Buffon’s Needle — Can Dropping Sticks Measure Pi?

Could you estimate pi without measuring a circle at all? Apparently you can — by dropping sticks onto a sheet of paper.

Best level: GCSE and A-level
Topics: Probability, geometry, experimental error, simulation and integration
Time: 30–60 minutes for an initial investigation, with extensions for larger datasets

A Surprisingly Mathematical Mess

Put a ruler, some sticks and a large sheet of paper on a table. Ask a student to estimate pi.

Their first instinct might be to draw a circle, measure its circumference and diameter, and divide one by the other.

But there is another approach: draw parallel lines, drop sticks onto them, and count how many cross a line.

There are no circles to measure. Yet pi appears in the answer.

This is Buffon’s needle problem, a beautiful connection between probability and geometry. Although the original problem uses a needle, blunt wooden sticks make a friendlier classroom version.

What appeals to me as a teacher is that students can begin the investigation before they understand the explanation. They can make a prediction, collect evidence and then ask why the result behaves as it does.

The mathematics grows out of something they have actually done.

Set Up Your Own Pi Experiment

You will need:

A large sheet of paper, taped flat to a table.
A ruler and a fine pen.
Identical straight sticks, preferably blunt-ended.
A tally sheet and calculator.
Optional: Excel or Python for the simulation extensions.

Short pieces of dowel or blunt craft sticks work well, with supervision appropriate to the group.

Measure the stick length carefully. Suppose each stick is 5 cm long.

Draw parallel lines 5 cm apart, measuring from the centre of one line to the centre of the next.

For this first version:

Stick length = distance between lines

Use a generously sized sheet. Mark a smaller rectangular area for the sticks’ centres, leaving at least half a stick length between this area and the paper’s edges. Its width across the parallel lines should span a whole number of line spacings.

This helps reduce edge problems and gives positions between the lines a fair chance.

Before starting, ask:

“Will more than half the sticks cross a line?”

Let everyone commit to a prediction.

Drop, Count and Repeat

Release one stick at a time from a modest height over the marked area.
Vary its release position and orientation, avoiding a habitual alignment.
Let it settle flat before counting.
Record whether its centreline reaches or crosses the centre of a drawn line.
Remove it before the next drop.
Record cumulative totals after 10, 100 and, using pooled results, 1,000 drops.

Agree the counting rule before starting. A stick lying close to a thick pen mark should not become an opportunity to choose whichever answer improves the estimate.

If a stick falls outside the intended setup, do not count it as a non-crossing. Follow a pre-agreed repeat rule and note how often this happens. Frequent misses suggest that the setup needs changing.

For group work, use the same stick length, line spacing and counting rules throughout. Add everyone’s crossings and drops together.

Calculate a fresh estimate from the combined totals, rather than simply averaging the groups’ estimates of pi.

The Formula That Turns Sticks into Pi

When stick length equals line spacing, the theoretical crossing probability is:

P = 2/pi

That is approximately 0.637: roughly 64 crossings in every 100 drops over the long run.

Let:

N = total number of valid drops.
C = number of sticks crossing a line.

The observed crossing proportion is:

P estimated = C/N

Rearranging gives:

pi estimated = 2 x N / C

This is the equal-length version of Buffon’s needle formula. math.utah.edu

If C = 0, the estimate is undefined. At the beginning of a short experiment, that can happen; record “not yet available” and continue.

Worked example: 1,000 drops

Suppose 641 sticks cross a line in 1,000 drops:

pi estimated = 2 x 1000 / 641

pi estimated = 2000 / 641

pi estimated = 3.120...

Rounded to two decimal places, that is 3.12.

Compared with 3.14159..., the estimate is about 0.7% low.

That is an encouraging result from a pile of wooden sticks. It is also an illustrative outcome, rather than a promise of what your own experiment will produce.

Where Is the Circle Hiding?

The lines are straight. The sticks are straight. So why does pi appear?

The hidden ingredient is rotation.

Imagine a stick lying almost parallel to the lines. It must land very close to a line to cross it.

Now turn it so that it lies at right angles to the lines. It reaches much further across the gap and has a greater chance of crossing.

The probability therefore depends on:

Where the centre of the stick lands.
The angle the stick makes with the parallel lines.

The calculation averages the crossing chance over all possible orientations. Pi enters through the mathematics of angles and rotation.

There is no circular object on the table, but circular geometry is still involved.

A-level extension: derive the result

Let L be the stick length and d the line spacing, with L <= d.

Let theta be the acute angle between the stick and the parallel lines. Let x be the perpendicular distance from the stick’s centre to the nearest line.

The stick crosses when:

x <= (L/2) x sin(theta)

For a fixed angle:

P(crossing at angle theta) = L x sin(theta) / d

Average this over uniformly distributed angles from 0 to pi/2 radians. Using plain-text notation:

P = (2/pi) x integral from 0 to pi/2 of [L x sin(theta)/d] dtheta

The integral of sin(theta) over that interval is 1, giving:

P = 2 x L / (pi x d)

When L = d:

P = 2/pi

This projection-based derivation connects the experiment directly to trigonometry and integration. Jonathan Davidson

GCSE students can explore the activity through relative frequency and rearranging a formula. A-level students can investigate why that formula works.

More Drops Help — But Progress Is Uneven

It is tempting to say:

“Keep dropping sticks and the estimate will get closer to pi.”

That needs a qualification.

With more independent, properly randomised trials, estimates tend to become more reliable. But the running estimate does not improve at every step.

It may move closer, then further away, then closer again.

