A Cambridge exam question
The puzzle is in the book of solutions to the 1854 Senate House examination, Cambridge's mathematics exam. There a rod is marked at random at two points, and you are asked to prove that the chance of a triangle is 1/4.
A stick 1 metre long is broken in two places, each chosen at random and independently of the other. That gives three pieces. You try to make a triangle out of them by joining their ends. What is the chance it works?
It feels as if three pieces will almost always fit, barring bad luck. But there is one condition: every piece must be shorter than half the stick. If one is longer, the other two together are too short to meet. That condition fails three times out of four.
Cambridge students got this puzzle in an exam as early as 1854. It is short, but it teaches two things: how to check whether a triangle will close, and how to count chances when the random thing is a point on a line.
A straight stick 1 metre, or 100 cm, long. Its thickness does not count: it is just a line segment.
Two points are picked on the stick at random: each one uniformly along its whole length, independently of the other. Every spot is equally likely. The stick is broken at these two points, giving three pieces.
You try to make a triangle from the three pieces: each piece becomes a side, and the ends of neighbouring sides are joined. No further breaking and no overlapping.
What is the chance that a triangle can be made?
"Three pieces will nearly always fit, well, say half the time"
A triangle forms only if every piece is shorter than 50 cm. And at least one piece is longer than half three times out of four. The chance of a triangle is 1/4.
Two random breaks, three pieces. Pick an answer before reading on.
A breakdown of your answer will appear here.
Hard to decide? Open the hints one at a time: the easy one first, then stronger ones.
First work out when three segments can make a triangle at all. Lay the longest one down as the base: what must the other two do?
The two shorter pieces together are the whole stick minus the longest. So how must the longest piece compare with half the stick?
Fix the first break, say at 30 cm. Which window must the second break hit so that no piece exceeds 50 cm? How wide is that window on average?
The stick is already broken in one place, and you can see where. Chance picks the second place. Place a bet: "it closes" pays +3 if you are right and −1 if not; "it won't close" pays +1 or −1. Two bots play alongside: the Optimist always bets on a triangle, the Sceptic always against. Who scores more over 10 sticks?
It gives the answer away quickly, so make your own guess first.
Each stick is one dot in each picture. On the left is a square: the first break across, the second up. On the right is Poincaré's equilateral triangle: the distances from the dot to its three sides equal the three piece lengths. Magenta dots are sticks that made a triangle, grey ones did not. The dashed line marks where all the magenta dots must land.
It shows the share of triangles at once, so guess first.
Break × break square. The dashes are two triangles, 1/8 of the square each.
Poincaré's triangle. The dashes are the middle triangle, 1/4 of the area.
First the rule for when a triangle closes. Then a count of how often that rule holds. Both steps fit into two pictures.
“In plain words” is hidden on all puzzles.
Lay the longest piece down as the base. The other two must reach from its ends to each other. If together they are shorter than the base, their ends will not meet, as on the right. And together they are the whole stick minus the base. So the base must be shorter than the rest of the stick, that is shorter than half, 50 cm. This card says when a triangle closes but not yet how often. The next one does.
Say the first break is at 30 cm. Then the left piece is already shorter than half. For the other two to be under 50 cm as well, the second break must land right of the middle, but no further than 30 + 50 = 80 cm. The window is 30 cm wide, as long as the left piece. The closer the first break is to an end, the narrower the window; even at best, in the middle, it is only 50 cm. Averaged over all first breaks the window is 25 cm wide, a quarter of the stick.
“Rigorous” is hidden on all puzzles.
Split the stick into ten equal parts and take the first break in the middle of each. The window width is the length of the shorter of the two pieces after the first break, and the chance of a triangle for that stick is the window width divided by 100 cm. The average of the ten rows is exactly 25%. Finer parts do not change the average: the window grows evenly from 0 to 50 cm and back, so on average it is half its maximum.
Each pair of breaks is a point in a 1 × 1 square, and all points are equally likely, so a chance equals an area. If the first break x is left of the second y, a triangle forms when x < 1/2, y > 1/2 and y − x < 1/2. That is a right triangle with legs 1/2, area 1/2 · 1/2 · 1/2 = 1/8. Its twin covers the case where y is on the left. Together 1/8 + 1/8 = 1/4.
Take an equilateral triangle of height 1. For any point inside, the distances to the three sides add up to the height, 1 (Viviani's theorem). So every point is a way to split the stick into three pieces, and vice versa. Two uniform breaks scatter points evenly over the whole triangle.
