Given that I have an ace of spades in my 5 card poker hand, what is the probability of having at least one more ace in my hand?

Question

My thought is that it would be $1-\frac{\binom {48}{4}}{\binom {51}{4}}$ because it is as if we are getting 1 minus the probability of having no aces in a 4 card hand for a 51 card deck with only 3 aces. Is this correct?

Answer 1

Another way is to do $$\frac{\text{ways to get ace of spades and at least one more ace}}{\text{ways to get ace of spades}}$$ The numerator is $\binom{51}{4} - \binom{48}{4}$ (first number is how many hands have ace of spades, second number is how many hands have ace of spades and no other aces) and the denominator is $\binom{51}{4}$. This returns the same thing as your answer.