A short experiment can produce a lucky answer. A much longer experiment can still be less accurate than that lucky result.

This is an excellent opportunity to distinguish accuracy in one run from reliability across repeated runs.

Record your own results:

Total drops	Cumulative crossings	Estimated pi
10	Record your result	20 / crossings
100	Record your result	200 / crossings
1,000	Record your result	2000 / crossings
10,000	Record your result	20000 / crossings

Leave the estimate blank if there have been no crossings.

Plot estimated pi against total drops and add a horizontal reference line at 3.14159. Look for fluctuations that generally become smaller, rather than a perfectly smooth approach.

For a stronger comparison, repeat whole experiments of the same size and examine the spread of their estimates.

Why extra accuracy takes so much work

For large samples under the ideal model, typical random error decreases approximately in proportion to:

1/sqrt(N)

Reducing typical error by a factor of 10 therefore takes roughly 100 times as many drops.

Buffon’s needle is a fascinating way to estimate pi. Obtaining many accurate decimal places requires considerable patience.

A Poor Result Can Be a Good Lesson

Suppose your estimate stubbornly stays near 3.5.

Before blaming bad luck, examine the method.

Are the sticks really the same length as the line spacing? Are they straight? Do they tend to land in one orientation? Are positions spread across the gaps? Are sticks bouncing into one another?

The mathematical model assumes:

Independent trials.
Uniformly random orientations.
Uniformly random centre positions across a line spacing.
Independence between position and angle.

A person dropping sticks does not automatically achieve all of that.

Repeating a biased method produces more evidence of the bias; it does not remove it.

For example, sticks that preferentially land parallel to the lines will generally produce fewer crossings. Since crossings appear in the denominator, the estimated pi will tend to be too high.

This is the kind of discussion I value in practical teaching. A disappointing number can prompt better scientific thinking than an impressive answer obtained without questioning the method.

Compare the Tabletop Experiment with Excel

A spreadsheet lets students simulate thousands of drops and inspect the rules behind each result.

Use a stick length of 1 unit and a line spacing of 1 unit.

Enter these headings and formulas:

Column	Heading	Formula in row 2
A	Trial	`=ROW()-1`
B	Distance to nearest line	`=RAND()/2`
C	Angle in radians	`=RAND()*PI()/2`
D	Crossing: 1 or 0	`=IF(B2<=0.5*SIN(C2),1,0)`
E	Cumulative crossings	`=SUM($D$2:D2)`
F	Estimated pi	`=IF(E2=0,"",2*A2/E2)`

Fill down to row 10001 for 10,000 trials.

Column B samples distances between 0 and 0.5. Column C samples angles between 0 and pi/2. Column D applies the crossing condition.

Plot columns A and F, then compare several runs.

RAND() changes when the workbook recalculates. To preserve a run, copy the results and paste them as values into a separate sheet.

There is an important question to discuss:

“If the spreadsheet already uses PI(), what are we demonstrating?”

We are demonstrating the probability model and the behaviour of repeated random trials. This version is not an independent discovery of pi.

Recognising that distinction is part of understanding the simulation.

Try a Python Simulation

The same model can be written in a short program:

import mathimport randomcheckpoints = {10, 100, 1000, 10000}crossings = 0for drops in range(1, 10001):    distance = random.random() / 2    angle = random.random() * math.pi / 2    if distance <= 0.5 * math.sin(angle):        crossings += 1    if drops in checkpoints:        if crossings == 0:            print(drops, "drops: no estimate yet")        else:            estimate = 2 * drops / crossings            print(                f"{drops:5d} drops | "                f"{crossings:5d} crossings | "                f"pi estimate = {estimate:.4f}"            )

Run it several times. Compare the results at each checkpoint.

For a more demanding investigation, repeat each sample size 100 times and examine how widely the estimates vary.

Like the spreadsheet, this program uses a known value of pi to sample angles. Its purpose is to explore the model and uncertainty.

The physical experiment remains particularly satisfying: its measuring equipment needs no numerical value of pi.

Change One Thing and Ask a Better Question

Once the basic experiment works, vary the ratio of stick length to line spacing.

For L <= d:

pi estimated = 2 x L x N / (d x C)

For example, if sticks are 5 cm long and lines are 10 cm apart:

P = 1/pi

pi estimated = N/C

Crossings are now less frequent.

Can students explain why doubling the spacing halves the theoretical crossing probability?

Keep sticks no longer than the spacing when using this formula. Longer sticks can cross more than one line, and the probability of at least one crossing needs a different calculation.

Why I Would Use This at Hemel Private Tuition

I like activities that let a student enter at a comfortable level and then discover there is more to explore.

One student can practise relative frequency and rearrangement. Another can investigate uncertainty. An A-level student can derive the formula. A student interested in computing can build and test the simulation.

They are working on the same problem, but asking increasingly demanding questions.

At Hemel Private Tuition, this practical approach connects mathematical ideas with something students can see, test and explain. It also develops an important habit for examinations: stating the assumptions behind a method and interpreting what an answer means.

Conclusion: A Number Hidden in Randomness

Buffon’s needle turns an ordinary tabletop activity into a surprising mathematical investigation.

A stick lands. You record a crossing or a miss. One result tells you very little. A large collection begins to reveal a relationship between geometry, rotation and probability.

The most valuable outcome is more than an estimate of 3.14. It is an understanding of why repeated trials help, why a model needs assumptions, and why a result deserves scrutiny even when it looks convincing.

Draw the lines. Drop the sticks. Count the crossings. Then ask the question that makes the activity worthwhile: why should this work at all?

04 November 2026