Given that at least $3$ cards in a hand are already aces, what is the probability that you are dealt all $4$ aces?

Question

I'd really appreciate a correct solution with explaination, but I'd also love to know what's wrong with my solution. I find I learn a lot from figuring out why my logic is wrong.

The correct answer is apparently $\frac{1}{95}$ or about $1\%$.

My answer is $\sim 4\%$, because I reasoned I could find the probability of getting all $4$ aces by finding the probability of not drawing all $4$ aces.

Therefore I thought the solution was $1-(\frac{^{48}C_2}{^{49}C_2})$.

Thank you in advance to anyone who answers!

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EDIT: If it helps, the wording of the original question was "You are dealt $5$ cards at random from a deck of poker cards, and are told at least $3$ of them are aces. What is the probability you have all $4$ aces in the dealt cards?"

Answer 1

The event of "at least three cards are aces" is not the same as the event of "the first three cards are aces". Your solution would work if they were, but they are not.

You should use Bayes' Rule: $$\mathsf P(A=4\mid A\geq 3)=\dfrac{\mathsf P(A=4)}{\mathsf P(A\geq 3)}=\dfrac{\mathsf P(A=4)}{\mathsf P(A=3)+\mathsf P(A=4)}$$

Where $A$ is the count for Aces in the hand.

Answer 2

By Bayes' Theorem, the desired conditional probability is $$\frac{\frac{\binom{4}{4}\binom{48}{1}}{\binom{52}{5}}}{\frac{\binom{4}{3}\binom{48}{2}+\binom{4}{4}\binom{48}{1}}{\binom{52}{5}}}=\frac{\binom{4}{4}\binom{48}{1}}{\binom{4}{3}\binom{48}{2}+\binom{4}{4}\binom{48}{1}}=\frac{1}{95}$$