Skip to content
Statistics & Probability · Mathematics

Buffon's Needle & Pen Through a Window: 3D Simulation

Throw needles at lines and estimate π, then throw pens at a window in 3D: what fraction pass through without touching a bar? Throw 1, 100 or 10 000 at once and watch the fraction settle on the formula — 2/(3π) ≈ 21 % for a pen as long as a pane is wide — and see why that number depends on what 'random' means.

Interactive virtual lab · no download · works on phones and laptops

Chapter
Probability
For
Class 11–12 · JEE
Time
~20 min

Buffon's Needle: A Pen Through a Window is a Pro simulation

Pro opens all 152 virtual labs — 693 simulations with guides and quizzes.

How to use it

  1. 1Pick a module from the list (4); set your level at the top right for more controls.
  2. 2Set the pen and pane sizes, then throw 1, 100, 1000 or 10 000 pens with the buttons on the stage.
  3. 3Use Predict before you look, then Derive, Questions, Practise and Compare.

What to notice

  • Buffon: a needle of length l on lines d apart crosses with probability 2l/(πd).
  • Flat throws at a square grid pass with 1 − (4la − l²)/(πa²).
  • Pens in any 3D direction pass more often: 1 − l/a + 2l²/(3πa²), 21.2 % at l = a.
  • The answer depends on the model of a random throw; the error shrinks like 1/√n.

For teachers

A ready-made teaching kit comes with this lab

  • Lesson plan

    25 min, 4 steps

  • Ask the class

    4 questions with answers

  • Quick quiz

    6 questions, printable

  • Key formulas

    4 formulas

About the Buffon's Needle: A Pen Through a Window simulation

Throw a pen at a window: what is the chance it goes through without touching a bar? This 3D Buffon's needle simulation builds the answer up from the classic problem. First, needles dropped on a floor ruled with parallel lines cross a line with probability 2l/(πd) — so counting crossings estimates π. Then the floor is ruled both ways like a window lying flat, and a pen must clear bars in both directions: Laplace's formula 1 − (4la − l²)/(πa²).

Then the window stands up and pens fly at it pointing in any direction in space. A pen tilted towards you casts a shorter shadow on the window, so it passes more often: averaging over directions uniform on the sphere gives P = 1 − l/a + 2l²/(3πa²), which is 2/(3π) ≈ 21.2 % when the pen is as long as a pane is wide — the result in the MathBiceps video. Throw 1, 100, 1000 or 10 000 pens with the buttons on the stage and watch the running fraction settle on the formula as the error shrinks like 1/√n, and see why the answer depends on exactly what a random throw means.

Also covers: buffon's needle, buffon needle problem, estimate pi by throwing needles, geometric probability, probability pen through window, laplace grid problem, monte carlo simulation, random orientation in 3d, uniform on a sphere, bertrand paradox, 2/3π, probability integration, jee probability, class 11 probability.

Frequently asked questions

What is Buffon's needle problem?

A needle of length l is dropped at random on a floor ruled with parallel lines d apart (l ≤ d). The probability that it crosses a line is 2l/(πd). It was posed by Buffon in 1777 and is the first problem in geometric probability; turning it round gives a way to estimate π by experiment.

What is the probability that a randomly thrown pen passes through a window?

It depends on the model. For a thin pen whose centre is uniform over a square pane, whose direction is uniform over all of 3D space, and which flies square to the window, P = 1 − l/a + 2l²/(3πa²). For a pen as long as the pane is wide that is 2/(3π) ≈ 21 %; if the pens lie flat it is only about 4.5 %.

Why does the answer depend on what 'random' means?

Different ways of choosing a random direction — uniform over the sphere, or a uniformly random tilt angle — give different averages of the pen's shadow length, and so different probabilities. This is Bertrand's paradox: a probability question must say how the random choice is made.

How many throws are needed to estimate π well?

The error of a random-sampling estimate falls only as 1/√n. A few hundred throws give π to about one decimal place; every extra correct digit needs about a hundred times as many throws.