The Broken Stick Puzzle, math that surprises you lesson

A stick of length $1$ is cut at two random points, which leaves three pieces. Sometimes the pieces can be joined end to end to make a triangle, and sometimes one piece is too long for the other two to reach around. How often does a triangle come out? This page finds the exact chance when the two cut points are chosen independently and uniformly along the stick. The triangle condition becomes a region in a square of possible cut positions, and the area of that region, one quarter, is the probability.

Key terms

Broken stick problem

Definition The broken stick problem asks for the probability that three pieces of a stick can be the sides of a triangle. In the version here, a stick of length $1$ is cut at two points $X$ and $Y$ , measured from its left end. The two points are independent, and each has the uniform distribution on the interval from $0$ to $1$ . The cuts produce three pieces, and a triangle with positive area can be built from them exactly when every piece is shorter than $1/2$ . The probability of this is

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

The answer belongs to this way of cutting, and a different procedure is a different experiment. For example, if the stick is cut once at a uniform point and the longer of the two pieces is then cut at a uniform point along it, the probability is $2\ln 2 - 1 \approx 0.386$ . If instead a fair coin decides which of the two pieces gets the second cut, the probability is half of that, about $0.193$ , because cutting the shorter piece leaves the longer one whole and too long. So the words “break a stick at random” do not fix an answer until they say how the cuts are chosen.

In plain words Two random cuts often leave one piece that is at least half the stick, and then the other two pieces together are too short to reach around it. That happens three times in four, so over many tries only about one stick in four gives a triangle.

Example Cuts at $0.3$ and $0.7$ give pieces of lengths $0.3$ , $0.4$ and $0.3$ , all shorter than $1/2$ , so they form a triangle. Cuts at $0.1$ and $0.9$ also fall on opposite sides of the midpoint, but they give pieces $0.1$ , $0.8$ and $0.1$ , and the middle piece is too long. The order of the cuts does not matter, so a first cut at $0.7$ and a second at $0.3$ give the same three pieces as cuts at $0.3$ and $0.7$ .

Triangle inequality

Definition The triangle inequality says that in every triangle each side is shorter than the sum of the other two. For side lengths $a$ , $b$ and $c$ ,

\begin{gathered} a < b + c, \quad b < a + c, \\[0.4em] c < a + b. \end{gathered}

Conversely, three positive lengths that satisfy all three strict inequalities are the side lengths of a triangle with positive area. It is enough to check the longest length, because the other two inequalities then hold automatically. If the longest length equals the sum of the other two, the three segments can only lie flat along one line, which is called a degenerate triangle and has area $0$ . For any three points $A$ , $B$ and $C$ , the same idea reads $AC \le AB + BC$ , with equality exactly when $B$ lies on the segment from $A$ to $C$ . When three lengths add up to $1$ and $L$ is the longest, the other two add up to $1 - L$ , so the condition $L < 1 - L$ becomes $L < 1/2$ .

In plain words Two sides of a triangle must be long enough together to reach around the third. If the longest piece is half the stick or more, the other two together are no longer than it, and they can only lie flat against it.

Example The lengths $0.2$ , $0.3$ and $0.5$ fail, since $0.2 + 0.3 = 0.5$ exactly, and the two short pieces lie flat along the long one. The lengths $0.25$ , $0.35$ and $0.40$ work, since the longest, $0.40$ , is less than $0.25 + 0.35 = 0.60$ . Both sets add up to $1$ , and the rule about halves agrees. Every piece of the second set is shorter than $1/2$ , while the longest piece of the first set equals $1/2$ .

Uniform distribution

Definition A random number $X$ has the uniform distribution on the interval $[0, 1]$ when, for all numbers $c$ and $d$ with $0 \le c \le d \le 1$ ,

P(c \le X \le d) = d - c,

the length of the interval. Intervals of equal length are equally likely, and a single value such as $X = 0.3$ has probability $0$ , so including or leaving out endpoints changes nothing. Two random numbers $X$ and $Y$ are independent when, for any intervals $I$ and $J$ , the probability that $X$ lies in $I$ and $Y$ lies in $J$ is the product of the two separate probabilities. For independent $X$ and $Y$ , each uniform on $[0, 1]$ , the probability that the point $(X, Y)$ lands in a rectangle inside the unit square is the area of the rectangle, and the same holds for every region of the square that has an area.

In plain words A uniform cut is equally likely to land in any stretch of the stick of a given length, wherever that stretch is. Independence means that the first cut tells you nothing about where the second one lands.

Example Both cuts land at $1/2$ or below with probability $\tfrac12 \cdot \tfrac12 = \tfrac14$ , and then the right piece has length at least $1/2$ , so no triangle forms. The uniform assumption matters. If each cut were instead uniform on the middle half of the stick, from $1/4$ to $3/4$ , the two cuts could never be more than $1/2$ apart. Apart from cases of probability $0$ , a triangle would then form exactly when the cuts fall on opposite sides of the midpoint, which has probability $\tfrac12 \cdot \tfrac12 + \tfrac12 \cdot \tfrac12 = \tfrac12$ , not $1/4$ .

