Chasing the Extremes

Before doing any algebra, let’s look at the two most extreme ways the needle can land.

Suppose the needle lands parallel to the cracks, running along the length of the plank. It lies entirely inside one plank and never touches a crack. The chance of crossing is 0.

Now suppose it lands perpendicular to the cracks, pointing straight across the planks. A needle of length lying across a plank of width cannot fit inside a single plank, so it always crosses. The chance of crossing is 1.

Every real toss lands somewhere between these two extremes, so the answer must lie somewhere between 0 and 1. To find out where, we need a way to describe a tilted needle.

The View from the Bottom

To solve this, let’s track just one point: the bottom tip of the needle, meaning whichever end lies lower down the floor in our picture.

When the needle lands, that bottom tip sits somewhere on a plank. Let’s call its distance from the lower edge of that plank . Because the tip can land anywhere on the wood, can be any value between and .

Because the bottom tip is the lowest point of the needle, the rest of the needle points upward from it. That means the needle can never cross the crack below the tip. The only crack it can cross is the next one up, at height .

Take the perpendicular case first. The top tip sits at height , and it crosses the next crack when , which simply means . Since landing exactly on the crack has zero chance, the needle crosses with probability exactly 1, just as we found above.

But most needles don’t land perpendicular. Let’s call the tilt : the angle between the needle and the perpendicular to the cracks. It runs from (perpendicular) to (parallel).

How far up does a tilted needle reach? Picture a right-angled triangle with the needle as its slanted side, of length . The vertical side of that triangle, the part of the needle’s length that actually points across the planks, is the side next to the angle . By basic trigonometry, its length is . The other side, , runs along the plank and doesn’t help the needle reach the crack at all.

So the top tip reaches a height of , and the needle crosses the next crack when that is more than the width of the plank:

Rearranging tells us exactly where the bottom tip needs to land:

Every Possible Drop in One Rectangle

Now let’s put this on a graph. The horizontal axis is the angle , from to . The vertical axis is the bottom tip’s height , from to .

Every possible way the needle can land is now a single point in a rectangle. Because we toss the needle without aiming, every position along the plank and every tilt is equally likely, so every point in the rectangle is equally likely too. That turns probability into area. The total area of the rectangle is its width times its height:

The winning drops, where the needle crosses a crack, sit above the curve and below the top of the plank at .

The winning region has a curved edge, so we find its area with calculus. Slice it into thin vertical strips. At angle , a strip starts at the curve, , and runs up to , so its height is

Adding up all the strips from to :

To get the probability, divide the winning area by the total area. The s cancel:

A Second View: The Centre Method

There is a second way to solve this, by looking at the centre of the needle instead of its tip.

Let be the distance from the centre of the needle to the nearest crack, whether that crack is above or below. The furthest the centre can ever be from a crack is the middle of the plank, so runs from to .

The centre sits halfway along the needle, so each half has length . Tilted at angle , each half reaches across the planks, by the same right-triangle argument as before.

The needle crosses when the nearest crack is within reach of the half pointing towards it:

We can graph this just as before. The angle still runs from to , but the height now only runs from to . The total area of this rectangle is

This time the winning region sits under the curve .

A strip at angle now runs from up to the curve, so its height is :

Divide the winning area by the total area, and the s and the s cancel again:

Two different pictures, two different rectangles, the same answer. The shaded regions look nothing alike, yet each one fills exactly the same fraction, , of its own rectangle.

Why No Paradox This Time?

If you read our article on Bertrand’s paradox, this might make you a little nervous. There, three reasonable ways of choosing “a random chord” gave three different answers. Here, two different methods agree. Why can we trust them?

The difference is that this time, “random” has only one sensible meaning. Bertrand’s question never said which quantity was chosen at random. Here we named our random quantities, a position and an angle, and the physical toss tells us how each one behaves. The floor looks exactly the same if you shift your view across by one plank, so the needle has no reason to prefer any spot within a plank: its position is uniform. And a tossed needle has no reason to prefer any direction: its angle is uniform. Both methods measure the same toss with these same two uniform quantities, which is why they cannot disagree.

Turning the Answer Around: Finding π with Needles

Here is where Buffon’s needle gets its fame. If the probability of crossing is , then we can turn the formula around. Toss many needles, count how many cross, and

You can estimate with nothing but a floor and a box of needles.

