Breaking a Stick Into a Triangle

Take a stick of length 1. Pick two points on it uniformly at random and break the stick at both, you now have three pieces. What's the probability that those three pieces can be arranged into a triangle?

Try it before reading on. The instinct is to guess something close to a half, or to reach for calculus. Neither is needed, the whole problem collapses once you stop thinking about "pieces of a stick" and start thinking about a single point on a two-dimensional map.

Setting up the picture

Call the two break points $x$ and $y$, both drawn uniformly from $[0,1]$. Every possible pair $(x, y)$ is a point inside the unit square, and because $x$ and $y$ are independent and uniform, that square is the sample space, every point in it is equally likely, so a probability is just an area.

Assume for a moment $x < y$ (you'll double back for the symmetric case). Then the three pieces have lengths $x$, $y - x$, and $1 - y$. Three segments form a triangle exactly when no one piece is longer than the sum of the other two, equivalently, no piece exceeds half the total length. With total length 1, that means all three pieces must be shorter than $\tfrac{1}{2}$.

Working the three inequalities

Write out the three no-piece-too-long conditions for $x < y$:

$x < \frac{1}{2}, \qquad y - x < \frac{1}{2}, \qquad 1 - y < \frac{1}{2}$

The first says the left piece is short. The third rearranges to $y > \tfrac{1}{2}$: the right piece is short. The middle one, $y < x + \tfrac{1}{2}$, says the middle piece is short.

Plot these three lines inside the triangle $0 < x < y < 1$ (half the unit square, since we assumed $x<y$). Shading the region where all three hold carves out a smaller triangle in the middle of that half-square. Working the vertices out, the lines $x=\tfrac12$, $y=\tfrac12$, and $y = x+\tfrac12$ meet the boundary $x<y<1$ at $(0,\tfrac12)$, $(\tfrac12,\tfrac12)$, and $(\tfrac12,1)$, that inner triangle has area exactly $\tfrac{1}{4}$ of the area of the $x<y$ half-triangle (which itself has area $\tfrac12$), so its area relative to the full unit square is $\tfrac18$.

Doubling for the symmetric case $y < x$ gives another region of area $\tfrac{1}{8}$, for a total good area of $\tfrac{1}{4}$ out of the full square's area of 1.

$P(\text{triangle}) = \frac{1}{4}$

Only one time in four does a randomly broken stick make a triangle, most breaks produce one piece that's simply too long relative to the other two, because a uniform break point has a real chance of landing very close to one end.

The transferable technique

Whenever a puzzle involves several independent uniform random quantities and a geometric condition on them (triangle formation, whether points fall within a distance, whether a sum stays under a bound), the move is the same: let the variables be coordinates, draw the region satisfying the condition, and read off a probability as a ratio of areas or volumes. This turns algebra-heavy conditional reasoning into a single picture, and it's the standard opening move for the whole family of stick-breaking, meeting-time, and random-point puzzles that show up in quant interviews.