Question

After shuffling a deck of 52 cards, a dealer deals out 5. let x = the number of suits represented in the five-card hand. (a) find the pmf of x. (hint: p(1) = 4p(all are spades), p(2) = 6p(only spades and hearts with at least one of each suit), and p(4) = 4p(2 spades ∩ one of each other suit).) (round your answers to five decimal places.)

Answer 1

The PMF is

$P(X=x)\approx\begin{cases}0.00198&\text{for }x=1\\0.14592&\text{for }x=2\\0.58836&\text{for }x=3\\0.26375&\text{for }x=4\\0&\text{otherwise}\end{cases}$

There is a total of $\dbinom{52}5=2,598,960$ possible 5-card hands. For reference, I'm using this notation for the binomial coefficient,

$\dbinom nk=\dfrac{n!}{k!(n-k)!}$

For $x=1$, suppose we only care about spades (S), like the hint suggests. Then there are $\dbinom{13}5=1287$ ways to get 5S and $\dbinom{39}0=1$ way to not draw cards of any other suit, so there are 1287 hands of 5S. The number of ways to do this is the same for any other suit of our choice, and with 4 suits to choose 1 from, we multiply 1287 by $\dbinom41=4$. Then the probability is

$P(X=1)=\dfrac{4(1287)}{\binom{52}5}=\dfrac{33}{16,660}$

The same by-suit breakdown applies for the other cases when $x>1$.

For $x=2$, suppose we want a hand with S and hearts (H) only. Then we can have 4S and 1H; 3S and 2H; 2S and 3H; or 1S and 4H. This translates to

$\dbinom{13}4\dbinom{13}1+\dbinom{13}3\dbinom{13}2+\dbinom{13}2\dbinom{13}3+\dbinom{13}1\dbinom{13}4$

or by symmetry,

$2\left(\dbinom{13}4\dbinom{13}1+\dbinom{13}3\dbinom{13}2\right)$

Now we're choosing 2 suits from the possible 4, so we multiply this by $\dbinom42=6$. Then

$P(X=2)=\dfrac{12(715\cdot13+286\cdot78)}{\binom{52}5}=\dfrac{143}{980}$

For $x=3$, suppose we want a hand with S, H, and clubs (C). Then we can have 3S, 1H, and 1C; 1S, 3H, and 1C; 1S, 1H, and 3C; 2S, 2H, and 1C; 2S, 1H, and 2C; or 1S, 2H, and 2C. By symmetry, this comes out to be

$3\left(\dbinom{13}3\dbinom{13}1\dbinom{13}1+\dbinom{13}2\dbinom{13}2\dbinom{13}1\right)$

and we're choosing 3 suits this time, so we multiply by $\dbinom43=6$. Then the probability is

$P(X=3)=\dfrac{18(286\cdot13^2+78^2\cdot13)}{\binom{52}5}=\dfrac{4901}{8330}$

Finally, for $x=4$, suppose we want 2S and 1 each of the remaining suits. This can be done in

$\dbinom{13}2\dbinom{13}1\dbinom{13}1\dbinom{13}1$

ways. Multiply by 4 to account for all possible choices of suit for which 2 cards show up in the hand so that the probability is

$P(X=4)=\dfrac{4(78\cdot13^3)}{\binom{52}5}=\dfrac{2197}{8330}$

With only 4 suits available, the probability of $X$ taking on any other value must be 0.