The Galton board
Each ball hits twelve pegs and bounces left or right at every one. Where it lands is just the number of rights. Drop a few and it looks random. Drop a thousand and the pile is a bell shape. The dark dots are the exact prediction for twelve pegs, the binomial distribution; the dashed orange curve is the normal distribution with the same mean and spread. More balls bring the pile closer to the dots. Only more rows of pegs would bring the dots closer to the curve.
Mission. Predict: if you drop 100 times as many balls (10 000 instead of 100), does the pile become exactly normal? Test it, then look for a gap between the dots and the curve that refuses to go away.
Answer
No. With more balls the sampling noise fades and the pile settles onto the binomial dots, not onto the curve; the gap between dots and curve stays, however many balls you drop. Two persistent ones: the curve is about 2% too tall at the middle bin (0.2303 against 924/4096 = 0.2256), and the share within one standard deviation of the mean (bins 5, 6 and 7) is 2508/4096 ≈ 61.2%, not the 68% of the normal rule. Extension: what would change the dots themselves? More rows, not more balls; that is the separate operation that improves the normal approximation.
What's going on
Each peg is a coin flip. Landing in bin k means exactly k of the twelve flips went right, so the chance of bin k is the binomial number C(12, k) / 2¹² when the coin is fair: 1, 12, 66, 220, 495, 792, 924, 792, 495, 220, 66, 12, 1 out of 4096. Those are the numbers in row 12 of Pascal's triangle. The middle bin is 924 times more likely than the edge.
The mean position is 12p and the standard deviation is √(12 p (1 − p)), which for a fair coin is √3 ≈ 1.73. For a normal curve, 68% of the area lies within one standard deviation of the mean and 95% within two. Twelve pegs do not quite deliver that. The bins within √3 of 6 are 5, 6 and 7, and they hold (792 + 924 + 792)/4096 = 2508/4096 ≈ 61.2% of the balls; within two standard deviations (bins 3 to 9) is 3938/4096 ≈ 96.1%. The readout shows both the exact figures and the normal rule, so you can watch the pile converge to 61%, not 68%.
Two different things are going on. Dropping more balls reduces sampling noise: the pile converges to the binomial dots, which it would do even with two rows of pegs. Adding more rows is what makes the binomial itself look more normal, and that is the central limit theorem: add up many small independent random nudges and the total is approximately normal, whatever the nudges look like individually. Heights, measurement errors and exam marks are bell-shaped for the same reason. Drag the bias slider: the pile shifts and squashes, and with only twelve rows the normal approximation gets worse as p moves away from ½. More rows, not more balls, would restore the bell.
Sheet: Standard deviation and the bell curve. Related demo: two dice, many rolls.