# Average length of the longest segment

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?

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There is a simple geometric answer: It is the first corrdinate of mass center of these points: (1,0,..) (1/2,1/2,0,..) .. (1/n,1/n,1/n,...1/n) =(1+1/2+...1/n)/n –  Yudong Tang Mar 23 '14 at 20:59

The answer to (B) is actually given in both Yuval Filmus' and my answers to the question about the average length of the shortest segment. It's $$\frac{1}{n} H_n,$$ where $H_n = \sum_{k=1}^n \frac{1}{k},$ i.e., the $n$th harmonic number.

"Clever" is of course subjective, but here's an argument for (A) in the $n$-piece case. At least there's only one (single-variable) integration in it. :)

If $X_1, X_2, \ldots, X_{n-1}$ denote the positions on the rope where the cuts are made, let $V_i = X_i - X_{i-1}$, where $X_0 = 0$ and $X_n = 1$. So the $V_i$'s are the lengths of the pieces of rope.

The key idea is that the probability that any particular $k$ of the $V_i$'s simultaneously have lengths longer than $c_1, c_2, \ldots, c_k$, respectively (where $\sum_{i=1}^k c_i \leq 1$), is $$(1-c_1-c_2-\ldots-c_k)^{n-1}.$$ This is proved formally in David and Nagaraja's Order Statistics, p. 135. Intuitively, the idea is that in order to have pieces of size at least $c_1, c_2, \ldots, c_k$, all $n-1$ of the cuts have to occur in intervals of the rope of total length $1 - c_1 - c_2 - \ldots - c_k$. For example, $P(V_1 > c_1)$ is the probability that all $n-1$ cuts occur in the interval $(c_1, 1]$, which, since the cuts are randomly distributed in $[0,1]$, is $(1-c_1)^{n-1}$.

If $V_{(n)}$ denotes the largest piece of rope, then $$P(V_{(n)} > x) = P(V_1 > x \text{ or } V_2 > x \text{ or } \cdots \text{ or } V_n > x).$$ This calls for the principle of inclusion/exclusion. Thus we have, using the "key idea" above, $$P(V_{(n)} > x) = n(1-x)^{n-1} - \binom{n}{2} (1 - 2x)^{n-1} + \cdots + (-1)^{k-1} \binom{n}{k} (1 - kx)^{n-1} + \cdots,$$ where the sum continues until $kx > 1$.

Therefore, $$E[V_{(n)}] = \int_0^{\infty} P(V_{(n)} > x) dx = \sum_{k=1}^n \binom{n}{k} (-1)^{k-1} \int_0^{1/k} (1 - kx)^{n-1} dx = \sum_{k=1}^n \binom{n}{k} (-1)^{k-1} \frac{1}{nk}$$ $$= \frac{1}{n} \sum_{k=1}^n \frac{\binom{n}{k}}{k} (-1)^{k-1} = \frac{H_n}{n},$$ where the last step applies a known binomial sum identity.

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You are right. It was my fault for not checking. –  TCL Dec 13 '10 at 16:55