Probability of getting an ace in the 2nd draw without knowing the first draw The question is: Given a 52-card deck, two cards are drawn without replacement. What is the probability that the second card is an ace?
So I think I already found the answer (correct me if my answer is incorrect), but I still have a question.
Here's my answer: $$\frac{48}{52}\cdot\frac{4}{51}+\frac{4}{52}\cdot\frac{3}{51}=0.07692$$
So in my answer, I'm just adding the cases. The first case is where the first draw isn't an ace and the second case is where the first draw is an ace. Now to my question. How come the expression below seem to overcount, despite seemingly doing the same thing? $$\frac{\binom{48}{1}\binom{4}{1}}{\binom{52}{2}} + \frac{\binom{4}{2}}{\binom{52}{2}} = 0.1493213$$
My guess is that the first term is double counting, but I really can't grasp why.
 A: You can avoid doing any work here with a symmetry argument. If you aren't given any information about the first draw then it doesn't matter what order you draw the cards in, or said another way it's the same as drawing the cards simultaneously. So the probability the second card is an ace is the same as the probability the first card is an ace, which is $\frac{4}{52} = \frac{1}{13}$.
(This argument generalizes - you can draw any fixed number of cards, up to and including the entire deck, and if you aren't given any information about any of the draws the probability that the last card you drew is an ace is still the same $\frac{1}{13}$.)
Your first calculation actually simplifies to this:
$$\frac{48}{52} \cdot \frac{4}{51} + \frac{4}{52} \cdot \frac{3}{51} = \frac{4}{52} \cdot \left( \frac{48}{51} + \frac{3}{51} \right) = \frac{4}{52}.$$
Empy's explanation of the problem with your second calculation is correct. If you just expand out the binomial coefficients you'll see that you've introduced an additional factor of $2$ in the first term but not in the second, and this corresponds exactly to overcounting cases where the first card is an ace. The ${52 \choose 2}$ denominator means you are choosing a set of cards and ignoring order which is only appropriate if the cards can be treated as indistinguishable, as in your second term where the cards are just "two aces."
A: In your first calculation, $4/51$ should be $3/51$ because there are only theee aces left.
In the second calculation, since the denominator is $52\choose2$, the order doesn't matter.  You are including an ace first as a win, not just an ace second.
