Problem:
What is the probability of getting exactly one pair in poker?
Answer:
Let $p$ be the probability we seek.
\begin{align*}
p &= \dfrac{ 13 {{4} \choose {2}} (48)(44)(40)}{ {{52} \choose {5}} } \\
{4} \choose {2} &= \dfrac{ 4(3) }{ 2} = 6 \\
{52} \choose {5} &= \dfrac{ 52(51)(50)(49)(48) }{ 5(4)(3)(2) } \\
{52} \choose {5} &= \dfrac{ 52(51)(5)(49)(48) }{ 4(3) } \\
{52} \choose {5} &= 52(51)(5)(49)(4) \\
\end{align*}
\begin{align*}
p &= \dfrac{ 13(6)(48)(44)(40) }{ 52(51)(5)(49)(4) } \\
p &= \dfrac{ 6(48)(44)(40) }{ 4(51)(5)(49)(4) } \\
p &= \dfrac{ 3(48)(44)(40) }{ 2(51)(5)(49)(4) } \\
p &= \dfrac{ 3(48)(44)(10) }{ 2(51)(5)(49) } \\
p &= \dfrac{ 48(44)(10) }{ 2(17)(5)(49) } \\
p &= \dfrac{ 48(44)(2) }{ 2(17)(49) } \\
p &= \dfrac{ 48(22)(2) }{ 17(49) } \\
p &= \dfrac{ 2112 }{ 833 } \\
p &\doteq fixme
\end{align*}
I believe the correct answer is: $ .422569 $.
I suspect this line is wrong:
$$ p = \dfrac{ 13 {{4} \choose {2}} (48)(44)(40)}{ {{52} \choose {5}} } $$
Here is why I think it is right. The first term, $13$ is there because there are
$13$ different ranks you can have. The next term is there because to have a pair, you select two cards of the same rank from four cards. Now for the last
three terms of the numerator. After selecting two of the same rank, such as a pair
of aces, there are $48$ cards left to be selected. After selecting another card,
such as a jack there are $44$ cards left to be selected.
In terms of the denominator, it is there because a poker hand is $5$ cards from a deck of $52$ cards.
Where did I go wrong?
You counted each hand $3!$ times, because the order in which the three singletons are selected does not matter. For instance, if your hand is $$5\clubsuit, \color{red}{5\diamondsuit}, \color{red}{8\diamondsuit}, \color{red}{9\heartsuit}, J\spadesuit$$ it does not matter whether you first select $\color{red}{8\diamondsuit}$, then $\color{red}{9\heartsuit}$, then $J\spadesuit$, or first select $J\spadesuit$, then $\color{red}{8\diamondsuit}$, and then $\color{red}{9\heartsuit}$.
There are $13$ ways to select the rank of the pair and $\binom{4}{2}$ ways to select two cards of that rank. There are $\binom{12}{3}$ ways to select three ranks from which one card each will be drawn and $4$ ways to select a card from each of those ranks. Since there are $\binom{52}{5}$ possible five-card hands, the probability of selecting exactly one pair is $$\frac{\dbinom{13}{1}\dbinom{4}{2}\dbinom{12}{3}\dbinom{4}{1}\dbinom{4}{1}\dbinom{4}{1}}{\dbinom{52}{5}}$$