Solution
Socially Distanced Points
Show the problem again
Three points are chosen independently and uniformly at random on the interval [0, 1]. What is the probability that every two of them are at least 1/4 apart? Give a reduced fraction.
Worked solution
The answer is 1/8. Sort the points as x₁ < x₂ < x₃. The condition is x₂ ≥ x₁ + 1/4 and x₃ ≥ x₂ + 1/4. Substitute y₁ = x₁, y₂ = x₂ − 1/4, y₃ = x₃ − 1/2. This is a volume-preserving bijection from the good sorted triples onto all sorted triples y₁ ≤ y₂ ≤ y₃ in [0, 1/2]. Sorted triples in [0, L] occupy volume L³/6, so the good region has volume (1/2)³/6 while all sorted triples in [0, 1] have volume 1/6. The probability is (1/2)³ = 1/8. In general, n points with minimum gap d have probability (1 − (n − 1)d)ⁿ.
Source: Classic geometric-probability exercise in wide circulation; no traceable original source. Statement written for AxiomIQ.