Three-of-a-kind Poker Hand Problem

Question

Problem:

• Three-of-a-kind poker hand: Three cards have one rank and the remaining two cards have two other ranks. e.g. {2♥, 2♠, 2♣, 5♣, K♦}

Calculate the probability of drawing this kind of poker hand.

My confusion: When choosing the three ranks, the explanation used $13 \choose 1$ and $12 \choose 2$. I used $13 \choose 3$ instead which ends up being wrong. I do not know why.

Answer 1

This mistake is made so often by beginners, that I advise that you consider it as reserving bags for chosen ranks from $13$ bags.

So one bag needs to be reserved for the triple, and two for the two singles, hence $\binom{13}1\binom{12}2$.

The order in which we reserve does not matter, so we could as well write $\binom{13}2\binom{11}1$, although the first, of course, will seem the more natural.

Answer 2

Actually, we can use $\binom{13}{3}$: it counts the number of ways of choosing 3 distinct ranks. Just don't forget to also choose which of those three ranks (i.e., $\binom{3}{1}$) is the special rank with 3 cards. It's another way of counting the same thing: $$\binom{13}{3} \binom{3}{1} = \frac{13!}{3!\ 10!} \times 3 = 13 \times \frac{12!}{2!\ 10!} = \binom{13}{1} \binom{12}{2}.$$

Afterwards, we also need to choose suits (hearts, diamonds, clubs, spades) for each rank.

Answer 3

Here is another way to solve it through unordered samples.

We are looking for hands of the kind $x_1$-$x_2$-$x_2$-$y$-$z$, where $x_1,x_2,x_3$ are all of the same face value (although of a different suit), whereas $y,z$ are different face values.

To work with unordered hands, let's fix the order of the cards as above, i.e. three of a kind are the first three cards followed by two other different kinds.

There are 13 possible face values (2, 3, $\ldots$, K, A), and for each face value, there are ${4\choose 3}$ ways to select 3 cards out of 4, disregarding order and without replacement. This fills $x_1$-$x_2$-$x_3$.

For $y$, there are 48 possibilities, since three face values have already been drawn and the one left cannot be used. For $z$, there are 44 possibilities since the other three remaining cards of the face values chosen in $y$ cannot be either.

However, we are not done yet, i.e. $13{4\choose 3}48\cdot44$ is not quite right because this number includes also poker hands s.t. 4s-4c-4h-2s-3h and 4s-4c-4h-3h-2s, which are obviously indistinguishable since order doesn't matter. But the last two cards can be ordered in 2! ways. Dividing by $2!$, we remove those hands that differ only in the ordering in the last two cards.

The right number of poker hands with a three-of-a-kind is thus

$$13{4\choose 3}\frac{48\cdot44}{2!},$$

and the required probability is

$$ \frac{13 {4\choose 3}\frac{48\cdot44}{2!}}{{52 \choose 5}}. $$