Probability Poker Question

Question

You are dealt $20$ cards. What is the probability you have all kings given that you hold at least one king?

So I set it up like $$ P(4\textrm{ Kings | at least one king}) =\frac{{4 \choose 4}{48 \choose 9}}{{52 \choose 20}-{48 \choose 20}} $$ Does this setup look correct?

Answer 1

$P(4 \text{ Kings | At least one King})=\frac{P(4 \text{ Kings and at least one King})}{P(\text{At least one King})}$

Now $P(4 \text{ Kings})=\frac{\binom{4}{4}\binom{48}{16}}{\binom{52}{20}}$ and $P(\text{At least one King})=1-P(\text{no King})=1-\frac{\binom{48}{20}}{\binom{52}{20}}$

So your final answer is $$\frac{\binom{4}{4}\binom{48}{16}}{\binom{52}{20}-\binom{48}{20}}$$