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.
Best Answer
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.