Can You Reach the Edge of the Square?

You start at the centre of a unit square and pick a random direction to move in, with all directions equally likely. On average, how far do you travel before you reach the perimeter?

The Fiddler, Zach Wissner-Gross, October 17, 2025(original post)

Solution

Put the centre at the origin, so the four walls are $x=\pm\tfrac12$ and $y=\pm\tfrac12$. A ray in direction $\theta$ first meets the wall that is perpendicular-closest, and the distance to that wall is $$r(\theta)=\frac{1}{2\max\!\bigl(|\cos\theta|,|\sin\theta|\bigr)}.$$ By the eightfold symmetry of the square it is enough to average over $\theta\in[0,\tfrac\pi4]$, where the right wall $x=\tfrac12$ is hit at distance $\tfrac12\sec\theta$. Hence $$\begin{aligned} E[r]&=\frac{1}{2\pi}\int_0^{2\pi} r(\theta)\,d\theta =\frac{2}{\pi}\int_0^{\pi/4}\!\sec\theta\,d\theta\\ &=\frac{2}{\pi}\,\ln\!\bigl(\sqrt2+1\bigr)\approx\boxed{0.5611}. \end{aligned}$$

The computation

Fire random rays from the centre: for each direction take the distance to the nearest of the four walls, $\tfrac12/\max(|\cos\theta|,|\sin\theta|)$, and average. The Monte Carlo matches the closed form $\tfrac2\pi\ln(\sqrt2+1)$.

import numpy as np
from math import pi, log, sqrt
rng = np.random.default_rng(0); th = rng.uniform(0, 2*pi, 4_000_000)
d = 0.5 / np.maximum(np.abs(np.cos(th)), np.abs(np.sin(th)))
print(round(d.mean(), 5), round(2/pi*log(1 + sqrt(2)), 5))   # 0.56102  0.5611

Extra Credit

Now you start at the centre of a unit cube and pick a uniformly random direction in three dimensions. On average, how far do you travel before reaching the surface?

Solution

By the cube’s symmetry, average over the $\tfrac{1}{24}$ of directions whose nearest face is the top and whose azimuth $\theta$ lies in $[0,\tfrac\pi4]$. With $\varphi$ the polar angle, the travel distance is $1/(2\cos\theta\sin\varphi)$ and the top face is nearest once $\varphi\ge\arctan(\sec\theta)$. Weighting by the solid-angle density $\tfrac1{4\pi}\sin\varphi$ and the $24$-fold symmetry, the $\sin\varphi$ cancels and $$E[r]=\frac{3}{\pi}\int_0^{\pi/4}\sec\theta\Bigl(\frac\pi2-\arctan(\sec\theta)\Bigr)\,d\theta \approx\boxed{0.6107}.$$ (The source’s value is paywalled; this closed form is my own.)

The computation

The same experiment in 3D: draw directions uniformly on the sphere (normalised Gaussians) and take the distance to the nearest cube face, $\tfrac12/\max(|v_x|,|v_y|,|v_z|)$. The Monte Carlo confirms the integral.

from scipy import integrate
val, _ = integrate.quad(lambda t: (1/np.cos(t))*(pi/2 - np.arctan(1/np.cos(t))), 0, pi/4)
rng = np.random.default_rng(0); v = rng.standard_normal((4_000_000, 3))
v /= np.linalg.norm(v, axis=1, keepdims=True)
print(round((0.5/np.max(np.abs(v), axis=1)).mean(), 5), round(3/pi*val, 5))   # 0.61068  0.61069