Citadel·Data Scientist·Technical Phone Screen·Senior

Senior

May 2026

Summary

Citadel data scientist interview with a conditional probability question that had a generalization twist. Pretty math-heavy and they clearly wanted to see your reasoning process, not just the answer.

Questions Asked (1)

You draw three cards without replacement from a standard 52-card deck. Given that at least one of the first two cards is an Ace, what is the probability the third card is also an Ace? Then generalize to a deck of N cards with A Aces, where the first m cards contain at least one Ace.

Algorithms & Data StructuresProduct Analytics & Metrics

Author's notes

The conditional probability part was manageable once I set up the sample space carefully.

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AI HintsAI Generated

Suggested Approach

First, solve the specific case by enumerating the possible Ace counts among the first two cards and using conditional probability. Then, generalize using combinatorics and the hypergeometric distribution, expressing the answer in terms of N, A, and m.

Pro tip: Emphasize that the condition 'at least one Ace in the first m cards' changes the distribution of Aces remaining in the deck, so you must account for the number of Aces already seen. A common mistake is to ignore this and assume independence.

1. Define the events and notation

Let E be the event that the third card is an Ace, and C be the condition that at least one of the first m cards is an Ace. For the specific case, m=2, N=52, A=4.

2. Compute the conditional probability for the specific case

Use P(E|C) = P(E ∩ C) / P(C). Compute P(C) = 1 - P(no Ace in first two) and P(E ∩ C) by summing over the number of Aces in the first two cards (1 or 2).

3. Generalize using combinatorics

For general N, A, m, compute P(C) = 1 - C(N-A, m)/C(N, m). Compute P(E ∩ C) by summing over k=1 to min(m,A) the probability of exactly k Aces in first m and an Ace third.

4. Simplify the expression

Use combinatorial identities to simplify the sum, possibly expressing the result in terms of hypergeometric probabilities or binomial coefficients.

5. Verify with edge cases

Check that the formula yields correct results for trivial cases (e.g., A=0, A=N, m=0) and that the probability is between 0 and 1.

Key Points to Mention

Conditional probability formula: P(A|B) = P(A ∩ B) / P(B)
Hypergeometric distribution for sampling without replacement
Complement rule: P(at least one Ace) = 1 - P(no Ace)
Law of total probability: summing over possible numbers of Aces in the first m cards
Combinatorial identities: Vandermonde's identity or binomial coefficient manipulations
Edge cases: A=0, A=N, m=0, m=N, etc.

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