# If cards are drawn one at a time from a shuffled deck, with replacement. Find the probability an ace appears before the first face card?

Relevance

INTUITIVE METHOD:

Imagine we threw away all the other cards and just had 4 aces and 12 face cards. We can do this because if we draw any other card we just ignore that draw.

In our newly reduced deck, draw a card. If it is an ace, we win. If it is a face card, we lose.

There are 4 aces out of 16 total cards.

P(ace before face card) = 1/4

It can be found with a more rigorous recursive probability method, but the intuitive method is just too obvious to ignore.

MATHEMATICAL METHOD:

If you draw an Ace, the probability of winning is 1.

If you draw a Face card, the probability of winning is 0.

If you draw any other card, the probability of winning is just the same as before.

Let A be the probability of drawing an Ace (4/52 = 1/13)

Let F be the probability of drawing a Face card (12/52 = 3/13)

Let N be the probability of drawing any other card (36/52 = 9/13)

Let W be the event that we win:

P(W) = P(A) * 1 + P(F) * 0 + P(N) * P(W)

P(W) = 1/13 * 1 + 3/13 * 0 + 9/13 * P(W)

P(W) = 1/13 + 9/13 * P(W)

P(W) - 9/13 * P(W) = 1/13

(1 - 9/13) P(W) = 1/13

4/13 * P(W) = 1/13

P(W) = 1/13 * 13/4

P(W) = 1/4