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Meta·Data Scientist·Technical Phone Screen·Intermediate

Intermediate
May 2026

Summary

Meta data scientist interview with a probability question about sampling. Pretty math-heavy, no fluff, just straight into the problem.

Questions Asked (1)

Q1

You have 1,000 people and each day you draw 10 names. If you sample without replacement (no one gets picked twice until everyone has been chosen), what day do you expect a specific person to be first selected? Then redo the problem assuming sampling with replacement.

Algorithms & Data StructuresProduct Analytics & Metrics
Author's notes

The without-replacement case clicked pretty fast once I thought about it as a uniform distribution over 100 days.

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AI HintsAI Generated

Suggested Approach

Clarify the sampling process and define the random variable for the day a specific person is first selected. For without replacement, use the fact that each person is equally likely to be in any position in a random permutation; for with replacement, model each day's selection as a Bernoulli trial with probability 10/1000.

Pro tip: State your assumptions explicitly (e.g., days are independent, draws are uniform) and connect the result to practical implications like expected wait time and fairness in sampling.

1. Clarify the problem

Confirm that each day 10 distinct names are drawn, and we want the expected day (1-indexed) when a specific person is first selected. For with replacement, each draw is independent and a person can be selected multiple times.

2. Without replacement: use symmetry

In a random permutation of 1000 people, each person is equally likely to be in any position. The specific person's position is uniformly distributed from 1 to 1000, so the expected position is (1000+1)/2 = 500.5. Since selections occur in blocks of 10 per day, the day is ceil(position/10). Compute the expected day by averaging over positions.

3. Compute expected day without replacement

Calculate E[day] = (1/1000) * sum_{k=1}^{1000} ceil(k/10). This sum equals 10*(1+2+...+100) = 10*5050 = 50500, so E[day] = 50500/1000 = 50.5 days.

4. With replacement: geometric distribution

Each day, the probability the person is selected at least once is 1 - (1 - 10/1000)^10? Wait, careful: each day 10 draws with replacement from 1000. The probability a specific person is not drawn in a single draw is 999/1000. Over 10 independent draws, probability not drawn is (999/1000)^10. So probability drawn at least once per day is p = 1 - (999/1000)^10. The number of days until first selection follows a geometric distribution with success probability p, so expected day = 1/p.

5. Compute numerical answer with replacement

Calculate p = 1 - (0.999)^10 ≈ 1 - 0.99004488 = 0.00995512. Then expected day = 1/p ≈ 100.45 days.

Key Points to Mention

  • Symmetry of random permutation for without replacement
  • Uniform distribution of positions and ceiling function for day
  • Geometric distribution for with replacement
  • Probability of at least one selection per day with replacement
  • Expected value calculation and numerical approximation
  • Comparison of results: without replacement ~50.5 days vs with replacement ~100.45 days

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