Lol.

Your calculation,

P(4 aces in top cards picked) = (4/52)^4 * (1 - 4/52)^(52-4) * 52!/(4!(52 - 4)!) = 0.203317

is entirely correct. That is, the left hand side is correct. (I haven't evaluated to confirm the r h s. I'll just take it at face value, doesn't matter.)

The flaws are that you require that there are 4 aces among the 52 top cards and assign

*that* as the probability of picking an A (two errors, the requirement and the assignment). Look again!

Please pick 52 top cards from 52 shuffled decks, leaving the values unseen. Shuffle the 52 cards, you think your chance of an ace is 4/52. This is incorrect, there is only a P=0.203317 that your 52 random's contain 4 aces, a P=0.8 approx there is not 4 aces.

There can be anything between 0 and 52 aces in the top cards you picked, each with a probability p(i) (with p(0) + p(1) + p(2) + ... + p(52) = 1). But this is irrelevant. What is relevant is that you take one card from a shuffled 52-deck shoe. The intermediate shuffling (pick top cards, blah, blah,) is just more shuffling. The probability that it is an ace is (4*52)/(52*52) = number of A/total number of cards = 1/13.