Citadel·Data Scientist·Technical Phone Screen·Intermediate

Intermediate

Apr 2026

Summary

First-round technical screen for a Data Scientist role at Citadel. One probability puzzle, conditional on card draws, and it was more involved than it looked.

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 that the third card is also an Ace?

Algorithms & Data Structures

Author's notes

I set it up correctly at first, conditioning on how many Aces could be in the first two draws (exactly one or exactly two) and weighting each case.

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AI HintsAI Generated

Suggested Approach

Use conditional probability and symmetry. Since the condition only involves the first two cards, the third card is equally likely to be any of the remaining 50 cards, so the probability is simply the expected number of Aces left divided by 50.

Pro tip: Leverage linearity of expectation: the expected number of Aces in the first two cards given at least one is 1 + 3/5, so the expected number of Aces left is 4 minus that, divided by 50. This avoids messy casework and shows strong probabilistic intuition.

1. Define the event and condition

Let A be the event that the third card is an Ace, and B be the event that at least one of the first two cards is an Ace. We need P(A|B).

2. Use symmetry to simplify

Given B, the third card is equally likely to be any of the remaining 50 cards. So P(A|B) = E[number of Aces left | B] / 50.

3. Compute expected Aces in first two given B

Let X be the number of Aces in the first two cards. Given B, X can be 1 or 2. By symmetry and the definition of conditional expectation, E[X|B] = 1 + P(both Aces | B). Compute P(both Aces | B) = (4/52 * 3/51) / P(B).

4. Calculate P(B) and E[X|B]

P(B) = 1 - P(no Aces in first two) = 1 - (48/52 * 47/51) = 1 - (2256/2652) = 396/2652 = 33/221. Then P(both Aces | B) = (12/2652) / (396/2652) = 12/396 = 1/33. So E[X|B] = 1 + 1/33 = 34/33.

5. Compute final probability

Expected Aces left = 4 - E[X|B] = 4 - 34/33 = (132-34)/33 = 98/33. Then P(A|B) = (98/33) / 50 = 98/1650 = 49/825.

Key Points to Mention

Conditional probability formula: P(A|B) = P(A ∩ B) / P(B).
Symmetry: given the condition, the third card is uniformly distributed among the remaining 50 cards.
Linearity of expectation: E[number of Aces left | B] = 4 - E[number of Aces in first two | B].
Complement rule: P(at least one Ace) = 1 - P(no Aces).
Avoid double-counting by using indicator variables or direct counting.
Final answer should be simplified fraction: 49/825.

AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.