True probability doesn't work like that. We are talking about individual draws in a 52 card pool coming up with one of 4 aces or other cards. So the probablility in a 52 card stack that you come up with and single set card is 4/52. But in probability it never is that simple. Since in this case we are going to treat each event as dependant on the previous event I will attempt here to describe the probability one draws 4 of any one card (Aces or 8s).
Probability (A) = p(A) which is to say that p(A) = 4/52
so the probability one draws a single Ace from a deck of
52 cards is 1/13th or 7.7% roughly.
Here is where it gets a bit more complicated though. Since we now have to find the exact probabilty that a SERIES of events will occur.
Event A is that the first card is an ace. Since 4 of the 52 cards are aces, p(A) = 4/52 = 1/13. Given that the first card is an ace, what is the probability that the second card will be an ace as well? Of the 51 remaining cards, 3 are aces. Therefore, p(B|A) = 3/51 = 1/17 and the probability of A and B is: 1/13 x 1/17 = 1/221. What is the probability the 3rd card in a series is an ace? Given the first two cards drawn were aces at 1/13 x 1/17 and that we have 50 cards remaining in which only two are aces the p(A|B|C) = 2/50 or 1/25. Again, we apply it to the equation and end up with 1/13 x 1/17 x 1/25 = 1/5525. Now finally we get to the probability that 4 cards will be aces. Following along the previous lines, there are now 49 cards left of which only 1 is an ace. Thus, p(A|B|C|D) = 1/49. And the final probability is going to be 1/13 x 1/17 x 1/25 x 1/49.
So the final probability of p(A)p(B)p(C)p(D) given that this final outcome is dependant on the three previous draws, making the variable dependant, is 1/270725.
My question here is, how the hell did you come up with 20%? Mathematically it is far less than 1%.
