Galton’s Board - Central Limit Theorem

Question

The reason for the question is study/learning. Know/apply programming techniques and concepts, transforming something "material" into "application".

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I received a video on Whatsapp, where mixed colored balls were separated by colors passing through the pins. Or it was a lot of mechanical technology or fake video. In the end, it was a fake video, where a graphic design played a Board of Galton for a study job.

Wanting to know more, I found in Wikipedia, and would like to turn it into a "program".

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The board basically simulates the "Central limit theorem" (link): The theorem describes the distribution of the mean of a random sample of a population with finite variance.

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Idea:

Starting from the board, I decided to create a matrix, where I have malhas and pinos, forming a line:

-------Malha------->
[P1][P2][P3][P4][P5]...

Pin values:

Like the odds on the balls going down is always bigger from the center to the sides, I created the pins with values, and the bigger, more chances to "continue" there. But also, as the pins are "interspersed" and not vertically aligned, also considered this fact.

Knitting:

// Definicoes
$qtdMalhas = 15;
$qtdPinos = 15;

// Auxiliares malhas/pinos
$am = $ap = 0;

// Criando as malhas e seus pinos
while ($am <= $qtdMalhas) {

    while ($ap < $qtdPinos) {

        if ($ap < ($qtdPinos/2)) {

            if ($am % 2 == 0) {
                $m[$am][$ap] = $ap;
            } else {
                $m[$am][$ap] = $ap+1;
            }
        } else {

            if ($am % 2 == 0) {
                $m[$am][$ap] = $qtdPinos-$ap-1;
            } else {
                $m[$am][$ap] = $qtdPinos-$ap;
            }
        }

        $ap++;      
    }

    $ap = 0;
    $am++;  
}

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Mesh structure:

Consider the matrix basically in the following format:

$m[0][0]-$m[0][1]-$m[0][2]-$m[0][3]-$m[0][4]...
$m[1][0]-$m[1][1]-$m[1][2]-$m[1][3]-$m[1][4]...
$m[2][0]-$m[2][1]-$m[2][2]-$m[2][3]-$m[2][4]...
$m[3][0]-$m[3][1]-$m[3][2]-$m[3][3]-$m[3][4]...
$m[4][0]-$m[4][1]-$m[4][2]-$m[4][3]-$m[4][4]...
...

Printing the fabric:

// Zera auxiliares malhas/pinos
$am = $ap = 0; 

// Imprime a malha
while ($am < $qtdMalhas) {

    echo '<br>|';

    while ($ap < $qtdPinos) {

        echo $m[$am][$ap].'|';

        $ap++;
    }

    $ap = 0;
    $am++;
}

Upshot:

|0|1|2|3|4|5|6|7|6|5|4|3|2|1|0|
|1|2|3|4|5|6|7|8|7|6|5|4|3|2|1|
|0|1|2|3|4|5|6|7|6|5|4|3|2|1|0|
|1|2|3|4|5|6|7|8|7|6|5|4|3|2|1|
|0|1|2|3|4|5|6|7|6|5|4|3|2|1|0|
|1|2|3|4|5|6|7|8|7|6|5|4|3|2|1|
|0|1|2|3|4|5|6|7|6|5|4|3|2|1|0|
|1|2|3|4|5|6|7|8|7|6|5|4|3|2|1|
|0|1|2|3|4|5|6|7|6|5|4|3|2|1|0|
|1|2|3|4|5|6|7|8|7|6|5|4|3|2|1|
|0|1|2|3|4|5|6|7|6|5|4|3|2|1|0|
|1|2|3|4|5|6|7|8|7|6|5|4|3|2|1|
|0|1|2|3|4|5|6|7|6|5|4|3|2|1|0|
|1|2|3|4|5|6|7|8|7|6|5|4|3|2|1|
|0|1|2|3|4|5|6|7|6|5|4|3|2|1|0|

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Could make the mesh in other ways ?

Yes, using zero and one for the pins but I believe I would have more work to draw according to the probability of each pin. Among other ways.

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With the mesh ready, how can I "pass the balls between her" ?

The idea would be to draw between the pin values below, whereas the odds of it staying ever closer to the higher value pin, that is, a draw with different odds, and thus obtain the result as the "Board of Galton".

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Feel free to post your way of doing.

Answer 1

I did with it:

<?php
    function cai_bolinha($linhas) {
        $sorteio = mt_rand() % (1 << $linhas);
        return substr_count(decbin($sorteio), '1');
    }

    $linhas = 15;
    $bolinhas = 100000;

    $casas = [];
    for ($i = 0; $i <= $linhas; $i++) {
        $casas[$i] = 0;
    }

    for ($i = 0; $i < $bolinhas; $i++) {
        $casa_resultante = cai_bolinha($linhas);
        $casas[$casa_resultante]++;
    }

    for ($i = 0; $i <= $linhas; $i++) {
        echo $i . "->" . $casas[$i] . "\n";
    }
?>

Here is the output (may vary, obviously):

0->5
1->35
2->310
3->1459
4->4079
5->9071
6->15089
7->19622
8->19771
9->15366
10->9265
11->4136
12->1409
13->334
14->46
15->3

See here working on ideone.

The trick is that the mt_rand() will generate any random number. This binary number is represented by zeros and ones, where zero means to fall to the left and one means to fall to the right. The % (1 << $linhas) is used so that the result has the amount of bits that interest us, represented by the variable $linhas, which corresponds to the amount of pin lines the ball has to cross, which is the same amount of "left or right" draws, with 50% probability for each side.

With that, the number $sorteio corresponds to a number where each bit describes to which side the ball went. If that number is zero (all bits are zero), then in all draws it went left. If all bits are 1, then in all draws he went right. Note that these two cases are quite rare and unlikely and the highest probabilities are that the resulting number is composed of a mixture of zeros and ones.

After that, having this $sorteio, by counting how many bits 1 exist in it, one discovers in which house the ball fell, where the zero house is that of the extreme left and the house $linhas that of the extreme right.

In the second loop for, make 100000 draws (the variable $bolinhas) with polka dots falling to observe the distribution. Note that the resulting distribution is an approximation of the normal distribution (in fact it is a binomial distribution), and as expected, the center houses (7 and 8) are the ones that have accumulated the largest numbers of balls, while the extreme houses (0 and 15) the smallest numbers.

Feel free to try other values for the variables $linhas and $bolinhas.