Probability of being dealt a pair in poker

Question

If it is assumed that all poker hands are equally likely, what is the probability of being dealt pair?

Here is what I did: First I choose $1$ out of $52$: ${52 \choose 1}$ - First Card

Second I choose $1$ out of $3$ to make a pair - ${3 \choose 2}$

Last I choose $3$ out of $50$ remaining card - ${50 \choose 3}$

So probability is $$ \frac{{52 \choose 1} \cdot {3 \choose 2} \cdot {50 \choose 3}}{{52 \choose 5}} $$

This seems to be wrong? Am I missing something? Please help.

Answer 1

You want to count the number of sets of five cards that are a pair, not the number of lists of five cards. So you don't just pick the first card in 52 ways.

Instead, how many ways are there to choose a pair? $13 \cdot \binom{4}{2}$. And how many ways to pick three other ranks? $\binom{12}{3}$. Then you can choose the suit of those three other ranks in $4^3$ ways.