 14:57
• @Karam the event that the suits split $3$ to $1$ can happen in one of two ways: $3$ from suit $A$, $1$ from suit $B$ OR $1$ from suit $A$, $3$ from suit $B$
– user801306
Commented Sep 2, 2021 at 15:06
• $2\times 3718$ is the number of ways to split $3,1$. $6084$ is the number of ways to split $2,2$.
– lulu
Commented Sep 2, 2021 at 15:06
• @ Matthew Pilling do you mean $3$ from suit $A$ and $1$ from suit $B$, or $1$ from suit $A$ and $3$ from suit $B$? Commented Sep 2, 2021 at 15:11
• Got it. That makes sense! Commented Sep 2, 2021 at 15:12

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

• why $2{13 \choose 3}{13 \choose 1}$? Commented Sep 2, 2021 at 15:03
• I am adding in the answer Commented Sep 2, 2021 at 15:04