Break a Stick Twice, and the Pieces Form a Triangle ¼ of the Time

Here is the question, in the form it usually arrives: a stick of length one is broken at two points, chosen independently and uniformly at random. What is the probability that the three pieces form a triangle? No coordinates are given, no distribution is written down, and you are expected to start talking within a few seconds.

It turns up in first-round quantitative interviews, most often for trading and market-making seats, and it is popular for a specific reason: the answer is a clean fraction, but almost nobody guesses it correctly. The interviewer is not testing whether you know it. They are watching whether you turn a vague physical story into a probability space without being told to, and whether you notice that three conditions collapse into one. Both of those happen in the first thirty seconds, which is why the problem survives as a screen.

Three inequalities that are secretly one

Call the pieces $a$, $b$ and $c$. They are positive and they sum to 1, because they are the whole stick. Three lengths form a triangle when each one is shorter than the other two put together, so the condition you are handed is a system:

Now use the constraint. Since $b + c = 1 - a$, the first inequality says $a < 1 - a$, which is just $a < \tfrac12$. The same substitution works on the other two. Three conditions become three copies of one condition, and since only the largest piece can possibly violate it, the whole system is a single statement about the maximum:

That is the entire mathematical content of the problem, and everything below is bookkeeping. A piece of length exactly one half is a degenerate case, a flat triangle, and it happens with probability zero, so strict and non-strict inequalities give the same answer and nobody needs to worry about the boundary.

Probability becomes area

Stop thinking about the stick. The experiment is two numbers, $X$ and $Y$, each uniform on $[0,1]$ and independent of the other. One run of the experiment is one point in the unit square, and because the pair is uniform on that square, the probability of any event is the area of the region it occupies. The problem is now a question about a picture.

Take the half where $X < Y$. There the pieces are $X$, $Y - X$ and $1 - Y$, and condition (1) applied to each of them gives three half-planes:

Each is a straight cut. What survives all three is one triangle, with corners at $(0,\tfrac12)$, $(\tfrac12,\tfrac12)$ and $(\tfrac12,1)$. Both of its legs have length $\tfrac12$, so its area is $\tfrac18$. The half where $X > Y$ is the mirror image and contributes the same.

Notice what that took: one reformulation, one square, two small triangles, and not a single integral. If you are asked this at a whiteboard, the square is the answer the interviewer is hoping to watch you draw.

The same quarter, from the other side

Areas are easy to get subtly wrong under pressure, so it is worth having a second route that shares no steps with the first. Condition on the first break instead of measuring a region.

Say the first break lands at $x$, somewhere in the left half. The right-hand piece must come in under a half, so the second break has to land past the midpoint. The piece between the two marks must also come in under a half, so the second break cannot land further than $\tfrac12$ beyond $x$. That is a window from $\tfrac12$ to $x + \tfrac12$, and its width is exactly $x$. Since the second break is uniform, the conditional probability is $x$ itself, and by symmetry the general answer is $\min(x, 1-x)$. Average it:

The same quarter, from a different direction. This version also hands you a fact the area argument hides: $\min(x,1-x)$ never exceeds $\tfrac12$. Even a first break placed perfectly at the midpoint, the luckiest one available, leaves you at a coin flip at best. There is no opening move that makes this experiment favourable.

Why a quarter still feels too low

Most people, told the answer is 25%, accept the algebra and keep a private suspicion that it should be higher. The suspicion is worth taking seriously, because the number that settles it is more interesting than the quarter.

Break the stick and look only at the longest piece. Call it $M$. It lives between $\tfrac13$ and 1, and its distribution function has two regimes. The upper one is the easy one: once $\ell > \tfrac12$, at most one piece can exceed $\ell$, since two of them would already be longer than the whole stick. The three events are therefore disjoint, and a single piece is a spacing with $\mathbb{P}(\text{piece} > \ell) = (1-\ell)^2$, so

Differentiating gives densities $6(3\ell - 1)$ and $6(1-\ell)$ on the two stretches, and the expectation splits along the same seam:

About 61%. The shortest piece works the same way, from $\mathbb{P}(\text{shortest} > s) = (1-3s)^2$ on $[0,\tfrac13]$, giving $\tfrac19$, and the middle piece follows for free, because the three averages have to add up to the stick: $1 - \tfrac{11}{18} - \tfrac19 = \tfrac{5}{18}$.

Now the quarter looks generous rather than stingy. A typical break does not produce three comparable pieces. It produces one long piece taking 61% of the stick, one middling piece at 28%, and a splinter at 11%. The longest piece is over budget on average, which means a typical outcome has already failed condition (1) before you look at anything else. Triangles are rare because lopsided cuts are normal.

There is a general version worth carrying around. For $n$ pieces the expected longest is $H_n / n$, the $n$th harmonic number over $n$, and the expected shortest is $1/n^2$. At $n = 3$ the first gives $(1 + \tfrac12 + \tfrac13)/3 = \tfrac{11}{18}$, which is a much faster way to the same number if you happen to remember it.

The follow-up that catches people

If the first answer goes well, the natural second question is a different experiment wearing the same clothes: break the stick once, then break the longer of the two pieces. It sounds like an improvement. You are attacking the piece that causes all the trouble, so the odds should go up.

They do, and by less than people expect. Let $L$ be the longer piece, so $L = \max(U, 1-U)$ is uniform on $[\tfrac12, 1]$ with density 2. The untouched short piece has length $1 - L$, which is at most a half, so it never causes a failure. The two halves of $L$ must both come in under $\tfrac12$, so the second cut has to land in a window of width $1 - L$ inside a piece of length $L$, and the conditional probability is $(1-L)/L$. Averaging:

So aiming at the long piece takes you from 25% to about 39%, and still leaves you short of a coin flip. There is a third variant that shows up in the same conversation and moves in the opposite direction: break once, then pick one of the two pieces at random and break that. Here the answer is $\ln 2 - \tfrac12 \approx 0.193$, and you can get it without a new integral. Breaking the shorter piece leaves the longer one untouched and therefore over a half, which fails outright. So the only way to win is to pick the longer piece, which happens half the time, and the answer is exactly half of (5). Choosing at random costs you more than the original problem did.

These three numbers are the real point of the exercise. The stick, the pieces and the triangle are identical in all three, and the answers are 0.25, 0.386 and 0.193. The protocol is not a detail of the setup, it is the problem, and a candidate who writes down the sample space before answering never mixes them up.

More than three pieces

Break the stick at $n-1$ uniform points and ask whether the $n$ pieces close into a polygon. Condition (1) generalises without effort: an $n$-gon exists exactly when no piece exceeds half the perimeter. A single spacing exceeds $\tfrac12$ with probability $(1/2)^{\,n-1}$, two pieces can never both do it, so the failures are disjoint and simply add:

At $n = 3$ it returns the quarter. At $n = 4$ it is a coin flip, at $n = 5$ it is $\tfrac{11}{16}$, and it climbs to 1 as the pieces get numerous and uniformly small. Halphen published this in 1873, along with the observation that the two natural ways of randomising the breaks agree.

Formula (6) may look familiar if you have met the problem of three random points on a circle catching the centre. That is not a coincidence and it is not an analogy: bend the stick into a loop and the breaks become uniform points on a circle, a piece longer than half the stick becomes an arc longer than half the circle, and both problems reduce to Wendel’s question of whether all the points fit inside one semicircle. Same count, same $n/2^{\,n-1}$, two problems that look nothing alike.