probability that a $9$ card hand has exactly two "three of a kind"

Question

Find the probability that a $9$ card hand has exactly $2$ three of a kind.

-probability is $\binom{52}{9}$

My approach to this problem started at the assumption that since there are $4$ suits with $13$ cards each, there are $13$ possibilities for the first pair of $3$.

-Since $4$ suits and $3$ cards, choose $3$ of the suits. $13\binom{4}{3}$

-Since another of the same would create a pair of $4$, $12\binom{4}{3}$

-Now we have $6$ cards, $9$ slots, so $\binom{9}{6}$. $13 \cdot 12\binom{4}{3}^2$

-$3$ slots left, $11~\text{cards} \cdot 10~\text{cards} \cdot 9~\text{cards}$, we have to account for the fact that order doesn't matter, so divide by $3!$.

So $13 \cdot 12 \cdot \binom{4}{3}^2 \cdot \binom{9}{6} \cdot \binom{11}{3}$, which is wrong. I'm not completely sure where my mistake was. Also a strange follow up question, if anyone used to be prone to these kinds of mistakes, are there any insights to make things more clear?

Answer 1

The number of ways we can select a hand of nine cards drawn from a $52$ card deck is $$\binom{52}{9}$$

Method 1: We consider two cases. We select two three of a kinds, a pair, and a single card or two three of a kinds and three single cards.

A deck of cards contains $13$ ranks (2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, A), each of which has four suits ($\color{red}{\heartsuit}, \color{red}{\diamondsuit}, \clubsuit, \spadesuit$).

Two three of a kinds, a pair, and a single card:

We select two ranks from which the three of a kinds are drawn. From each of these ranks, we select three of the four suits. From the remaining eleven suits, we select one for the pair. For this rank, we select two of the four suits. From the remaining ranks, we select one for the single card. We select one of the four suits from this rank.

$$\binom{13}{2}\binom{4}{3}^2\binom{11}{1}\binom{4}{2}\binom{10}{1}\binom{4}{1}$$

Two three of a kinds and three single cards:

We select two ranks from which the three of a kinds are drawn. From each of these ranks, we choose three of the four suits. From the remaining eleven suits, we select three from which a single card will be drawn. From each of the selected ranks, we choose one of the four suits.
$$\binom{13}{2}\binom{4}{3}^2\binom{11}{3}\binom{4}{1}^3$$

Total: Since the cases are mutually disjoint, the number of ways we can select nine-card hands with exactly two three of a kinds is $$\binom{13}{2}\binom{4}{3}^2\binom{11}{1}\binom{4}{2}\binom{10}{1}\binom{4}{1} + \binom{13}{2}\binom{4}{3}^2\binom{11}{3}\binom{4}{1}^3$$
Hence, the probability of selecting a nine-card hand with exactly two three of a kind is $$\frac{\dbinom{13}{2}\dbinom{4}{3}^2\dbinom{11}{1}\dbinom{4}{2}\dbinom{10}{1}\dbinom{4}{1} + \dbinom{13}{2}\dbinom{4}{3}^2\dbinom{11}{3}\dbinom{4}{1}^3}{\dbinom{52}{9}}$$

Method 2: We count the number of nine-card hands in which we select two three of a kinds without selecting a third three of a kind.

We select two of thirteen ranks. From each of these ranks, we choose three of the four suits. Since we cannot choose any more cards from these ranks, we have $52 - 2 \cdot 4 = 44$ cards from which we can select the remaining cards. However, we must exclude the possibility that we select three cards from one of the remaining eleven ranks. Thus, the number of nine-card hands with exactly two three of a kinds is $$\binom{13}{2}\binom{4}{3}^2\left[\binom{44}{3} - \binom{11}{1}\binom{4}{3}\right]$$ Hence, the probability that a nine-card hand contains exactly two three of a kinds is $$\frac{\dbinom{13}{2}\dbinom{4}{3}^2\left[\dbinom{44}{3} - \dbinom{11}{1}\dbinom{4}{3}\right]}{\dbinom{52}{9}}$$

Method 3: We subtract the number of hands with three three of a kinds from those with at least two three of a kinds.

This is an adaptation of a method proposed by @HughEntwistle.

Hands with at least two three of a kinds:

We select two ranks from which three cards will be selected. We select three of the four suits of each selected rank. Since we cannot select any more cards from these two ranks, we select the remaining three cards in the hand from the remaining $52 - 2 \cdot 4 = 44$ cards.
$$\binom{13}{2}\binom{4}{3}^2\binom{44}{3}$$ However, we have counted hands with three three of a kinds three times, once for each of the $\binom{3}{2}$ ways we can designate two of the three ranks as our two three of a kinds. We must exclude these.

Hands with three three of a kinds:

We choose three of the thirteen ranks. For each rank we select, we choose three of the four suits.
$$\binom{13}{3}\binom{4}{3}^3$$

Hence, the number of nine-card hands with exactly two three of a kinds is $$\binom{13}{2}\binom{4}{3}^2\binom{44}{3} - 3\binom{13}{3}\binom{4}{3}^3$$ Therefore, the probability that a nine-card hand contains exactly two three of a kinds is $$\frac{\dbinom{13}{2}\dbinom{4}{3}^2\dbinom{44}{3} - \dbinom{3}{2}\dbinom{13}{3}\dbinom{4}{3}^3}{\dbinom{52}{9}}$$

Answer 2

Here is a solution via generating functions and the Principle of Inclusion / Exclusion (PIE).

There are $\binom{52}{9}$ nine-card hands, all of which we assume are equally likely. Say a hand has "Property j" if it has exactly three cards of rank $j$, for $j = 1, 2, 3, \dots ,13$. For PIE, we count $S_i$, the number of hands with $i$ of the properties, for $i = 0, 1, 2, 3$. Usually we don't count $S_0$ for PIE, but in this case we will let $S_0$ be the total number of possible hands. So $$\begin{align} S_0 &= \binom{52}{9} \\ S_1 &= \binom{13}{1} \binom{4}{3} \binom{48}{6} \\ S_2 &= \binom{13}{2} \binom{4}{3}^2 \binom{44}{3} \\ S_3 &= \binom{13}{3} \binom{4}{3}^3 \end{align}$$ Now let $f(x)$ be the generating function of the $S_i$'s, i.e. $$f(x) = S_0 + S_1 x + S_2 x^2 + S_3 x^3$$ The reason we do this is that $f(x-1)$ is the generating function of the number of hands with exactly $i$ of the properties, without over-counting (see the reference below). Supplying numeric values for the $S_i$'s and expanding $f(x-1)$, we find $$f(x-1)=3,057,466,984 + 605,116,512 \; x + 16,473,600 \; x^2 + 18,304 \; x^3$$ This means there are 3,057,466,984 hands with no three-of-a-kinds, 605,116,512 hands with exactly one three-of-a-kind, 16,473,600 hands with exactly two three-of-a-kinds, and 18,304 hands with three three-of-a-kinds.

So the probability of drawing a hand with exactly two three-of-a-kinds is $$16,473,600 / \binom{52}{9} \approx 0.0044776$$

Reference: generatingfunctionology by Herbert S. Wilf, section 4.2.