Problem
A circle is randomly generated by sampling two points uniformly and independently from the interior of a square and using these points to determine its diameter. What is the probability that the circle has a part of it that is off the square? Give your answer in exact terms.
Solution
Let \(P_1,P_2\) be the picked points and \(M\) be the midpoint of \(P_1 P_2\). Our random circle intersects the square iff the distance of \(M\) from the boundary of the square is less than the length of \(MP_1\) or \(MP_2\). Thus, assuming that the square is given by \([−1,1]^2\) and \(P_1=(x_1,y_1)\), \(P_2=(x_2,y_)\), we want the probability of the event
\[ \min \left(1-\lvert \frac{x_1 + x_2}{2} \rvert, 1-\lvert \frac{y_1 + y_2}{2} \rvert \right) \leq \frac{1}{2} \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2} \]
with \(x_1, x_2, y_1, y_2\) being independent and uniformly distributed random variables over the interval \([-1,1]\).
Computational Method
We use Monte Carlo simulation to estimate the probability. Using the Python code below, we see that the required probability is \(\mathbf{0.476}\).
from random import uniform
from math import sqrt
runs = 10000000
cnt = 0
for _ in range(runs):
x_1, x_2, y_1, y_2 = uniform(-1,1), uniform(-1,1), uniform(-1,1), uniform(-1,1)
if min(2-abs(x_1+x_2), 2-abs(y_1+y_2))<= sqrt((x_1-x_2)**2+(y_1-y_2)**2):
cnt += 1
print(cnt/runs)