A Probability Variation of the Broken Stick Problem If we have a stick of length one and we break it into five places with the four breaks being chosen randomly down the stick. What is the probability that you can find some combination of the three pieces whose sum is longer than the longest piece?
 A: Let $L=[l_1,l_2,...,l_5]$ the random vector with the lengths of the five places of the stick. We suppose that $L\sim Dir(a_1,a_2,...,a_5)$, where $Dir$ denotes the Dirichlet distribution with pdf $f_D(X|a)=\frac{1}{B(a)}\prod_{i=1}^K x_i^{a_i-1} $ which is used when $\sum_{i=1}^K x_i= 1$.
Because length of sticks have equal mean value we have that $a_i=1$ $\forall i$.
Without the loss of generality we could suppose that you always choose the first three pieces $l_1,l_2,l_3$. The probability of winning is $P(l_1+l_2+l_3\geq l_4+l_5)$
From Direchlet's property of aggregation we have that  if $(X_1,X_2,X_3,X_4,X_5)\sim Dir(a_1,a_2,a_3,a_4,a_5)$ then $(X_1+X_2+X_3,X_4+X_5)\sim Dir(a_1+a_2+a_3,a_4+a_5)$, so in our case we have that $(l_1+l_2+l_3,l_4+l_5)\sim Dir(3,2)$
The winning probability is $\int\int_{\{x_1\geq 1/2\}} f_D(x_1,x_2|3,2)dx_1 dx_2 $
Edited:
Where $f_D(x_1,x_2|3,2)=f_D(x_1,1-x_1|3,2)=\frac{1}{B(3,2)}x_1^2(1-x_1)=12 x_1^2(1-x_1)$
(Dirichlet distribution of 2 random variables is equal to Beta distribution.)
So the winning probability is 
$12\int_{1/2}^{1} x_1^2(1-x1)dx_1=12\frac{11}{192} =0.6875$
