At the end of the 19th century, the English polymath Sir Francis C. Galton (1822–1911) developed an arrangement to demonstrate the so-called binomial distribution. This arrangement was later called the Galton board in his honor. A version of it is realized in MATHEMATICS ADVENTURE LAND:
Between two glass plates, several 50-cent coins are fixed with three pins each and arranged evenly so that — as an overall structure — an equilateral triangle of twelve “cascades” results (see Figure 1 below):
If you now let a coin fall in vertically from above, each of these obstacles (i.e., the locked 50-cent coins) decides at random whether it falls to the right or to the left. The probability for this is in each case. Below the Galton board there are several compartments in which the inserted coins are stacked on top of each other. That is, in this way a kind of bar chart is created, which approximates the density of a normal distribution as the number of inserted coins increases. The shape of this density is a bell-shaped curve (“Gaussian bell curve”), as shown in Figure 2 below:
1-cent, 2-cent, … or 50-cent coins are suitable for carrying out the experiment in MATHEMATICS ADVENTURE LAND.
Note: The coins thrown in for an experiment at the Galton board will later benefit as a contribution to the work of the non-profit(!) association for the promotion of the work of the MATHEMATICS ADVENTURE LAND. This is how the name “money clay board” was created for this exhibit.
And now … the mathematics of it:
The Galton board in MATHEMATICS ADVENTURE LAND has a total of steps, i.e. the inserted coin has to “decide” for right or left exactly twelve times on its way down. It is reasonable to assume that all these decisions are stochastically independent of each other: Whether the coin decides one way or the other at an obstacle — it does not affect its behavior below. Thus, essentially the same random trial is run twelve times in a row. Accordingly, then, there are exactly
Possibilities of how the coin can behave. Each of these possibilities corresponds to exactly one path in the associated tree diagram. However, what is counted in the experiment is not how often a ball uses each of these (equally probable) paths, but how often it chooses right and left.
For this purpose, there are compartments below the Galton board which reflect this. The coin lands exactly in the compartment if it decides exactly times for the right (and accordingly times for the left).
If one now performs this experiment with a sufficiently large number of coins, it can be observed that most of them collect in the middle compartments. The outer compartments, on the other hand, are rarely reached.
This, in turn, has a simple mathematical reason: the number of paths leading into the compartment is exactly , because it is necessary to select exactly of the steps where the coin decides to go to the right. Accordingly, the probability of landing in the compartment is exactly
This probability distribution is called a binomial distribution with parameters (the probability of a coin falling to the right at an obstacle) and (the number of stages of the random experiment). The is also called their individual probabilities.
The fact that the binomial distribution for “large” values is approximated by the normal distribution, describes the so-called limit theorem of Moivre-Laplace:
for , fixed and arbitrary, and .
We sketch another proof of this theorem below:
First, we observe that the density function
has the derivative , thus satisfying the differential equation . If we add that the integral
we obtain the density function
of the standard normal distribution as a unique solution of this differential equation. The assertion of Moivre-Laplace’s limit theorem is now that the (suitably rescaled) “density function” of the binomial distribution with parameters and converges to this function . By “rescaling” we mean here that the coordinate is first shifted to the left by and then the axis is compressed by and the axis is stretched by . We thus obtain from the binomially-distributed random variable with values in , the random variable which takes values in
assumes. This is a discrete random variable with distribution
Where is the Dirac measure in . Since the random variable is discrete, it has no density function in the proper sense. However, we can replace the Dirac measure in the formula just noted by a measure of mass one uniformly distributed on (where denotes the Lebesgue measure). We obtain in this way another random variable , which has a density function. It is easy to see that the distributions of and are weakly equivalent in the limit, i.e., their difference weakly converges to for . Thus, we can work with instead of . To show that their distribution function weakly goes against ( the Lebesgue measure), we now simply show that it satisfies in the limit of the above differential equation . So let us write for the density function of and assume that for a . This is certainly reasonable, since such values fill the whole axis as increases, and are increasingly dense. Then we have
But what is . In fact, is constant in a neighborhood of , so should actually hold. But in the intercept of , suddenly makes a jump; so there would be. To account for both, we use a difference quotient instead of a differential quotient:
Thus, for the left side of the differential equation
and for the right side
Thus, to check the above differential equation in the limit, it suffices — taking into account and — the equation
for to be proved. We leave this calculation to the reader and refer to the proof on Wikipedia by carrying it out. Since also , we are done. Of course, this sketch is not a rigorous proof — but it can be converted into one with some effort.
 Henze, N.: Stochastik für Einsteiger. Eine Einführung in die faszinierende Welt des Zufalls, Springer Spektrum, 10. Auflage, Wiesbaden, 2013.