Probability as area

Definition Suppose a random point is uniformly distributed in a region $S$ of the plane with finite positive area, which means that regions of equal area inside $S$ are equally likely to contain it. Then the probability that it lands in a region $A$ inside $S$ is

P(A) = \frac{\text{area of } A}{\text{area of } S}.

For two independent cuts, each uniform on $[0, 1]$ , the point $(X, Y)$ is uniform in the unit square, which has area $1$ , so every probability about the cuts is an area. A line or a single point has area $0$ , so events such as $X = Y$ or a piece of length exactly $1/2$ have probability $0$ .

In plain words Draw every possible pair of cuts as a point in a square. The pairs that give a triangle fill part of the square, and the share of the square they fill is the probability.

Example Take the half of the square where $x < y$ , so the pieces are $x$ , $y - x$ and $1 - y$ . All three are shorter than $1/2$ exactly when

\begin{gathered} x < \tfrac12, \quad y > \tfrac12, \\[0.4em] y - x < \tfrac12. \end{gathered}

On the stick these say that the left cut comes before the midpoint, the right cut comes after it, and the two cuts are less than half the stick apart. Together they cut out the right triangle with corners $(0, \tfrac12)$ , $(\tfrac12, \tfrac12)$ and $(\tfrac12, 1)$ . Both of its legs have length $1/2$ , so its area is $\tfrac12 \cdot \tfrac12 \cdot \tfrac12 = \tfrac18$ . Swapping the names of the two cuts gives a matching triangle in the half where $y < x$ , so the whole success region has area $1/8 + 1/8 = 1/4$ , and the probability is $1/4$ . Sorting the cuts first and working only in the half where $x < y$ is the same experiment drawn differently, and then the denominator is the area of that half,

\frac{1/8}{1/2} = \frac14.

Counting only one of the two triangles while dividing by the area $1$ of the whole square gives the wrong answer $1/8$ .

Check yourself

The cuts land at $0.1$ and $0.6$ . Can the three pieces form a triangle? Would the answer to the puzzle change if cases like this one were counted as successes?

Show the answer

The pieces have lengths $0.1$ , $0.5$ and $0.4$ . The longest piece equals the sum of the other two, since $0.1 + 0.4 = 0.5$ , so the pieces can only lie flat along a line. They give a degenerate triangle with area $0$ , not a genuine triangle, because the condition that every piece is shorter than $1/2$ is strict.

Counting such cases as successes would not change the probability. A piece of length exactly $1/2$ happens only on the lines $x = \tfrac12$ , $y = \tfrac12$ and $y - x = \pm \tfrac12$ in the square, and lines have area $0$ . The probability stays exactly $1/4$ either way, but the geometry still says these pieces do not make a triangle.

Make the rule stricter. What is the probability that all three pieces are shorter than $0.4$ ?

Show the answer

In the half of the square where $x < y$ , the conditions become $x < 0.4$ , $y > 0.6$ and $y - x < 0.4$ . The lines $y = 0.6$ and $y = x + 0.4$ meet at $x = 0.2$ , so the region is the right triangle with corners $(0.2, 0.6)$ , $(0.4, 0.6)$ and $(0.4, 0.8)$ . Both legs have length $0.2$ , so its area is $\tfrac12 \cdot 0.2 \cdot 0.2 = 0.02$ . The mirror triangle in the other half doubles this, so

\begin{aligned} &P(\text{all pieces} < 0.4) \\[0.4em] &\quad = 0.04 = \frac{1}{25}. \end{aligned}

The same steps for any bound $t$ with $\tfrac13 \le t \le \tfrac12$ give legs of length $3t - 1$ and probability $(3t - 1)^2$ . This equals $1/4$ at $t = 1/2$ , as in the puzzle, and $0$ at $t = 1/3$ , since three pieces that add up to $1$ cannot all be shorter than $1/3$ .

Change the procedure. Cut the stick once, and suppose the cut lands at $0.2$ . Then cut the longer piece, of length $0.8$ , at a point chosen uniformly along it. What is the chance that the three pieces form a triangle? What is the chance if a fair coin decides which of the two pieces gets the second cut?

Show the answer

Let the second cut split the longer piece into lengths $v$ and $0.8 - v$ , where $v$ is uniform between $0$ and $0.8$ . The piece of length $0.2$ is already shorter than $1/2$ , so a triangle needs $v < 0.5$ and $0.8 - v < 0.5$ , which means $0.3 < v < 0.5$ . That interval has length $0.2$ out of $0.8$ , so the chance is

\frac{0.2}{0.8} = \frac14.

If the coin picks the short piece instead, the piece of length $0.8$ stays whole and no triangle is possible. With a fair coin the chance is $\tfrac12 \cdot \tfrac14 + \tfrac12 \cdot 0 = \tfrac18$ .

The value $1/4$ here is a coincidence of this first cut. For a first cut at a point $u$ with $0 < u < 1/2$ , the same reasoning gives the chance $u/(1 - u)$ under the longer piece rule, and averaging over a uniform first cut gives $2\ln 2 - 1 \approx 0.386$ , not $1/4$ .