The Galton board and the bell curve

On a Galton board with 10 rows of pegs, a ball lands in the middle bin with probability 252/1024, about 0.246.

Where do the balls land on a board with 10 rows of pegs?

Start with the board below. Two hundred balls have already gone through 10 rows of pegs. Before reading anything else, press Drop 100 and watch where the new balls go. Then press Add 1,000 at once. Which bin wins? Roughly what share of all the balls does it hold? Write both down; the rest of the lesson explains them.

Drop a ball into the top of a board studded with pegs. At each peg it bounces left or right, and after the last row it falls into one of the bins at the bottom. One ball goes anywhere. Hundreds of balls build the same shape every time: a heap in the middle, falling away on both sides. The device is called a Galton board, a bean machine or a quincunx, and it turns the rules from the earlier lessons into a picture.

Bin 5 can be reached by 252 paths, bin 0 by one

Number the bins from 0 on the left. A ball that bounces right k times out of n rows lands in bin k, no matter in which order the rights and lefts happened. Every path of n bounces has probability (1/2)ⁿ when left and right are equally likely, so the chance of bin k is (number of paths with k rights) ÷ 2ⁿ. With 10 rows there are 2¹⁰ = 1,024 paths. Only one path reaches bin 0 (left every time), but 252 different paths reach bin 5. So P(bin 5) = 252/1024, about 0.246, and P(bin 0) = 1/1024, less than 0.001. That is the share you saw in the lab: close to a quarter of the balls in bin 5, and about one ball in a thousand in bin 0.

The path counts are the numbers in Pascal's triangle, where each number is the sum of the two above it, just as each peg can be reached from the two pegs above it. The count for k rights out of n is written C(n, k), "n choose k". When the chance of going right is p instead of 1/2, each path with k rights has probability pᵏ(1 − p)ⁿ⁻ᵏ, and P(bin k) = C(n, k) pᵏ (1 − p)ⁿ⁻ᵏ. This is the binomial distribution, which OpenStax covers in chapter 4. Its mean is n × p and its standard deviation is √(n p (1 − p)). For 10 rows and p = 0.5 the mean is bin 5 and the standard deviation is √2.5, about 1.58 bins.

Ten rows already put 65.6% of the balls within one standard deviation

As the number of rows grows, the heap looks more and more like the smooth, symmetric bell of the normal distribution. For a bell shape, about 68% of values fall within one standard deviation of the mean and about 95% within two. On a 10-row board, bins 4 to 6 (within 1.58 of bin 5) hold about 65.6% of the balls and bins 2 to 8 about 97.9%, already close. The reason is the same one that makes sums of three dice rounder than sums of two: many small independent pushes, added up, tend towards the bell. Set the chance of going right to 0.7 and the heap moves to bin 7 and leans: the binomial is only symmetric when p = 0.5.