It works, but slowly. The estimate wanders wildly for the first hundred or so tosses and only creeps towards after that. In this simulation, 20,000 tosses got three decimal places right. Random experiments like this improve with the square root of the number of tries, so every extra correct digit costs about a hundred times more tosses.

Which makes one historical result very strange. In 1901, the Italian mathematician Mario Lazzarini reported tossing a needle 3,408 times, using a needle five-sixths as long as the plank width. He counted 1,808 crossings, which gives

correct to six decimal places. With 3,408 honest tosses, an estimate typically misses by a few hundredths. And look closely at his numbers: and . Those are exactly the multiples needed to produce the famous fraction . It looks less like luck and more like knowing exactly when to stop.

The Hidden Circle

The answer is roughly 0.6366, about a 64% chance. But the decimal isn’t the interesting part. Look at the exact answer, .

We dropped a straight needle onto straight cracks. There are no circles on the floor and no circles on the needle. So where did come from?

Look back at the calculation. The winning area was plain , with no in it at all. All of the came from the width of the rectangle: the range of possible angles, a quarter-turn, which measures in radians. A needle that can point in any direction is quietly picking a point on a circle. The circle was never on the floor. It was hiding in the word “random”.

In an earlier puzzle, a logarithm in the answer was the fingerprint of a curved region. Here, is the fingerprint of a random direction. Whenever an answer depends on an angle that could point anywhere, a circle is hiding somewhere in the setup, even if none is drawn.

Your Turn

1. What if the needle is shorter than the plank is wide, with length ? Only one thing changes in the bottom-tip method. Once you have your formula, check it against Lazzarini’s numbers: with , 3,408 tosses and 1,808 crossings, it should give exactly .

2. What if the needle is longer than the plank is wide? Try drawing the rectangle again. What goes wrong with the curve, and how would you fix it?

A straight needle, straight cracks, and still a circle hiding in the answer.

Three Programmers, Three Pictures

Suppose three programmers are each asked to “draw a random chord” and each writes a different, perfectly reasonable program. Here is what 450 chords from each program look like, with the long chords in orange.

The three pictures are not the same. Let us see where each answer comes from.

A Useful Trick: Rotate the Triangle

Each method below starts by picking something at random: a point on the edge, or a direction for a radius. That random choice will almost never line up neatly with the triangle. But the triangle can be turned. Rotating it about the centre changes no lengths at all, so after each random choice we are free to rotate the triangle into whatever position makes the picture easiest to read. We will use this trick twice.

Method 1: Random Endpoints

The first programmer picks two random points on the edge of the circle and joins them.

Wherever the first point lands, rotate the triangle so that one of its corners sits exactly there. Now look at the chord from that corner to the second point. It is longer than a side of the triangle only if the second point lands on the far arc, between the other two corners. The three corners split the circle into three equal arcs, so that far arc is exactly one-third of the edge.

Method 2: Random Radius

The second programmer draws a radius in a random direction, picks a random point along it, and draws the chord through that point at right angles to the radius.

This time, rotate the triangle so that one of its sides is perpendicular to the chosen radius. That side then crosses the radius at exactly its halfway mark. (The centre of the circle is also the centre of the triangle, which sits two-thirds of the way down each altitude. The altitude is long, so the centre is from each corner and from each side.)

A chord crossing the radius closer to the centre than that side is longer than the side; a chord crossing farther out is shorter. So the chord is long exactly when the random point lands on the inner half of the radius.

Method 3: Random Midpoint

The third programmer says a chord is determined by its midpoint. They pick a random point anywhere in the area of the circle and draw the one chord that has that point as its middle.

The chord’s distance from the centre is simply its midpoint’s distance from the centre. By the same reasoning as Method 2, the chord is long exactly when that distance is less than , that is, when the midpoint falls inside a smaller circle of radius . (That small circle is exactly the circle that fits snugly inside the triangle.) Its area compared with the big circle is

The Hidden Assumption

So which programmer is right?

Each of them is right about their own program. None of the arithmetic is wrong. Notice, too, that all three methods use the same test: a chord is long exactly when it passes within of the centre. They disagree only about how often chords pass close to the centre, and that depends entirely on how each program generates its chords.

