Getting exactly one pair in a poker hand

Question

I am not understanding this problem:

In a deck of 52 cards, of 13 ranks, and 4 suits, how many different 5 card hand can we get such that, there is always exactly one pair.

There is a similar example here . But I don't understand the logic behind it.

My thoughts are:

assume we have the hand ${a_{1},a_{2},b,c,d}$ where $a_{1}$ and $a_{2}$ are two cards of the same rank and $b,c,d$ are distinct cards from each other and $a$.

So my logic:

there are $\binom{52}{1}$ ways to pick $a_{1}$ and $ \binom{3}{1}$ ways to pick $a_{2}$.

and then $ \binom{50}{1}$ for $b$, $\binom{49}{1}$ for $c$, and $ \binom{48}{1}$ for $d$

which gives us $\binom{52}{1}\binom{3}{1}\binom{50}{1}\binom{49}{1}\binom{48}{1}$ = $18345600$ ways to get a pair...

...which is the wrong answer.

Where have I gone wrong?

Answer 1

There are $\binom{13}{1}$ ways to pick the kind we have a pair in. For each such way, we have $\binom{4}{2}$ ways to pick the actual cards.

For each of these ways, there are $\binom{12}{3}$ ways to pick the kinds we will have one each of. For each of these kinds, there are $\binom{4}{1}$ ways to pick the actual cards, for a total of $\binom{13}{1}\binom{4}{2}\binom{12}{3}\binom{4}{1}^3$ ways.

Remark: One of common problems in counting is inadvertent double (or multiple) counting. For example, your $\binom{52}{1}\binom{3}{1}$ includes the Ace of $\spadesuit$ together with the Ace of $\heartsuit$. But it also counts the Ace of $\heartsuit$ together with the Ace of $\spadesuit$. These give the same one-pair hand.

Answer 2

Hint: When you pick $b,c,d$ you need to make sure they don't match the pair, so there are less than $50$ choices for $b$.

Answer 3

there are $52\choose 1$ ways to pick $a_1$ and $3\choose 1$ ways to pick $a_2$ .

So far so good, but you have over counted.   For example: you have counted picking a diamond in for $a_1$ and a heart for $a_2$ as well as picking a heart for $a_1$ and a diamond for $a_2$, but both are the same selection.   So, as order does not matter, you need to divide the count by the $2!$ ways to rearrange those two cards.

Alternatively I'd suggest counting ${13\choose 1}{4\choose 2}$ ways to pick the one face and two suits of the paired cards.

and then $50\choose 1$ for b , $49\choose 1$ for c , and $48\choose 1$ for d

You have to avoid picking cards that have the same face as a, or each other. You want only the one pair. So by your way you need to count $48\choose 1$ ways for b, $44\choose 1$ for c, and then $40 \choose 1$ for d, and likewise deal with overcounting the arrangement of three singletons by dividing by $3!$.

I'd count the way to pick three different faces and a suit for each as: ${12\choose 3}{4\choose 1}^3$

So your result should be $$\frac 1 {2! 3!}{52\choose 1}{3\choose 1}{48\choose 1}{44\choose 1}{40\choose 1} = {13\choose 1}{4\choose 2}{12\choose 3}{4\choose 1}^3 $$