Drop $n$ points $X_1,\dotsc, X_n$ at random in the unit interval, uniformly and independently. These points divide the interval into $n+1$ sub-intervals. Their lengths are called the spacings of the subdivision. What is the expected length of the longest of these intervals? What is its asymptotic behavior for $n$ large?
I was surprised to find out that the problem of finding the expectation of the largest segment was solved in late 19th century. What follows is a more detailed presentation of the arguments in Sec. 6.3 of the book Order Statistics by H.A. David and H.N. Nagaraja.
If you read French, you can click here to see what P. Levy had to say about this problem back in 1939.
Relabel by $Y_1,\dotsc, Y_n$, the points $\{X_1,\dotsc, X_n\}\subset [0,1]$ so that $Y_1\leq \cdots \leq Y_n$. We set
$$ Y_0=0,\;\;Y_{n+1}=1,\;\;\;\;L_n= \max_{1\leq i\leq n+1} Y_i-Y_{i-1},.$$
We want to determine the expectation $\newcommand{\bE}{\mathbb{E}}$ $E_n$ of $L_n$.
The probability distribution of $\vec{X}=(X_1,\dotsc, X_n)$ $\newcommand{\bone}{\boldsymbol{1}}$ is
$$ p_{\vec{X}}(dx)=\bone_{C_n}(x_1,\dotsc, x_n) dx_1\cdots dx_n, \;\;C_n=[0,1]^n, $$
where $\bone_A$ denotes the indicator function of a subset $A$ of a given set $S$.
The probability distribution of $\vec{Y}=(Y_1,\dotsc, Y_n)$ is
$$p_{\vec{Y}}(dy)= n! \bone_{T_n}(y_1,\dotsc, y_n) dy_1\cdots dy_n, $$
where $T_n$ is the simplex $\newcommand{\bR}{\mathbb{R}}$
$$ T_n=\bigl\{ y\in\bR^n;\;\;0\leq y_1\leq \cdots \leq y_n\leq 1\,\bigr\}. $$
The spacings are the lengths of the $n+1$ segments that the points $Y_1,\dotsc, Y_n$ determine on $[0,1]$
$$ S_i= Y_{i}-Y_{i-1},\;\; i=1,\dotsc, n+1, $$
where $Y_{n+1}:=1$. The resulting map $\vec{Y}\to\vec{S}=\vec{S}(\vec{Y})$ induces a linear bijection
$$ T_n\to \Delta_n=\bigl\{\, (s_1,\dotsc, s_{n+1})\in\bR_{\geq 0}^{n+1};\;\;s_1+\cdots+s_{n+1}=1\,\bigr\}. $$
This shows that the probability density is $\DeclareMathOperator{\vol}{vol}$
$$p_{\vec{S}}(ds) =\frac{1}{\vol(\Delta_n)} dV_{\Delta_n}, $$
where $dV_{\Delta_n}$ denote the usual Euclidean volume density on the standard simplex $\Delta_n$. This description shows that the random vector of spacings $\vec{S}$ is exchangeable, i.e., the probability density of $\vec{S}$ is invariant under permutation of the components. Thus the joint probability distribution of any group of $k$ components of $\vec{S}$, $k\leq n$, is the same as the joint distribution of the first $k$-components $S_1,\dotsc, S_k$.
