Question 1

Why does the Galton board produce a normal distribution?

Accepted Answer

Each ball makes N independent binary decisions (left or right at each peg). The total displacement from center equals the number of right-deflections minus left-deflections, which is a sum of N independent ±1 random variables. By the Central Limit Theorem, the sum of many independent random variables with any distribution converges to a Gaussian. For the Galton board specifically, with N rows and equal probabilities, the result is exactly binomial, which approaches Gaussian as N grows large due to Stirling's approximation of factorials.

Question 2

What is the Central Limit Theorem?

Accepted Answer

The Central Limit Theorem (CLT) states that if you sum (or average) a large number of independent, identically distributed random variables with finite mean μ and variance σ², the resulting distribution approaches a normal distribution with mean Nμ and variance Nσ², regardless of the original distribution's shape. This is why Gaussian distributions are ubiquitous in nature: any measured quantity that results from many independent additive contributions (thermal noise, measurement errors, biological traits influenced by many genes) will be approximately normally distributed.

Question 3

What happens if the pegs are biased (unequal left/right probability)?

Accepted Answer

If the probability of deflecting right at each peg is p ≠ 0.5, the resulting distribution is a binomial with mean Np and variance Np(1−p). The distribution shifts from center: with p > 0.5, balls accumulate to the right; p < 0.5 shifts the peak left. The shape remains bell-like (still converges to Gaussian by CLT) but is skewed for small N and becomes symmetric only in the limit. Biased Galton boards model processes where outcomes are not equally likely—like a weighted coin or a genetic trait with dominance.

Question 4

How is the Galton board related to Pascal's triangle?

Accepted Answer

The number of paths leading to bin k after N rows equals the binomial coefficient C(N,k), the entries of Pascal's triangle. Row 4 is 1,4,6,4,1, meaning with 4 pegs the center bin is 6x more likely than the edge bins, following Pascal's rule C(N,k) = C(N-1,k-1) + C(N-1,k).

Question 5

What practical applications rely on the normal distribution that Galton's board demonstrates?

Accepted Answer

The normal distribution underpins statistics and quality control throughout science: measurement uncertainties, z-tests and t-tests, manufacturing control charts, standardized test scores, and approximately Gaussian daily asset returns in finance (Black-Scholes, Value-at-Risk).

Question 6

What does the Rows (n) slider do, and why does the distribution look smoother with more rows?

Accepted Answer

Rows (5-16) sets the number of independent binary left/right decisions N. More rows means more possible bins (N+1) and finer discretisation, so the histogram increasingly resembles a smooth Gaussian rather than a blocky staircase, as the Central Limit Theorem kicks in.

Question 7

Why does increasing the Rows slider also widen the spread of the final histogram?

Accepted Answer

Variance is Np(1-p), which for p=0.5 is N/4, so spread grows proportionally with N even though standard deviation grows only as sqrt(N). More rows means more distinct bins spread further from center, widening the visible histogram.

Question 8

Does the Ball Radius slider change the physics or only the appearance of the simulation?

Accepted Answer

Ball Radius (3-8 pixels) is purely a rendering setting for how large each ball appears. The peg-bounce logic deciding left/right deflection at each row is unaffected, so the statistical distribution across bins stays the same.

Question 9

Why does dropping many balls at once produce a more recognisable bell shape than dropping just one or two?

Accepted Answer

A single ball's final bin is one random sample, telling you almost nothing about the shape. Dropping thousands of balls builds a histogram that, by the law of large numbers, converges toward the true binomial probability distribution.

Question 10

Is the outcome at each peg in this simulation truly 50/50, and does it matter if it isn't exactly?

Accepted Answer

The default simulation uses equal left/right probability, centring the distribution on the middle bin. Small departures from 50/50 shift the mean toward one side, but the bell shape survives because the Central Limit Theorem applies to any fixed per-peg probability, not just p=0.5.

🎯 Ball Cascade Patterns in a Peg Board

What did you find?