Probability of a Triangle

A stick of length $1$ is broken at two points chosen independently and uniformly along its length. What is the probability that the three resulting pieces can form a triangle?

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#The triangle condition

Three lengths that sum to $1$ form a triangle exactly when each is shorter than $\tfrac{1}{2}$ . The triangle inequality only bites on the longest piece, which must be under the sum of the other two, namely $1$ minus itself, so the longest must be below $\tfrac{1}{2}$ , and then the other two automatically clear it.

#Map the two cuts

Let $U$ and $V$ be the cut positions, independent and uniform on $[0,1]$ , so $(U,V)$ is uniform on the unit square and probability is area. Split on which cut sits to the left.

If $U < V$ the pieces are $U$ , $V - U$ , and $1 - V$ . Keeping all three below $\tfrac{1}{2}$ asks for $U < \tfrac{1}{2}$ , $V > \tfrac{1}{2}$ , and $V - U < \tfrac{1}{2}$ , a triangle with vertices $(0,\tfrac{1}{2})$ , $(\tfrac{1}{2},\tfrac{1}{2})$ , $(\tfrac{1}{2},1)$ of area $\tfrac{1}{8}$ . The case $U > V$ is its mirror image with the same area.

Each cut point is uniform, so (U, V) is uniform on the square. A triangle needs every piece shorter than 1/2, which leaves only the two shaded triangles, a quarter of the area in all.

#Read it off

\PP(\text{triangle}) = \frac{1}{8} + \frac{1}{8} = \frac{1}{4}. \tag{1}

A triangle is the exception, not the rule. Three quarters of the time one piece swallows more than half the stick and the other two cannot reach across it.