This is the real lesson. The word “uniform” only means something once you say which quantity is uniform. A chord can be described by its two endpoints, by its distance from the centre, or by its midpoint. Spreading one of these descriptions evenly does not spread the others evenly. Choose a different description, and you have quietly asked a different question.

Probability is not just a property of the shapes on the page. It is a property of the process that generates them. Say “at random” without naming that process, and the answer bends to whatever assumption you made without noticing.

If this feels familiar, it should. In Break the Longer Piece, describing the second break in the wrong way gave the wrong answer of . The fix was to describe it in the way the stated process actually made uniform. Bertrand’s paradox is the same trap, with nothing in the problem to tell us which description to use.

Is There a Best Answer?

The story does not quite end at “everyone is right.” In 1973 the physicist E. T. Jaynes asked a more physical question: what happens if you actually toss long straws onto a circle drawn on the floor? He argued that the answer should not change if the circle is moved slightly or drawn a little larger or smaller, and that this requirement singles out Method 2. Tossed straws, he predicted, would give . The debate over whether his argument settles the paradox continues, and his paper, The Well-Posed Problem, is well worth looking up.

Why Did Our Earlier Puzzles Work?

In The Geometry of a Missed Connection, we drew a square and simply declared that every point in it was equally likely. In the broken-stick problem, we did the same. Why was that safe there, when it falls apart here?

Because in those problems, the question itself named the random quantity. Alice and Bob each arrive at a uniformly random time, and the two axes of our square were those arrival times. The stick was broken at uniformly random positions, and the square’s axes were those positions. “Every point is equally likely” was not an extra assumption we slipped in; it was the problem statement, redrawn as a picture.

Bertrand’s question names no such quantity. “A random chord” does not say whether the endpoints, the distance, or the midpoint should be spread evenly, so the problem has no single answer until we decide.

Before measuring any area, ask which quantity the problem says is random. Once that is fixed, the geometry takes care of the rest.

The Trap

At first, this sounds like a nightmare of moving targets. Alice’s and Bob’s arrival times are completely independent of each other; neither one knows or cares when the other shows up. But whether they meet depends on both times at once, and the 15-minute window slides along with whoever happens to arrive first. Trying to track that sliding window with brute-force arithmetic or case-by-case calculus feels like trying to nail jelly to a wall.

But if we stop treating this as a problem about time and start treating it as a problem about space, the answer simply draws itself.

Mapping Time to Space

Imagine a square graph. Let the horizontal axis be Alice’s arrival time, measured in minutes after 12:00, from 0 to 60. Let the vertical axis be Bob’s arrival time, also from 0 to 60.

Every possible pair of arrival times is now just a single point inside this square. If Alice arrives at 12:20 and Bob at 12:30, that whole lunch hour becomes the point . Because both arrivals are independent and uniformly random, every point in the square is equally likely, so probability becomes a matter of measuring area. The total area is

When do they actually meet? Exactly when the gap between their arrival times is 15 minutes or less:

where is Alice’s arrival and is Bob’s. On our graph, this “safe zone” is a diagonal strip running from the bottom-left corner to the top-right, bounded by the lines and .

The Corners of Failure

Measuring that diagonal strip directly is a bit awkward. Measuring the region where they fail to meet, however, is easy.

The parts of the square outside the strip form two right triangles, one in the top-left corner and one in the bottom-right:

• The top-left triangle is where : Bob arrives more than 15 minutes after Alice. Bob is too late, and Alice has already left. Both of its legs are 45 minutes long.
• The bottom-right triangle is where : Alice arrives more than 15 minutes after Bob. Alice is too late. It also has two legs of 45 minutes.

The area of each triangle is half its base times its height:

Together, the total “failure area” is .

The Answer

To find the success area, subtract the failures from the whole square:

Divide that favorable area by the total area:

By turning time into a square, a moving-target probability problem reduces to finding the area of two triangles. No calculus required.

Chasing the Extremes

Before trusting any answer, it helps to push the problem to its limits and see whether the picture still makes sense.

What if neither friend waits at all? Then they meet only if they arrive at exactly the same instant. The strip shrinks to the thin diagonal line , which has no area. The two failure triangles swell until they fill the entire square. The probability of meeting is .

What if each friend is willing to wait the full 60 minutes? Then whoever arrives first is still sitting there when the other walks in, no matter what. The strip widens until it swallows the whole square, the failure triangles shrink to nothing, and the probability is .

