A rope of 1m is divided into three pieces by two random points. Find the average length of the largest segment... with an algorithm!

Question

About the title, I've read here that the answer is 11/18.

I've tried with a simple algorithm:

def cut(first_len:int) -> List[Union[float,float]]:
    c1 = random.uniform(0,first_len)
    return [c1, first_len-c1]
if name=='main':
    max_val = 0
num_iter = 1000000
m_len = 1
for _ in range(num_iter):
    #first random cut
    cut_vals = cut(m_len)

    # second random cut on the longest piece
    max_cut2_vals = cut(max(cut_vals))

    # get the max
    m2 = max(max_cut2_vals)

    max_val = max_val+m2

print(max_val/num_iter)

that returns 0.563, which is a bit lower than 11/18. I assume that I'm missing some fundamental property.

Can someone help me?

Answer 1

A rope is divided into three pieces simultaneously: you pick two random real numbers $x, y$ from the $(0, 1)$ interval and then pick the longest piece: either from $0$ to $x$ (which has length $x$), from $x$ to $y$ (which has length $y-x$) or from $y$ to $1$ (which has length $1 - y$). This is assuming that $0 \le x \le y \le 1$ which we can simulate easily in Python:

#!/usr/bin/env python3
from random import random
from statistics import mean
def two_cuts():
    x, y = random(), random()
    if x > y
        x, y = y, x # to make sure that 0 <= x <= y <= 1
    return (x, y)
def longest(x, y):
    return max(x, 1-y, y-x)
print(mean([longest(two_cuts()) for _ in range(10*5)]))

I've got $0.611285852578838 \approx 0.611111111111111 \ldots = 11/18$. Everything agrees.