A problem I solved yesterday during my 10-minute subway ride.

Given a line segment of unit length, we put two points into it randomly so that it is divided into three segments. What is the probability that these three segments can form a triangle?

Let the two points be \(x\) and \(y\), and assume them to be uniformly distributed on [0:1]. If the three segments can form a triangle, it must satisfy the triangle inequality, i.e. no one side is longer than 0.5. Assume \(x>y\), then we must have \(y<0.5\) and \(0.5<x<0.5+y\). The probability of such is

\[\begin{aligned} & \int_{0.5}^{0.5+y}\int_0^{0.5} dy dx \\ =& \int_0^{0.5} \int_{0.5}^{0.5+y} dx dy \\ =& \int_0^{0.5} y dy \\ =& [\frac{1}{2}y^2]_0^{0.5} = \frac{1}{8} \end{aligned}\]

Similarly for the case \(y>x\), we must have \(x<0.5\) and \(0.5<y<0.5+x\), and the probability is also \(\frac{1}{8}\). So the total probability is \(\frac{1}{4}\).

To verify the result, below is the Python code:

import random

def triangle():
	(x,y) = (random.random(), random.random())
	return 1 if max(min(x,y), abs(x-y), max(x,y)) < 0.5 else 0

def main():
	N = 1000000
	print "%f\n" % (sum(triangle() for i in xrange(N))/float(N))

if __name__ == '__main__':

It prints 0.25