So as patience grows from 0 to 60 minutes, the chance of meeting climbs steadily from certain failure to certain success, and a 15-minute wait lands them at , just short of a coin flip.

Your Turn

Replace the 15 minutes with any waiting time , anywhere from 0 to 60. What is the probability that Alice and Bob meet now? Your two failure triangles still do all the work, and your formula should agree with both extremes above.

And once you have it, one more question. How patient do Alice and Bob need to be to have a better than even chance of getting that coffee?

Why you might expect a lot

There is a good reason to think this should help, and it is worth saying out loud before we compute anything.

A triangle fails when one piece is too long — at least half the stick. In the original problem nothing stops that happening. Both breaks land wherever they like, and quite often they land close together, leaving one enormous piece and two fragments.

The new rule attacks exactly that failure. By always breaking the longer piece, we are deliberately going after the piece most likely to be the culprit. It feels like it ought to fix most of the trouble.

Hold on to that intuition. We will come back to it.

Setting it up

Let the stick have length and let the first break leave a shorter piece of length . Since it is the shorter piece, . The longer piece has length .

Now for the second break, and here is where the problem gets interesting.

The trap

The obvious move is to copy what worked last time: call the second break point , write down the three piece lengths, and draw the picture.

Try it. You get pieces , and , the triangle inequality gives you three tidy conditions, and you can shade a region and measure it. It is quick, and it gives an answer of one half.

That answer is wrong, and the reason is worth more than the problem.

Comparing areas only measures probability when every point in the picture is equally likely. In the original problem it was: both breaks were uniform on the same interval, independently. Here they are not. The second break lives on the longer piece, and how long that piece is depends on where the first break landed. A given value of is not equally available for every — for small the long piece is nearly the whole stick, for near it is only half of it.

So the two variables are tangled together, and the area of a region no longer tells us anything about how often we land in it.

Untangling them

There is a fix, and it is a small one. Instead of measuring the second break as a length, measure it as a fraction.

Let be the proportion of the way along the longer piece where the second break falls, so . Whatever the long piece happens to measure, the break is equally likely to fall anywhere along it — so is uniform on , and it no longer cares what is.

The two variables are independent again. The picture will work again.

The three pieces are now

A quick check: they sum to .

Applying the triangle inequality

Same rule as last time. A triangle forms exactly when every piece is under half. Taking them one at a time, and dividing through by where it helps:

Piece . From we get , that is which we already knew — the first break always leaves a shorter piece.

Piece . From :

Piece . From :

Look at those last two. They are the same statement twice, once for each sign. A number beats both a quantity and its negative exactly when it beats the absolute value, so together they say

Unwrapping the absolute value and tidying up:

So has to land in a band centred on — cutting the long piece exactly in half — and that band has width

This is already telling us the story. If the first break barely nicked the end, is tiny, the band is almost nothing, and you would have to halve the long piece with near-perfect accuracy to get a triangle. If the first break landed near the middle, is close to , the band is the full height, and every second break works.

The picture

Plot horizontally from to , and vertically from to . Every possible outcome is a point in that rectangle, and now — with measured as a fraction — every point really is equally likely.

The green rectangle is everything that can happen. The orange band is everything that gives a triangle: pinched to a point at the left, opening to full height at the right.

The rectangle has area , and that is what we divide by — the same move as last time. Then, the first break could equally have landed on either side of the middle, and we chose to call the shorter piece . That convention throws away half of the possibilities, exactly as insisting threw away half the square in the previous problem. So

Where the method finally strains

Last time the orange region was a triangle, and the area of a triangle is . Anyone can finish from there.

This time the edges are curves. The band width is a hyperbola, and no amount of clever chopping into triangles will give its area. Adding up the widths of infinitely many infinitely thin vertical strips is precisely what an integral does:

Substituting , so that and , and noting that gives while gives :

Dividing by the sample space:

About two in five

So the answer moved from to roughly .

It did help. Deliberately attacking the piece most likely to ruin things does improve your odds, by more than half again. But look how modest that is against the intuition we started with. The new rule felt like it should fix most of the failures. It fixes a little over a third of them, and the majority of attempts still end with three pieces that will not close.

