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?
 A: Rather than breaking a stick of length 1 into six parts with five breaks, break a circle of circumference 1 into six parts with six breaks and then pick one from those breaks as the "original ends".  
So, what is the probability that all six breaks are not on the same semicircle?
Well wolog we pick one point as reference. The other five points will be uniformly distributed over $[-1/2,1/2]$ relative to that point. You want the probability that the distance between the least and most order statistic from those five points is more than 1/2.
$$\int_{1/2}^0\int_{x+1/2}^1 f_{X_{(1)},X_{(5)}}(x,y)~\mathsf d y~\mathsf d x$$
A: similar to what @hardmath said 
you can break the line into predetermined equal segments for example (10,20,....,40,...100 etc)
Example
lets break the line into 20 segments, the total number of ways 6 line segments can be selected is 
total number of positive integers in the equation
$x_1+x_2+x_3+x_4+x_5+x_6 = 20$
= $\binom{19}{5}$
To get our event E: one segment is greater >= 10
let x_1 be that segment and  possible cases for X_1 can be (10|11|12|13|14|15) 
It can not be more than 15 because in that case value of other segment has to be 0 which is not possible.
when $x_1 = 10 : x_2+x_3+x_4+x_5+x_6 =10 : \binom{9}{4}$
$x_1 = 11 : x_2+x_3+x_4+x_5+x_6 =9 : \binom{8}{4}$
$x_1 = 12 : x_2+x_3+x_4+x_5+x_6 =8 : \binom{7}{4}$
$x_1 = 13 : x_2+x_3+x_4+x_5+x_6 =7 : \binom{6}{4}$
$x_1 = 14 : x_2+x_3+x_4+x_5+x_6 =6 : \binom{5}{4}$
$x_1 = 15 : x_2+x_3+x_4+x_5+x_6 =5 : \binom{4}{4}$
Total number of cases for event (E): 6*($\binom{4}{4}$+$\binom{5}{4}$+$\binom{6}{4}$+$\binom{7}{4}$+$\binom{8}{4}$+$\binom{9}{4}$) = 1512
Probability of event E: $\frac{1512}{11628}$ = .13
P(E)  = .152 in case of 30 segments
P(E)  = .167 in case of 50 segments
We can see that the probaility of E converges to some number A
our answer = 1 - A
A: 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.)

