This post is related to a previous SE post If a 1 meter rope …. concerning average length of a smallest segment.
A rope of 1m is divided into three pieces by two random points. Find the average length of the largest segment. My answer is 11/18. Here is how I do it:
Here we have two independent random variables $X,Y$, both uniform on $[0,1]$. Let $A=\min (X,Y), B=\max (X,Y)$ and $C=\max (A, 1-B, B-A)$. First we want to find the probability density function $f_C(a)$ of $C$. Let $F_C(a)$ be the cumulative distribution function. Then $$ F_C(a) = P(C\le a)=P(A\le a, 1-B\le a, B-A\le a).$$ By rewriting this probability as area in the unit square, I get $$F_C(a)=\left\{\begin{array}{ll} (3a-1)^2 & \frac{1}{3}\le a\le \frac{1}{2}\\ 1-3(1-a)^2 & \frac{1}{2}\le a\le 1\end{array}\right.$$ from which it follows that $$f_C(a)=\left\{\begin{array}{ll} 6(3a-1) & \frac{1}{3}\le a\le \frac{1}{2}\\ 6(1-a) & \frac{1}{2}\le a\le 1\end{array}\right.$$ Therefore the expected value of $C$ is $$\int_{1/3} ^{1/2}6a(3a-1) da+\int_{1/2} ^{1}6a(1-a) da= \frac{11}{18}.$$
My questions are:
(A) Is there a "clever" way to figure out this number 11/18?
(B) What is the answer if the rope is divided into $n>3$ segments?