The reason is visible in the picture. Breaking the longer piece protects you against one failure — the long piece staying long — but it does nothing about the other. If the first break leaves a very short piece, that piece is already too short to reach across, and no amount of care with the second break can rescue it. That whole left-hand region of the rectangle is lost before the second break is even made.

The fingerprint

One last thing, and it is the part I like most.

The answer contains a logarithm. Snap a stick, snap it again, and falls out.

Logarithms do not come out of the areas of triangles. They cannot. If your region has straight edges, you can always finish with school geometry and you will always get an answer built from ordinary arithmetic. The moment a logarithm appears, it is proof that something curved was involved.

So that is not decoration. It is a fingerprint — evidence, left at the scene, that this problem was never going to yield to the method that solved the first one. And you could have predicted it before computing anything, simply by noticing that the width of the band was rather than something straight.

That is worth carrying forward. The shape of an answer often tells you what kind of argument had to produce it.

The perfect break

When most people first meet this problem, the mind reaches for the most symmetric answer: break the stick into three equal pieces.

If the stick is 3 metres long and you snap it at the 1-metre and 2-metre marks, you get three pieces of 1 metre each. Do they form a triangle? Absolutely — a perfect equilateral one.

So at least one break works. Which raises the real question: what else works?

To find out, it helps to look at the opposite extreme — a break that fails completely.

The terrible break

Snap the stick so that one piece is enormous — say 90% of the total length — leaving two tiny fragments.

Lay the long piece flat and attach the two fragments to its ends. They dangle. They can never reach across the gap to meet each other.

Put the two breaks side by side and a rule starts to emerge:

For the pieces to connect, no single piece can be too long.

“Too long” is doing a lot of work in that sentence. Let’s make it precise.

The detour rule

Think of two points, and , as two towns.

The edge is the straight road between them — and the straight road is always the shortest one. The path from to and then to is a detour. It has to be longer.

That is the whole idea:

Two sides of a triangle, added together, must exceed the third. And since any of the three sides could play the role of “the straight road,” this has to hold for all three pairings at once.

Mathematics has a name for this: the triangle inequality.

Testing it on the terrible break

Take a 3-metre stick. The huge piece is 2.7 metres, and the two fragments are 15 cm each.

Here cm, cm and cm. Check the pairing that involves the two short pieces:

It fails. And one failure is enough — the triangle inequality has to hold for every pairing, not just a convenient one. The other two pairings do hold here ( is comfortably more than ), but that rescues nothing. A chain breaks at its weakest link.

The breaking point

So where exactly is the boundary between “works” and “fails”?

The three pieces always add up to the whole stick — 100% of it. So ask: what if the longest piece is exactly half?

Then the other two pieces add up to exactly half as well. Try to build the triangle and they lie perfectly flat along the long piece. Two sides collapse onto the third. You get a straight line, not a triangle.

That is the edge. And it gives us the golden rule:

A triangle forms if and only if no single piece is half the stick or longer.

Now we can count.

The complete solution

Let the stick have length .

Drop two random break points on it. Call the one nearer the left end , and the other . So — not because we have restricted anything, but because we are free to choose which label goes where. Both points are chosen uniformly on .

This gives three pieces:

And of course .

Turning probability into area

Plot the two break points on a graph: on the horizontal axis, on the vertical. Both run from to , so every possible pair of breaks is a single point somewhere inside a square.

But we agreed to always call the smaller number . So always — and the line , running diagonally from to , splits the square in two:

• Below the line lie the cases where . Our labelling choice threw this half away.
• Above the line lie the cases where . This is our entire universe of possibilities.

That upper-left triangle, shaded green, is our sample space:

Now shade only the points that satisfy all three conditions — , , and . They carve out a smaller triangle with corners at , and , shown in orange. Its two legs are each long:

And geometric probability is nothing more than a ratio of areas:

One in four

Snap a stick at two random points, and one time in four you can build a triangle from the pieces.

Not one in two. Not one in three. One in four — and notice where the answer came from. We compared the size of two shaded regions. No integrals. No counting. Just area.

That is the quiet power of geometric probability: it takes a question about randomness and turns it into a question about shape.

Before you go

Here is a variant worth sitting with.

Instead of dropping both break points at once, do it in two stages. Break the stick at one random point. Then look at the two pieces, pick the longer one, and break that at a random point.

Same stick. Same three pieces. Same triangle question.

Is the answer still ?

It is not. And the reason why is worth a post of its own.