Two cards drawn without replacement — probability both are aces

Question

Two cards are drawn at random from a standard deck of 52 cards, one after another, without replacement. Find the probability that both cards are aces.

Solution — Step by Step

A standard deck has 52 cards, of which 4 are aces (ace of spades, hearts, diamonds, clubs).

We draw two cards without replacement — meaning the first card is not returned to the deck before the second is drawn.

Let event $A_1$ = first card is an ace, and $A_2$ = second card is an ace.

We want $P(A_1 \cap A_2) = P(A_1) \times P(A_2 | A_1)$ .

$P(A_1)$ : 4 aces out of 52 cards:

P(A_1) = \frac{4}{52} = \frac{1}{13}

$P(A_2 | A_1)$ : Given the first card was an ace, there are now 3 aces left in 51 remaining cards:

P(A_2 | A_1) = \frac{3}{51} = \frac{1}{17}

Combined probability:

P(A_1 \cap A_2) = \frac{1}{13} \times \frac{1}{17} = \frac{1}{221}

Total ways to choose 2 cards from 52:

\binom{52}{2} = \frac{52!}{2! \cdot 50!} = \frac{52 \times 51}{2} = 1326

Favourable ways to choose 2 aces from 4:

\binom{4}{2} = \frac{4!}{2! \cdot 2!} = 6

P(\text{both aces}) = \frac{6}{1326} = \frac{1}{221}

Both methods give $\boxed{\dfrac{1}{221}}$ .

Why This Works

Without replacement means the events are dependent — what happens on the first draw affects the probabilities for the second draw. This is why we use conditional probability $P(A_2 | A_1)$ .

After drawing one ace, there are only 3 aces left in 51 cards (not 4 in 52). If we incorrectly assumed independence (as in drawing with replacement), we’d get:

P_{\text{wrong}} = \frac{4}{52} \times \frac{4}{52} = \frac{1}{169}

This is larger than the correct answer $1/221$ — because drawing with replacement gives more chances of getting two aces than drawing without.

The combinations method automatically handles dependence because it counts pairs directly.

Alternative Method

For “without replacement” problems, the combinations method is often cleaner:

P = \frac{\binom{k}{r}}{\binom{n}{r}}

where $n$ = total cards, $k$ = number of favourable cards, $r$ = number drawn.

P = \frac{\binom{4}{2}}{\binom{52}{2}} = \frac{6}{1326} = \frac{1}{221}

Common Mistake

The most common error is ignoring “without replacement” and treating the two draws as independent:

P_{\text{incorrect}} = \frac{4}{52} \times \frac{4}{52} = \frac{16}{2704} = \frac{1}{169}

This is wrong. After drawing the first ace, only 51 cards remain and only 3 are aces. Always ask: “Is this with or without replacement?” — and adjust the second draw’s denominator and numerator accordingly.

The answer $1/221$ can be remembered as roughly a 0.45% chance — less than 1 in 200. This makes intuitive sense: drawing one ace is about 1 in 13; drawing another ace from the remaining deck is about 1 in 17. So the combined chance is about 1 in 221.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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