Bean machine

A Galtonbrett (after Francis Galton ) is a mechanical model for the demonstration and illustration of the binomial distribution, a probability distribution that plays a role in many random experiments.

Concept

The Galtonbrett consists of a regular array of obstacles that an inserted from the top ball can bounce off each left or right. After passing the obstacle, the balls are collected in trays, to be counted therein.

The Galtonbrett simulates a physical instrument whose measurement is noisy. The horizontal position of the ball is the value to be measured, the more accurate is present at the upper entrance while he was changed later in one of the compartments by a noise signal. The obstacles symbolize small disturbances that can influence the measured value is positive or negative. In sum, they can grow to a larger disorder, but also add up to zero. The fill level of the subjects end up giving information on the frequency distribution of the different strengths of the summed interference. In real measurements, the corresponding for example to the noise distribution of an electrical signal caused by a large number of very small noise that can contribute as positive as negative. A fundamental mathematical law, the central limit theorem guarantees that a almost any composite distribution of such very small and very numerous individual faults in the sum converges to the bell-shaped Gaussian normal distribution. If the requirements for such a noise distribution satisfies, one speaks of Gaussian noise. In a finite number of disorders, such as when Galtonbrett, we obtain the binomial distribution, which also converges in the limit of many disorders, and many subjects against the normal distribution. The statistical regularities of such random measurement noise can be studied and checked against the Galtonbretts in a clear way.

Mathematical viewing

Each bouncing a ball on one of the obstacles is a Bernoulli trial. The two possible outputs are " ball falls to the right" (X = 1) and " ball falls to the left" (X = 0).

For a symmetrical structure, the probability to fall to the right, and the probability to fall to the left. By unbalanced structure or by skewing of the board but you can also reach a different value, which of course still is, because the balls that fall to the right, fall to the left. This case will be discussed below.

By the ball after passing the first obstacle to a new hits, in which the same prerequisites apply, here's another Bernoulli experiment is carried out; the passage of the entire device is thus a multi-stage Bernoulli chain, wherein the number of horizontal rows of obstacles, the length of this chain. In the illustrated image is therefore a 4- fold repetition of a Bernoulli trial, ie a Bernoulli chain of length 4

One can now calculate the probability with which a ball falls into a particular one of the subjects. If only one obstacle (A ) is the probability 1/2 for the left and right, or, in other words, in the central half of the balls to the right and falls to the left half. Thus, half of the balls on B and the other half meets C where it again divided with equal probabilities to the left and right. But that falls only 1/ 4 of the balls in B to the left, 1/4 to C to the right, and 1/4 from the left and from the right in the space between B and C. Here, the probabilities add up then, and 1 / 4 1 /4 = 2/4 (= 1/ 2, but not reduced intentionally ) falling in the middle between B and C throughout.

Based on the picture you can pursue, as the flow of balls splits at each obstacle number ( at the next one is, therefore, with eighth notes, have to expect at the next but with sixteenths of the total) and, on the other hand, associated in any space between two adjacent obstacles again.

The thus resulting probabilities after the last division and unification at the lowest obstacle row ( G, H, I, J) are the probabilities with which the balls fall into the compartments (R, S, T, U, V).

In the example, all these probabilities have the denominator of 16, since there are 4 rows of obstacles ( 16 = 24). The counters are obtained by adding the counter in the row above, which is the union of the streams of balls in the gaps. Thus, the following scheme for the probabilities:

Numerator: Denominator:      Number 0: 1 = 1 → = 2 ^ 0 = 1 11 ^ 0 ( The power series of the number 11            1: 1 1 = 2 → = 2 ^ 1 = 11 11 ^ 1 does not break off when the            2: 1 2 1 = 4 → = 2 ^ 2 121 = 11 ^ 2 meters from the 6th line in            3: 1 3 3 1 = 8 → = 2 ^ 3 1331 = 11 ^ 3, another place-value            4: 1 4 6 4 1 = 16 → = 2 ^ 4 = 14641 11 ^ 4 system are translated ). One can see that the counters are the binomial coefficients, because they occur after the Pascal's triangle scheme. The denominators are powers of 2, they follow from the probability 1/2 to fall to the right or left.

The subjects R, S, T, U, V can be numbered according to how often must drop a ball to the right to reach the relevant subject. This gives professional R the number 0, because a ball that lands in R, not once fallen to the right, but always to the left. Trade S has the number 1, because a ball that lands here is just like once to the right ( in the first, second, third or fourth row, but in any case, just this once ). According to the other subjects get the numbers 2, 3, and 4

Likewise follows:

As a general rule for the subject:

Herein is the binomial coefficient in row 4, column of Pascal's Triangle (Note: the columns are the numbers of subjects, so start with 0).

Generalizing the formula further to a Galtonbrett with obstacle rows (instead of 4 as in the example), so on a Bernoulli chain length, so will apply:

To further generalization now imagines a unbalanced Galtonbrett ago, in which the probabilities for left and right are not the same size. The balls fall to the probability to the right and to the left.

Then they end up in Box 0 with probability q4 instead of (1/2 ) 4, because they must still falling four times to the left, but do now, each with its probability.

In the rightmost compartment ( for the right they have to fall four times what they do with each of probability), they end up with.

For the other subjects they need the number ( ) corresponding to the left side to the right times and the other times (). This they do with the probabilities or. What is not changing, the associations of the sphere currents in the gaps. Thus, the binomial coefficients remain unaffected. The general formula for a 4-row Galtonbrett with the probability to fall to the right, is therefore:

And for the general formula of an appropriately -row board eventually follows:

This is exactly the binomial distribution a Bernoulli chain length with and.