Oct 08, 2026 by Offer

SIG Expected Value Interview 2027: Two Points on a Circle

The expected distance is 4/π, or about 1.273. Fix one point; with angular separation θ uniform on [0, 2π), distance = 2 sin(θ/2), and integrating gives 4/π. This "sig expected distance two points circle" question is commonly reported by candidates as a test of setting up a clean integral.

SIG Expected Distance Two Points Circle: What This Question Assesses

The question tests whether you reduce a two-variable problem to one variable using symmetry. Fixing one point and parameterizing by the angle between the points is the key move — candidates who try to work in x-y coordinates drown. SIG wants the instinct to exploit symmetry before computing.

SIG Expected Distance Two Points Circle: How to Answer

Use symmetry to fix one point. By rotation symmetry, fix the first point at angle 0. The second point's angle θ is uniform on [0, 2π).
Write distance as a function of θ. The chord length for a unit circle is 2 sin(θ/2).
Integrate. E = (1/2π) ∫₀²π 2 sin(θ/2) dθ. Substitute u = θ/2: the integral equals 8/2π... carefully: ∫₀²π sin(θ/2) dθ = [−2cos(θ/2)]₀²π = 4, so E = (1/2π) × 2 × 4 = 4/π ≈ 1.273.
Sanity-check. The maximum distance is 2 (diameter) and the minimum is 0; 1.273 sits plausibly between, closer to the upper-middle since small angles are relatively unlikely to produce tiny distances... actually just confirm it is between 0 and 2.

Sample line: "Fixing one point by symmetry, the distance is 2 sin(θ/2) with θ uniform, and the integral gives 4/π — about 1.27."

Common Mistakes

Setting up a double integral over both points' coordinates instead of fixing one by symmetry.
Using arc length instead of chord length — the question asks for distance, i.e. the straight-line chord.
Forgetting the 1/2π normalization, which turns the integral into an expectation.

If the symmetry step was not your first move, practice geometric-probability problems — SIG's harder rounds chain exactly this skill into multi-stage setups. Question formats may vary by role and region; check SIG's official careers page.

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FAQ

What is the expected distance between two random points on a unit circle? 4/π ≈ 1.273 — derived by integrating the chord length 2 sin(θ/2) over a uniform angle.

Why can you fix one point? Rotation symmetry: the distribution of the distance depends only on the relative angle, so fixing one point loses no generality.

Is this a real SIG interview question? Geometric expected-value problems are commonly reported by candidates in SIG interviews, though exact problems may vary by role and region.

What is the chord length formula? For a circle of radius R and central angle θ, chord length = 2R sin(θ/2). Here R = 1.

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