Two Aces from Deck (No Replacement)
What is the probability of drawing two Aces in a row from a standard 52-card deck if the cards are drawn without replacement?
Related Concepts
Hint
There are two common ways to approach this:
- Sequential Probability:
- What's the probability the first card is an Ace?
- Given the first was an Ace (and not replaced), how many Aces are left? How many total cards are left? What's the probability the second card is an Ace?
- Multiply these probabilities.
- Combinations:
- How many ways can you choose 2 Aces from the 4 Aces available in the deck? (Order doesn't matter here for the group of two Aces).
- How many ways can you choose any 2 cards from the 52 cards in the deck?
- Divide the number of ways to get two Aces by the total number of ways to get two cards.
Explanation: Probability of Drawing Two Aces
Imagine you're picking two cards from a shuffled deck, one after the other, without putting the first one back.
We want to find the chance that both cards are Aces.
- First Card: There are 4 Aces in a deck of 52 cards. So, the chance of the first card being an Ace is 4 out of 52.
- Second Card (Important!): Let's say your first card was an Ace. You didn't put it back. Now, there are only 3 Aces left, and only 51 total cards left in the deck. So, the chance of the second card also being an Ace (after you already picked one Ace) is 3 out of 51.
To get the probability of both these things happening in order, we multiply these chances together.
A standard deck has 52 cards, with 4 Aces. We are drawing two cards without replacement, meaning the first card drawn is not put back into the deck before the second card is drawn. This makes the events dependent.
Method 1: Sequential Probability (Conditional Probability)
Let A₁ be the event that the first card drawn is an Ace, and A₂ be the event that the second card drawn is an Ace.
- Probability of the first card being an Ace:
There are 4 Aces in 52 cards.
P(A₁) = 4/52 - Probability of the second card being an Ace, given the first was an Ace (P(A₂|A₁)):
If the first card drawn was an Ace and not replaced, there are now 3 Aces left in the deck, and a total of 51 cards remaining.
P(A₂|A₁) = 3/51 - Probability of both events happening (drawing two Aces in a row):
We use the multiplication rule for dependent events:P(A₁ and A₂) = P(A₁) × P(A₂|A₁)
P(two Aces) = (4/52) × (3/51)
P(two Aces) = 12 / 2652
Simplifying the fraction:12 / 2652 = 1 / 221
Method 2: Using Combinations
This method considers the problem as choosing a set of 2 cards where order doesn't matter, and then finding the proportion of those sets that consist of 2 Aces.
- Number of ways to choose 2 Aces from the 4 Aces available:
This is a combination problem, denoted as C(n, k) or "n choose k". Here, n=4 (total Aces), k=2 (Aces to choose).
C(4, 2) = 4! / (2! × (4-2)!) = 4! / (2! × 2!) = (4 × 3 × 2 × 1) / ((2 × 1) × (2 × 1)) = 24 / 4 = 6
There are 6 ways to choose 2 Aces from 4. - Total number of ways to choose any 2 cards from the 52 cards in the deck:
Here, n=52 (total cards), k=2 (cards to choose).
C(52, 2) = 52! / (2! × (52-2)!) = 52! / (2! × 50!) = (52 × 51) / (2 × 1) = 2652 / 2 = 1326
There are 1,326 ways to choose 2 cards from a 52-card deck. - Probability of drawing two Aces:
P(two Aces) = (Ways to choose 2 Aces) / (Total ways to choose 2 cards)
P(two Aces) = C(4, 2) / C(52, 2) = 6 / 1326
Simplifying the fraction:6 / 1326 = 1 / 221
Final Result
Both methods yield the same probability:
P(drawing two Aces without replacement) = 1/221
As a decimal, this is approximately:
1/221 ≈ 0.00452488... or about 0.45%
Why both methods work: The sequential method explicitly considers the order of draws and how probabilities change. The combinations method implicitly accounts for order by considering all possible unordered pairs, ensuring that the "favorable" pairs are counted in the same "unordered" way as the total pairs. Both are valid ways to solve problems involving sampling without replacement.
Extend Your Thinking: What's the probability of drawing three Kings in a row without replacement from a standard deck? Try solving it using both methods!