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Galton board

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Galton board

Balls dropped0
Mean—np = 5.00
Std deviation—√(npq) = 1.581
Every ball makes an independent left/right choice at each of the 10 rows. Drag to orbit, scroll or pinch to zoom.

Show

bars = balls actually collected
outline = the binomial expectation
bell curve through the same peaks
10 0.50

What's the maths?

Where's the maths? This is probability and statistics — the branch of mathematics that measures chance and then describes what a whole pile of random results looks like. Each ball meets a row of pegs, and at every row it independently goes left or right, going right with probability p. One ball is unpredictable, so the maths of a single ball is a tree diagram: you multiply the probabilities along a branch to get the chance of one particular path. Ten thousand balls are extremely predictable, and the maths of the pile is a histogram with a mean and a standard deviation. The numbers that turn one into the other are the entries of Pascal's triangle, because they count how many different paths land in the same bin. So the same board shows you probability (going down) and statistics (looking across), and the bell shape at the bottom is not put there by hand — it is what counting paths forces to happen.

📘 On the Sec 1–4 syllabus
  • Sec 3/4 · S2 Probability — 2.3 combined events with tree diagrams; 2.4 multiplication of probabilities
  • Sec 2 · S1 — 1.5 histograms; 1.8 mean
  • Sec 3/4 · S1 — 1.12/1.15 standard deviation as a measure of spread
  • Sec 3/4 (A-Math) · A5 — Binomial Theorem, nCr notation (Pascal's triangle)

Beyond the syllabus: the normal distribution is not examined.

Follow one ball down a board with n rows. It makes n independent choices, so one particular route — say R, L, R, R, L — has probability p·q·p·p·q, where q = 1 − p. Multiplying along a branch of a tree diagram is exactly the Sec 3/4 rule for independent events.

The ball lands in bin k if it went right k times, in any order. Every such route has the same probability pkqn−k, so all that is left is to count the routes — and that count is the Pascal's triangle number nCk. Turn the peg numbers on: each one is the number of ways of reaching that peg, and it is the sum of the two above it, because a ball can only arrive from the left or from the right.

P(X = k) = nCk pk(1 − p)n−k

  • With p = 0.5 the counts alone decide the shape, so the bins copy row n of Pascal's triangle: 1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1 for 10 rows.
  • The mean of the pile is np and its standard deviation is √(np(1 − p)). Ten rows at p = 0.5 give a mean of 5 and a spread of about 1.58 bins — check the readouts against the theory as the balls pile up.
  • Doubling the rows doubles the mean but multiplies the spread by only √2. Randomness "averages out": the pile gets relatively narrower the longer the board.
  • Bias the board (p ≠ 0.5) and the whole heap slides to np and gets slightly narrower — p(1 − p) is largest at p = 0.5.
Try this: set 10 rows and p = 0.5 and drop 10 000 balls — the teal bars should sit inside the gold outlines. Now predict the mean and the standard deviation for 14 rows at p = 0.3 before you move the sliders (np = 4.2 and √(np(1−p)) = ?), then drop 10 000 and see how close you were.

Drag to orbit the board, scroll or pinch to zoom. Sliders change the board and start a fresh count.