Drawing 2 cards from a shuffled deck, what is the probability the second one is an Ace?

There are $52\times 51$ outcomes when you draw 2 cards one after another without returning the first drawn card to the deck.

There are $48\times 4$ outcomes when the first card is not Ace and the second card is Ace. There are $4\times 3$ outcomes when both cards are Ace. The total outcomes are $48\times 4 + 4 \times 3 = 51\times 4$.

The probability is $\frac{51\times 4}{52\times 51}=\frac{51}{52}\times\frac{4}{51}=\frac{4}{52}=\frac{1}{13}$.

For any given position $n$ in the deck (here $n=2$, meaning the second card,) if the deck is shuffled then the probability that the card in position $n$ is an ace is $4/52 = 1/13$, because $4$ out of the $52$ cards in the deck are aces.

Break the answer into two scenarios:

1. If the first card is an ace and the second card is also an ace

The probability, in this case, is $$\dfrac{4}{52}\dfrac{3}{51}.$$

2. If the first card is not an ace but the second card is an ace

   The probability, in this case, is $$\dfrac{48}{52}\dfrac{4}{51}.$$

Since these two scenarios are mutually exclusive, the final probability is the sum of these two probabilities: $$\dfrac{4}{52}\dfrac{3}{51} + \dfrac{48}{52}\dfrac{4}{51}.$$