On a straight stick of length $1$, choose $n-1$ independent uniformly random points. Cut the stick at those points, yielding $n$ fragments.
Let $\mathbb{E}_n$ be the expectation of the product of lengths of the fragments.
Simulations suggest that $\mathbb{E}_n=\prod\limits_{k=n}^{2n-1}\frac1k$. For example, $\mathbb{E}_5=\frac{1}{5\cdot6\cdot7\cdot8\cdot9}$.
This is an elegant result, so I suspect there is an intuitive explanation, but it eludes me.
Is there an intuitive explanation for $\mathbb{E}_n=\prod\limits_{k=n}^{2n-1}\frac1k$ ?
My thoughts:
I have only been able to prove the result for $n=2$ and $n=3$, and my proofs are not intuitive.
$\color{red}{n=2}$:
Hold the stick horizontally.
Let the distance of the random point from the left end be $x$.
$$\mathbb{E}_2=\int_0^1x(1-x)\mathrm dx=\frac16=\prod\limits_{k=2}^{2(2)-1}\frac1k$$
$\color{red}{n=3}$:
Hold the stick horizontally.
Let the distance of the 1st chosen point from the left end be $x$.
Let the distance of the 2nd chosen point from the left end be $y$.
(We make no assumption about which of $x$ and $y$ is greater.)
For a fixed $x$ value, the expectation of the product of fragment lengths is $$P(y<x)\times\int_0^xy(x-y)(1-x)\mathrm dy+P(y>x)\times\int_x^1x(y-x)(1-y)\mathrm dy$$
where $P(y<x)=x$ and $P(y>x)=1-x$.
Then we integrate this expectation from $x=0$ to $x=1$:
$$\mathbb{E}_3=\int_0^1\left(x\int_0^xy(x-y)(1-x)\mathrm dy+(1-x)\int_x^1x(y-x)(1-y)\mathrm dy\right)\mathrm dx=\frac{1}{60}=\prod\limits_{k=3}^{2(3)-1}\frac1k$$
Context:
I recently answered a question involving products of lengths of random line segments. That made me think of this question.
Best Answer
You first pick $n-1$ random points which splits the interval $[0,1]$ into $n$ segments $U_1,\ldots,U_n$ of lengths $L_i=|U_i|$. You wish to compute the expected value $\mu=E[L_1\cdots L_n]$. We can assume the segments $U_i$ are sorted in increasing order, although that doesn't really matter.
Next, similarly pick $n$ points $X_1,\ldots,X_n$ randomly in $[0,1]$. The probability that $X_i\in U_i$ is $L_i$, and since the $X_i$ are picked independently, the probability that $X_i\in U_i$ for all $i$ equals $\mu=E[L_1\cdots L_n]$.
The above is the same as first picking $2n-1$ points randomly, and then selecting at random the $n$ points $X_1,\ldots,X_n$ from these. The remaining $n-1$ points split the interval into segments $U_i$, and the probability that $X_i\in U_i$ for all $i$ is the same as the points $X_1,\ldots,X_n$ being every second point out of the $2n-1$ in increasing order starting at the lowest. The probability of this, ie of $X_i$ being $n$ specific points out of the $2n-1$ picked in increasing order, is $$ \mu=\frac{1}{\binom{2n-1}{n}\cdot n!}=\frac{1}{(2n-1)\cdots n}. $$