The joint probability distribution of the first $n$ components is
$$ p_n(ds)= n! \bone_{D_n}(s_1,\dotsc,s_n) ds, $$
where $D_n$ is the simplex
$$ D_n=\{\, (s_1,\dotsc, s_n)\in\bR^n_{\geq 0};\;\;s_1+\cdots +s_n\leq 1\,\}. $$
Indeed, $(S_1,\dotsc, S_n)\in D_n$ and
$$ ds=|ds_1\cdots ds_n|=|d y_1\wedge d(y_2-y_{1})\wedge \cdots \wedge d(y_{n}-y_{n-1})|=|dy_1\cdots d y_n|. $$
The joint density of the first $n-1$ components $(S_1, \dotsc, S_{n-1})$ is
$$ p_{n-1}(s_1,\dotsc, s_{n-1}) = n! \int_0^\infty \bone_{D_n}(s_1,\dotsc, s_n) ds_n $$
$$= n! \bone_{D_{n-1}}(s_1,\dotsc, s_{n-1})\int_0^{1-s+1-\cdots-s_{n-1}} ds_n= n! \bone_{D_{n-1}}(s_1,\dotsc, s_{n-1})(1-s_1-\cdots-s_{n-1}) $$
The joint density of the first $(n-2)$ components is
$$ p_{n-2}(s_1,\dots, s_{n-2})=\int_0^\infty p_{n-1}(s_1,\dotsc, s_{n-1}) ds_{n-1} $$
$$ = n!\bone_{D_{n-2}}(s_1,\dotsc, s_{n-2})\int_0^{1-s_1-\cdots-s_{n-2}} (1-s_1-\cdots -s_{n-1}) d s_{n-1} = \frac{n!}{2!} \bone_{D_{n-2}}(s_1,\dotsc, s_{n-2})(1-s_1-\cdots -s_{n-2})^2. $$
Iterating, we deduce that the joint probability density of the the first $k$ variables is
$$ p_k(s_1,\dotsc, s_k) =\frac{n!}{(n-k)!}p_{D_k}(s_1,\dotsc, s_k) (1-s_1-\cdots-s_k)^{n-k}. $$
If $1\leq k\leq n$, then$\newcommand{\bP}{\mathbb{P}}$
$$ \bP[S_1>s,\dotsc, S_k>s]=\frac{n!}{(n-k)!}\int_s^\infty\cdots\int_s^\infty p_{D_k}(s_1,\dotsc, s_k) (1-s_1-\cdots-s_k)^{n-k} ds_1\dotsc ds_k $$
$$=\frac{n!}{(n-k)!}\int_s^1ds_1\int_s^{1-s_1}ds_2\cdots\int_s^{1-s_1-\cdots-s_{k-1}} (1-s_1-\cdots-s_k)_+^{n-k} ds_k =(1-ks)_+^n,\;\;x_+:=\max(x,0).$$
Also
$$ \bP[S_1>s,\dotsc, S_n>s, S_{n+1}>s] = \bP[S_1>s,\dotsc, S_n>s, S_1+S_2+\cdots +S_n\leq 1-s]$$
$$=n!\int_{\substack{s_1,\dotsc s_{n}>s\\ s_1+\cdots +s_{n}\leq 1-s}}ds_1\cdots d s_{n} $$
$$=n!\int_{\substack{s_1,\dotsc s_{n-1}>s\\ s_1+\cdots +s_{n-1}\leq 1-2s}}ds_1\cdots d s_{n-1}\int_s^{1-s-s_1-\cdots -s_{n-1}} d s_n $$
$$= n!\int_{\substack{s_1,\dotsc s_{n-1}>s\\ s_1+\cdots +s_{n-1}\leq 1-2s}} (1-2s-s_1-\cdots -s_{n-1}) ds_1\cdots ds_{n-1} $$
$$=n!\int_{\substack{s_1,\dotsc s_{n-2}>s\\ s_1+\cdots +s_{n-2}\leq 1-3s}}ds_1\cdots ds_{n-2}\int_s^{1-2s-s_1-\cdots -s_{n-2}} (1-2s-s_1-\cdots -s_{n-1}) d s_{n-1} $$
$$=\frac{n!}{2!} \int_{\substack{s_1,\dotsc s_{n-2}>s\\ s_1+\cdots +s_{n-2}\leq 1-3s}} (1-3s-s_1-\cdots -s_{n-2})^2 ds_1\cdots ds_{n-2} $$
