If a segment of length 1 is randomly divided into n intervals, with what probability are all intervals are less than 1/k? If $n-1$ points are chosen at random on a line segment of length $1$ (with uniform distribution), thus dividing it into $n$ segments, what is the probability that no segment has a length greater than $1/k$?
I've gotten this far as of now-
For $k=2$, only one segment can be greater than $1/2$, so, the probability is just 1 - n times the probability of first segment being of length greater than $1/2$.
so $P = 1-(n/2^{n-1})$ 
 A: Let us write
$$F_n(x) = \mathbb{P}\left(\max_{1\leq i \leq n} L_{n,i} \leq x \right), $$
where $L_{n,i}$ is the length of the $i$-th gap created by $n-1$ points chosen uniformly at random on $[0, 1]$, independent of each other.
Now let $U_1, \cdots, U_n \sim \mathcal{U}[0,1]$ be independently chosen points on $[0, 1]$ and $L_{n+1,i}$ be the length of the corresponding $i$-th gap. Conditioning on $L_{n+1,1} = \min\{U_1,\cdots,U_n\}$, we easily check that
\begin{align*}
F_{n+1}(x)
&= \mathbb{P}\left( \{ L_{n+1,1} \leq x \} \cap \Big\{ \max_{2\leq i \leq n+1} L_{n+1,i} \leq x \Big\} \right) \\
&= \sum_{k=1}^{n} \mathbb{P}\left( \{ U_k \leq x \} \cap \{ \forall l \neq k \ : \ U_l > U_k \} \cap \Big\{ \max_{2\leq i \leq n+1} L_{n+1,i} \leq x \Big\} \right) \\
&= \sum_{k=1}^{n} \mathbb{E} \left[ \mathbb{P}\left( \{ \forall l \neq k \ : \ U_l > U_k \} \cap \Big\{ \max_{2\leq i \leq n+1} L_{n+1,i} \leq x \Big\} \, \middle| \, U_k \right) \mathbf{1}_{\{ U_k \leq x \}} \right] \\
&= \sum_{k=1}^{n} \mathbb{E} \left[ F_n\left(\frac{x}{1-U_k}\right) (1 - U_k)^{n-1} \mathbf{1}_{\{ U_k \leq x \}} \right] \\
&= \int_{0}^{x \wedge 1} F_n\left(\frac{x}{1-u}\right) n(1-u)^{n-1} \, du
\end{align*}
Here, the last line follows from the fact that, given the value of $U_k$ and $U_l > U_k$ for $l \neq k$, points $\{U_l : l \neq k\}$ are i.i.d. and uniformly distributed over $[U_k, 1]$.
With the initial condition $F_1(x) = \mathbf{1}_{[1,\infty)}(x)$, this completely determines $F_n$ at least theoretically. As to an exact formula, we claim that

Claim. We have
$$ F_n(x) = \sum_{k=0}^{n} (-1)^k \binom{n}{k}(1-kx)_{+}^{n-1}, \tag{*} $$
where we interpret $x_+^0 = \mathbf{1}_{\{x > 0\}}$ when $n = 1$.

This easily follows from the recursive formula of $(F_n)$ together with the integration formula
$$\int_{a}^{b} n x_+^{n-1} \, dx = b_+^n - a_+^n$$
for $a \leq b$ and $n \geq 1$. In particular, this tells that


*

*$F_n(\frac{1}{2}) = 1 - n(\frac{1}{2})^{n-1}$,

*$F_n(\frac{1}{3}) = 1 - n(\frac{2}{3})^{n-1} + \frac{n(n-1)}{2}(\frac{1}{3})^{n-1}$,

*$F_n(\frac{1}{4}) = 1 - n(\frac{3}{4})^{n-1} + \frac{n(n-1)}{2}(\frac{2}{4})^{n-1} - \frac{n(n-1)(n-2)}{6}(\frac{3}{4})^{n-1}$
and so forth.

Addendum. The formula $\text{(*)}$ seems to suggest an inclusion-exclusion argument but I haven't tried pursuing this direction.
A: Let $L_i$ be the length of some subsegment. We have that:
$\mathbb{P}[L_i\leq1/k:\forall i]=1-\mathbb{P}[L_i>1/k:\exists i]$
Now $\mathbb{P}[L_i>1/k:\exists i]\implies$there are all $n-1$ points in a subsegment $(1-\frac{1}{k},1]$. Keep in mind that since $L_i>1/k$ the point forming $L_i$ is also in this length.
Since points are uniformly distributed and $\mathbb{P}[\textrm{point in }(1-\frac{1}{k},1]]=1-\frac{1}{k}$, we get:
$\mathbb{P}[n-1\textrm{ points in }(1-\frac{1}{k},1]]=\mathbb{P}[\textrm{point in }(1-\frac{1}{k},1]]^{n-1}=(1-\frac{1}{k})^{n-1}$
Hence
$\mathbb{P}[L_i\leq1/k:\forall i]=1-(1-\frac{1}{k})^{n-1}$
