How random bounces add up to a bell
2026-05-13
Drop a ball at the top. It hits a peg. It goes left or right - a coin flip, no preference. It hits another peg. Another coin flip. After twelve rows it lands in one of thirteen bins. That one ball is a sequence of twelve random choices. There is no reason it should land anywhere in particular.
Drop another. Then a hundred. Then a thousand. The bins underneath fill up. And the shape they fill up into is not random.
Balls fall continuously, each one's path decided by a coin flipped twelve times. The bars grow. The faint tan curve is the prediction - where each bin should be after this many drops. The bars chase the curve. Press pause to look closely; press reset to start over. The rows buttons change how many peg rows the balls fall through.
Notice the numbers floating above the bins. Those are not bin counts; those are the numbers from Pascal's triangle, row N. With twelve rows, the row reads: 1, 12, 66, 220, 495, 792, 924, 792, 495, 220, 66, 12, 1. The bars converge on those ratios.
Why Pascal's triangle
Think of one ball's journey. At each of the N rows it flips a coin: L or R. The whole journey is a string of N letters - say LRLLRRLR... - and the bin it lands in is determined by one number: how many R's there are. k rights and N-k lefts puts the ball in bin k.
How many different journeys end in bin k? Exactly the number of N-letter strings with k of one letter and N-k of the other. That number is the binomial coefficient - "N choose k", written C(N, k). Pascal's triangle is the table of all those numbers.
Each individual journey is equally likely - the same N coin flips, just in a different order. So the chance a ball ends in bin k is C(N, k) / 2N. Multiply by however many balls you drop, and you get the expected count. That is the curve the bars chase.
Why the bell
Look at the bins at the edges. Only one journey lands in bin 0: all left, every time. Only one journey lands in bin N: all right, every time. To land there, every coin flip has to cooperate. The chance is 1 in 2N.
Look at the bin in the middle. To land in bin N/2, you need exactly N/2 lefts and N/2 rights, in any order. With twelve rows, that's the number of ways to arrange six R's among twelve slots - 924 different orderings. The middle bin is 924 times more likely than either edge bin. The bell isn't a designed shape. It's a counting argument.
What it becomes
Crank the rows up to fourteen. The histogram smooths out. Crank it down to six. It's chunky and uneven. The more rows, the smoother the bell. In the limit - infinitely many rows, infinitely many balls - the histogram becomes a continuous curve, and the curve is the normal distribution, the same bell that shows up in heights, in test scores, in noise, in anything that is itself a sum of many small random things.
That convergence has a name - the Central Limit Theorem. It is one of the most surprising statements in mathematics: that almost no matter what little random thing you add up many times, you get the same bell. The Galton board is the simplest possible proof of it that you can watch fall.
A thousand coin flips, played out one ball at a time. Pure randomness on the way down, perfect order at the bottom. The bell was already in Pascal's triangle. The balls just drew it.