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?

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
