DRW·Data Scientist·Technical Phone Screen·Intermediate

Intermediate

Jun 2026

Summary

Got a probability/geometry question at DRW for a Data Scientist role. Pretty clean problem but the setup trips you up if you're not careful about what you're actually computing.

Questions Asked (1)

Three points are chosen uniformly at random on the circumference of a unit circle. They divide the circle into three arcs. What is the expected length of the arc that contains the point (1,0)?

Algorithms & Data Structures

Author's notes

My first instinct was to just say 2π/3 because three arcs, circle splits evenly on average, done.

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AI HintsAI Generated

Suggested Approach

Use symmetry and linearity of expectation: fix the point (1,0) and consider the arc containing it. The expected length can be derived by integrating over the positions of the other two points, or by recognizing that the arc length is the minimum of two gaps that cover (1,0). Alternatively, use the known result that the expected length of the arc containing a fixed point is 2/3 of the circumference.

Pro tip: Mention that the answer is independent of the circle's radius and that the same logic applies to any fixed point. Also, note that the expected length is 2/3 of the circumference, which for a unit circle is 4π/3.

1. Understand the problem

Clarify that three random points divide the circle into three arcs, and we need the expected length of the arc that contains the fixed point (1,0).

2. Use symmetry

Recognize that by rotational symmetry, the expected length is the same for any fixed point on the circle, so we can fix (1,0) without loss of generality.

3. Model the arc containing the point

The arc containing (1,0) is bounded by the two nearest points on either side of (1,0) along the circumference. Its length is the sum of the two adjacent gaps.

4. Compute expected length

Use linearity of expectation or integrate the joint distribution of the two nearest points. The expected length of each adjacent gap is 1/3 of the circumference, so the sum is 2/3 of the circumference.

5. Calculate final answer

For a unit circle, circumference is 2π, so expected length = (2/3)*2π = 4π/3.

Key Points to Mention

Symmetry: the expected length is independent of the fixed point's location.
The arc containing (1,0) is the sum of two adjacent gaps between consecutive points.
Each gap between consecutive random points has expected length 1/3 of the circumference.
Linearity of expectation: expected sum of two gaps = sum of expected gaps.
The circumference of a unit circle is 2π, so the expected length is 4π/3.
The result generalizes to any number of points: for n points, expected length is 2/n of the circumference.

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