What is the probability of drawing a card from an ordinary deck of cards and getting an ace?

The probability of drawing a diamond-faced card from a pack of 52 playing cards is easy to determine. Since there are 13 diamond-faced cards in the deck, the probability becomes 13/52 = 1/4 = 0.25.

The probability of drawing an ace from a pack of 52 playing cards is also easy to determine. There are 4 aces in the deck of 52 cards; thus, the probability becomes 4/52 = 1/13 = 0.076923. This represents a much lower probability than drawing a card in a specific suit, illustrated in the preceding example.

You can figure this out without any knowledge of conditional probabilities:

There are 52 choices for the first pick and 51 choice for the second pick. So, in total, there are $52 \times 51 = 2652$ possible "hands" that consist of two cards.

Now, the question is, how many of these 2652 contain a single ace. Let's do some counting of the possible one-ace hands:

There are $4 \times 48 = 192$ hands that have an ace as their first pick and a non-ace as the second pick (because there are 4 aces and 48 non-aces). Similarly, there are $48 \times 4 = 192$ hands that have a non-ace as their first pick and an ace as the second pick. So the total number of hands that have a single ace is $192+192=384$.

So, the probability of a one-ace hand is $384/2652$. Dividing top and bottom by $12$ gives us $384/2652 = 32/221$.