This is a follow up to
Probability of poker pair - what's wrong with this?
The aim is to compute the probability of a pair in poker.
Consider a solution that initially counts the number of ways of picking a pair followed, in that order, by three cards that still conform to the fact that the pair will exist and be unique.
As under the original problem the pair does not need to be the first two cards to be picked, we need to multiply your answer by $C(5,2)=10$.
How to formalize the above argument using conditional probabilities?
$$P(\textrm{one pair}) = \sum P(\textrm{one pair}|\textrm{first card})P(\textrm{first card}) = P(\textrm{one pair}|\textrm{first card})$$
Now, how to continue the formal derivation?
From the above post we know that the answer is
$$ P(\textrm{one pair}|\textrm{first card}) = C(5,2) \frac{3}{51} \cdot \frac{48}{50} \cdot \frac{44}{49} \cdot \frac{40}{48} $$
Clearly, this is also
$$ P(\textrm{one pair}|\textrm{first card}) = C(5,2) \cdot $$
$$ \cdot P(\textrm{2nd card forms pair with 1st card}|\textrm{first card}) $$
$$ \cdot P(\textrm{3nd card does not pair with 1st and 2nd card}|\textrm{first 2 cards}) $$
$$ \cdot P(\textrm{4th card does not pair with 1st, 2nd, 3rd card}|\textrm{first 3 cards}) $$
$$ \cdot P(\textrm{5th card does not pair with 1st, 2nd, 3rd and 4th card}|\textrm{all other cards}) $$
However, I was wondering if there is a more elegant way to write the above expression, making more explicit the symmetry in the problem and why $C(5,2)$ appears in the solution.
First, let's choose which rank our pair will be. There are $13$ ranks, and we will choose $1$. Now let's choose which suits to select for our pair. Each rank has $4$ suits, of which we will choose $2$.
This gives us $13\choose1$$4\choose2$
Now, let's choose the non-matching cards. Out of the $12$ remaining ranks, we want to choose $3$ distinct ranks. From each of the $3$ ranks, we will pick $1$ card.
This gives us $12\choose3$${4\choose1}^3$
Putting everything together with the total number of ways to choose a $5$ card poker hand from a $52$ card deck we get: $\frac{{13\choose1}{4\choose2}{12\choose3}{4\choose1}^3}{52\choose5}$