First card an ace, will the second be an ace too?

Question

You draw two cards one after another from a well-shuffled standard $52$ -card deck, without putting the first card back. The first card you drew is an ace.

Given that, what is the probability the second card is also an ace?

Accepted Answer

The clue tells you one ace has already been removed and set aside. That leaves: - 52 - 1 = 51 cards remaining in the deck, - 4 - 1 = 3 aces still among them. Since the remaining cards are equally likely to be drawn next: P(	ext{second ace} \mid 	ext{first ace}) = \frac{3}{51} = \frac{1}{17} \approx 0.0588. Without replacement, taking one ace out lowers both the count of aces and the size of the deck, so the second draw is conditioned on the smaller, ace-depleted deck. Takeaway. For draws without replacement, condition by literally updating the counts: subtract what has already been removed …

First card an ace, will the second be an ace too?

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