A piece longer than half means the point lies closer to a corner than the midline. The three midlines cut the triangle into four equal triangles. The three corner ones mean "one piece too long", the middle one means "a triangle closes". It takes up exactly 1/4 of the area.
A stick of length 1, the first break x uniform on [0, 1]. Given x, the chance of a triangle equals the window width, min(x, 1 − x).
Another rule: break once, then break the longer piece (length 1 − x for x < 1/2) at a uniform point. The window is the same, x, but is now divided by the longer piece's length.
n pieces from n − 1 uniform breaks make an n-gon when the longest is under half. The event "piece i is longer than half" has probability (1/2)ⁿ⁻¹, and no two such events can happen together.
“Deeper” is hidden on all puzzles.
Break the stick at n − 1 random spots. The left piece is longer than half if all n − 1 breaks fall in the right half: a chance of (1/2)ⁿ⁻¹. The piece lengths are interchangeable: shuffling the pieces does not change any probability (mathematicians say the split is uniform on the simplex, or Dirichlet with all parameters 1). So any piece is longer than half with the same chance.
Two pieces cannot both be longer than half, since together they would be longer than the stick. So the chances simply add up: some piece is too long with probability n · (1/2)ⁿ⁻¹. For three pieces that is 3/4, and the triangle gets the rest, 1/4.
The oldest known statement is in the book of solutions to the Senate House examination, Cambridge's mathematics exam, for 1854: a rod is marked at random at two points and divided into three parts; show that the chance of forming a triangle is 1/4. Henri Poincaré, in his textbook "Calcul des probabilités" (1896), treated it with the equilateral triangle you saw above.
Martin Gardner brought it to a wider audience in his "Mathematical Games" column in Scientific American (October 1959, "Probability and Ambiguity"). The column's point: the answer depends on how exactly the stick is broken "at random".
Mathematically: what happens if you break the stick into n pieces and build a polygon. Logically: the same stick gives different answers if "at random" means different things. Practically: a random split is a simple baseline for market shares, seats in a parliament or species counts.
Mathematics
n segments make an n-gon by the same rule: the longest must be shorter than the sum of the rest, that is shorter than half the stick. The chance is 1 − n/2ⁿ⁻¹: 1/2 for four pieces, 11/16 for five, already 251/256 for ten, over 98%. The more breaks, the smaller the pieces, and a piece that is too long becomes rare.
The triangle is the worst case: only three pieces, and one of them very often takes more than half. On average the longest of three takes 11/18 of the stick, about 61%.
Logic
The bar is the simulated share of triangles, the tick the exact value.
The words "at random" do not define the problem on their own: you have to say what exactly is uniform. It is the same lesson as in Bertrand's paradox with the random chord. That is why the puzzle says outright that both points are uniform and independent.
Practice
Imagine parliament seats split among parties completely at random: like a stick broken at n − 1 uniform points. "Nobody has a majority" is exactly "the pieces make a polygon". With three parties a coalition is needed only 1/4 of the time, and 3/4 of the time one party holds over half the seats. The average shares of three parties from largest to smallest are 11/18, 5/18 and 2/18, that is 61%, 28% and 11%. Check them against the bars on the left.
Real elections are not random, so this model is a reference point, not a forecast: it says how much inequality to expect "for no reason at all". That is how the ecologist Robert MacArthur used it in 1957: he compared the abundances of bird species with the shares of a broken stick to see where nature departs from a random split.
The puzzle is in the book of solutions to the 1854 Senate House examination, Cambridge's mathematics exam. There a rod is marked at random at two points, and you are asked to prove that the chance of a triangle is 1/4.
In his textbook "Calcul des probabilités" (1896) Henri Poincaré showed that the three pieces can be seen as the distances from a point to the sides of an equilateral triangle. Since then that picture appears in almost every solution.
Martin Gardner used the broken stick in his column "Probability and Ambiguity" (Scientific American, October 1959) as an example of an answer that depends on how you break.
Even when a triangle closes, it is usually obtuse. An acute one comes up with probability 3 ln 2 − 2 ≈ 7.9%, only about a third of all the triangles that close at all (as computed by Ionaşcu and Prăjitură, 2010).
Each of the three pieces is a third of the stick on average. But the longest takes 11/18 on average, about 61%, and the shortest only 1/9, about 11%. That is why the longest so often exceeds half.
In "On the relative abundance of bird species" (PNAS, 1957) the ecologist Robert MacArthur proposed the broken-stick model: an environment's resources are split among species like a stick at random points. The model has no parameter to fit at all.
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Is “School maths” the right level for this puzzle?
Thank you! This helps set the levels right.
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