Breaking up a Stick into different pieces Assume you have a stick of length 1. Choose 5 breaking points randomly along the stick such that the stick is divided into 6 parts. What is the probability that no part is greater than 1/2?

For the case where the breaks are five independent observations from
$\mathsf{Unif}(0,1),$ an argument similar to @GrahamKemp's can be used
to show that the longest piece is longer than $1/2$ with probability $6/32 = 3/16 = 0.1875.$ 
If the intervals are arranged in a circle, the probability that the clockwise
180-degree gap following any one initial point is empty is $1/32,$ so the
total probability of a piece greater than .5 in length is $6/32.$

In the following simulation, the vector x contains points $0, 1,$ and the
five breakpoints, sorted in order. Taking the maximum difference, we get
the length big of the longest piece. The result  0.187218 of mwan(big > .5)
is consistent with $6/32$ within simulation error.  
set.seed(318)  # retain this statement for exact same simulation; delete for frexh run
m = 10^6;  cuts=5; big = numeric(m)
for(i in 1:m) {
  x = c(0, sort(runif(cuts)), 1)
  big[i] = max(diff(x)) }
mean(big);  sd(big);  mean(big > .5)
## 0.4082853
## 0.1081648
## 0.187218

A histogram of the lengths of the one million such longest pieces simulated
is shown below:

Note: By contrast, if the bathtub-shaped distribution $\mathsf{BETA}(.5,.5)$ is used
to make the breaks, then there will be relatively few cuts near the middle of $(0,1),$
and the probability that the length of the biggest piece exceeds $1/2$ increases
to about 0.41. (This illustrates @hardmath's Comment.)

