Is the following argument correct?
"Without loss of generality, assume the stick is of unit length and let X and Y be the distance along from one end of the stick (denoted 0) that the cuts are made. Clearly, X and Y have independent, continuous uniform distributions on the interval from 0 to 1. Suppose X < Y; the lengths of the three pieces are then given by X, (Y - X) and (1 - Y), and so the triangle inequality yields:
X =< (Y - X) + (1 - Y) = 1 - X, and so X =< 1/2;
(Y - X) =< X + (1 - Y), and so Y =< X + 1/2;
(1 - Y) =< X + (Y - X), and so 1/2 =< Y.
Hence, we can find the probability that the pieces form a triangle when X < Y by integrating the joint pdf of X and Y over the region given by
0 =< X =< 1/2 and 1/2 =< Y =< X + 1/2,
which is a right triangle in the xy-plane with vertices (0, 1/2), (1/2, 1/2) and (1/2, 1). Since the joint pdf is 1 in this region, the integral is the area of this triangle, and so is equal to 1/8.
When Y < X, a symmetric argument shows that the probability that the pieces form a triangle is also 1/8. Since Y = X has probability 0, we can thus conclude that the total probability the pieces form a triangle is
1/8 + 1/8 = 1/4. "
