Meta·Data Scientist·Technical Phone Screen·Senior

Senior

Jun 2026

Summary

Meta Data Scientist interview with a probability/statistics heavy question about ad insertion strategies. Single question but it went pretty deep into distribution math and system risk.

Questions Asked (1)

Two teams are debating ad insertion strategies: one wants a 4% random replacement rate, the other wants a fixed every-25th-slot approach. Walk through the expected ad count, variance, and probability of hitting more than double the expected load under each method.

A/B Testing & ExperimentationProduct Analytics & MetricsTechnical Trade-offs

Author's notes

This one is basically a binomial vs deterministic comparison dressed up as a product conflict.

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AI HintsAI Generated

Suggested Approach

Compare the two strategies by modeling each as a random process: the 4% random replacement as a Binomial distribution and the every-25th-slot as a deterministic pattern with periodic randomness. Calculate expected ad count, variance, and tail probability (P(X > 2*E[X])) for each, then discuss the practical implications for ad load stability and user experience.

Pro tip: Emphasize that while both methods have the same expected ad count, the random method has higher variance and a non-negligible probability of exceeding double the expected load, which could degrade user experience. The fixed method provides predictable ad load, which is often preferred in production systems.

1. Define the models and assumptions

Clarify that the 4% random replacement means each slot independently has a 4% chance of being an ad, modeled as Binomial(n, p=0.04). The every-25th-slot approach places an ad deterministically every 25 slots, so for n slots, the number of ads is floor(n/25) or ceil(n/25) depending on alignment.

2. Compute expected ad count

For random: E[X] = n * 0.04. For fixed: E[X] = n / 25 = n * 0.04. Thus, both have the same expected ad count, which is a key insight.

3. Compute variance

For random: Var(X) = n * 0.04 * 0.96 = 0.0384n. For fixed: Var(X) is 0 if n is a multiple of 25, otherwise it's at most 0.25 (due to rounding). So the random method has significantly higher variance.

4. Calculate probability of exceeding double the expected load

For random, use normal approximation or exact binomial to find P(X > 2 * 0.04n) = P(X > 0.08n). For large n, this probability is extremely small but non-zero. For fixed, the probability is 0 because the ad count never exceeds ceil(n/25) which is at most n/25 + 1, far below 2*(n/25) for large n.

5. Interpret results and discuss trade-offs

Conclude that the random method introduces variability and a small risk of ad overload, while the fixed method ensures a consistent ad load. Discuss how this impacts user experience, system load, and A/B testing power.

Key Points to Mention

Binomial distribution for random replacement and deterministic pattern for fixed interval.
Equal expected ad count (4% of slots) for both methods.
Variance: random has variance n*p*(1-p), fixed has near-zero variance.
Probability of exceeding double expected load: for random, use tail probability (e.g., Chebyshev or normal approximation); for fixed, it's zero.
Practical implications: random may cause unpredictable user experience and system load spikes; fixed provides predictability but may introduce periodicity artifacts.
Consideration of sample size and time horizon: for small n, the random method's tail risk is more pronounced.

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