The question is:
If I pick 4 cards out of a 52 cards deck and replace them, what is the possibility that I will get all 4 by picking 8 cards again?
I just used the hypergeometric formula, it's a little unintuitive, especially why the numerator is as such. I am not sure if my answer is right, as well.
My answer: $$\frac{(\binom{4}{4})(\binom{48}{4})}{(\binom{52}{8})}=0.535$$
The reasoning is that the numerator is all the desired possibilities, for all 4 picks, I want to get all those cards that were picked beforehand, thus 4 choose 4; out of the remaining 48 cards that do not contained the afore-picked cards, comes the other 4 cards out of 8. The denominator is all the possbilities.
EDIT: Ok I think I discovered a flaw in this formula. Because say you have 5 balls, and you picked 2 of them on the first round. Say you want to pick 4. The total number of possible scenarios is 5 choose 4. But the total number of favorable outcomes is 4 choose 2=6. There are 6 ways to place those 2 previously picked balls into 4 slots. But the hypergeometric formula has the numerator 2 choose 2 times 3 choose 2=3. Why?
Your answer is correct. However, the first part of the question has no bearing on the second part other than knowing which four cards to be included in the next $8$.
For the numerator, think of it as deliberately choosing the four special cards (in one way). Then the remaining can be chosen in $\binom{48}{4}$ ways.