Day 1 · Race to Ace¶
Edited by Charles
| Study record | Value |
|---|---|
| Question | 2 of 4 |
| Topic | Probability |
| Difficulty | Medium |
| Attempts | 1 |
| Mastery (Notion notation) | ❚ |
| Latest attempt | October 22, 2026 |
| Source | OpenQuant: Race to Ace |
Expected Number of Draws Until the First Ace¶
Problem¶
What is the expected number of cards drawn from a standard 52-card deck until the first ace appears (including the ace)?
Solution: Symmetry + Linearity of Expectation¶
A deck contains 4 aces and 48 non-ace cards.
The 4 aces divide the 48 non-ace cards into 5 gaps:
Gap 1 | A₁ | Gap 2 | A₂ | Gap 3 | A₃ | Gap 4 | A₄ | Gap 5
Let Gᵢ denote the number of non-ace cards in the i-th gap.
Step 1: Apply Symmetry
Since the deck is randomly shuffled, all five gaps have the same expected length:
E[G₁] = E[G₂] = ... = E[G₅]
Step 2: Apply Linearity of Expectation
The total number of non-ace cards is 48:
G₁ + G₂ + G₃ + G₄ + G₅ = 48
Therefore:
E[G₁] = 48/5 = 9.6
Step 3: Include the First Ace
The number of draws includes all cards in the first gap plus the first ace.
E[X] = E[G₁] + 1 = 10.6
General Formula¶
For N cards containing K special cards, the expected number of draws until the first special card is:
E[X] = (N + 1)/(K + 1)
For the first ace:
E[X] = 53/5 = 10.6
Follow-up: Expected Draws Until the r-th Ace¶
To reach the r-th ace, we pass through r gaps and draw r aces.
E[Xᵣ] = r × (N − K)/(K + 1) + r
Therefore:
E[Xᵣ] = r(N + 1)/(K + 1)
For the second ace:
E[X₂] = 2 × 53/5 = 21.2
Key Takeaways¶
- Technique: Symmetry + Linearity of Expectation
- Concept: Order Statistics / Expected Waiting Time
- General Formula: E[Xᵣ] = r(N + 1)/(K + 1)
- Interview Tip: Use the gap argument instead of calculating the full probability distribution.
- Important: Excluding the first ace gives 9.6; including it gives 10.6.