$$ =\frac{n!}{2!} \int_{\substack{s_1,\dotsc s_{n-3}>s\\ s_1+\cdots +s_{n-3}\leq 1-4s}} ds_1\cdots ds_{n-3} \int_s^{1-3s -s_1-\cdots -s_{n-3}} (1-3s-s_1-\cdots -s_{n-2})^2ds_{n-2}$$
$$=\frac{n!}{3!} \int_{\substack{s_1,\dotsc s_{n-3}>s\\ s_1+\cdots +s_{n-3}\leq 1-4s}} (1-4s-s_1-\cdots -s_{n-3})^3ds_1\cdots ds_{n-3} $$
$$=\cdots = \bigl(\, 1-(n+1)s\,\bigr)_+^n. $$
With
$$ L_n:=\max_{1\leq i\leq n+1} S_i $$
we have
$$\bP[L_n>s]=\bP\left[\,\bigcup_{i=1}^{n+1}\bigl\lbrace\, S_i>s\,\bigr\rbrace\,\right] $$
(use inclusion-exclusion exchangeability)
$$ = \sum_{k=1}^{n+1}(-1)^{k+1}\binom{n+1}{k} \bP\bigl[\, S_1>s,\dotsc, S_k>s\,\bigr] =\sum_{k=1}^{n+1} \binom{n+1}{k}(1-ks)_+^n .$$
The last equality is sometime known as Whitworth formula and it goes all the way back to 1897. We have
$$ E_n=\bE[L_n]=\int_0^\infty \bP[L_n>s] ds =\sum_{k=1}^{n+1}(-1)^{k+1} \binom{n+1}{k}\int_0^\infty (1-ks)_+^n ds = \sum_{k=1}^{n+1}(-1)^{k+1} \binom{n+1}{k}\int_0^{1/k}(1-ks)^n ds $$
$$=\frac{1}{n+1} \sum_{k=1}^{n+1}(-1)^{k+1} \frac{1}{k} \binom{n+1}{k}=\sum_{k=1}^{n+1}(-1)^{k+1} \frac{1}{k^2} \binom{n}{k-1}. $$
Here is $E_n$ answers for small $n$ (thank you MAPLE)
$$ E_1=\frac{3}{4},\;\;E_2=\frac{11}{18},\;\;E_3=\frac{25}{48},\;\;E_4=\frac{137}{300},\;\; E_5=\frac{49}{120}. $$
It turn out that $L_n$ is highly concentrated around its mean. In the work of P. Levy I mentioned earlier he showed that the random variables $nL_n-\log n$ converges in probability to a random variable $X_\infty$ with probability distribution $\mu$ given by
$$ \mu\bigl(\,(-\infty,x])= \exp\bigl(\,-e^{-x}\,\bigr),\;\;x\geq 0.$$
Moreover
$$\lim_{n\to \infty} \bE[\, nL_n-\log n\,]=\gamma=\bE[X_\infty],\;\;\lim_{n\to \infty}\mathrm{Var}[ nL_n\,]=\frac{\pi^2}{6}=\mathrm{Var}[X_\infty]. $$
$$\frac{n}{\log n} L_n \to 1\;\; \mbox{a.s.} $$
Note that this shows that
$$E_n\sim\frac{\log n}{n}\;\; \mbox{as $n\to \infty$}. $$
I'll stop here, but you can learn more from this paper of Luc Devroye and its copious list of references.
Denote by $A_n$ the smallest spacing
$$ A_n:=\min_{1\leq k\leq n+1} S_k. $$.
Note that
$$\bP[ A_n>a]=\bP[ S_1,S_2,\dotsc, S_{n+1}>a]=\big( 1-(n+1)a\big)^n_+. $$
Hence
$$ \bE[ A_n]=\int_0^\infty\bP[A_n>a] da= \int_0^\infty \big( 1-(n+1)a\big)^n_+ da=\int_0^{\frac{1}{n+1}} (1-(n+1)a)^n da$$
$$ = \frac{1}{n+1}\int_0^1 (1-t)^n dt= \frac{1}{(n+1)^2}. $$
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