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.