Probability that the smallest side of the line forming triangle will be smaller than $\frac L3$

There is a line the length of which $$L$$. We throw $$2$$ random points on it, hence $$3$$ segments are formed. I've denoted the length of the first segment $$x$$, the other one $$y$$, hence the 3rd one $$L-x-y$$.

I've found that the probability that these $$3$$ segments would form a triangle is $$\frac 14$$. Now I need to find what is the probability that the length of the smallest side will be not bigger(<=) than $$\frac L3$$ if these $$3$$ segments form a triangle.

The answer is $$1$$ but I can't figure out why.

Thanks.

In any division of the stick into three pieces, if no piece is shorter than $$L/3$$ then all sticks must be $$L/3$$ in length, else they would form a stick of length greater than $$L$$, and this event has measure zero (i.e. it almost never happens). Thus there is probability 1 that the shortest stick is shorter than $$L/3$$, irrespective of whether the pieces form a triangle or not.
• If they are all equal to $\frac L3$ then the shortest side is not less than that? Oct 26 '18 at 7:42