Galtonboard / Galtonbrett Simulation (or Bean machine or quincunx or Galton box)

[@davidwbulger]

Watching the small balls fall, you can see that they tend to gather momentum and run in diagonal paths to the left or right-that is, they don't change direction much. This illustrates the importance of the independence assumption in the central limit theorem: if the individual random variables are not independent, then their sum may not tend to a normal distribution.

[@michaelwestern11]

Could someone please verify if this is indeed true? I saw some other comments which said that it is cause there just weren't enough small balls.

[@PrayBehuman]

It is amazing that the distribution varies depending on the size of the ball. Thanks for the experiment.

[@gnomishmath]

Lack of independence seems to be causing the drift from the normal distribution. The large balls are processed relatively independently as only 1 can fit through each condition at a time.

[@Thyagohills]

Doesn't the fact that some balls are hitting each other interfere with the independence assumption? I mean, the probability of going right or left is biased by the interactions between the balls.

[@ArtOfTheProblem]

Could I use this for an educational video?

[@3D-PHASE]

Art of the Problem feel free to use it for whatever you like.

[@Jcknight7996]

Will having small balls in a real Galtonboard live you the roughly even spread across the slots?

[@brandomiranda6703]

Maybe this is a misconception that I have, but if you throw them all from the middle, of course they will pie most in the middle and die off to the sides. How is this a fair experiment to show the CLT (Central Limit Theorem)?

[@gorgolyt]

Why not? This is part of why the central limit theorem works. There are not many things in the tails because they constantly have to move in the same direction, which is unlikely. Just because it's intuitive, why should it be wrong?

[@brandomiranda6703]

I never implied that intuitive explanation have to be wrong. In fact, Im usually a fan of them. My question was addressed at the fact that the distribution from this experiment seems biased because all the balls are thrown from the middle. I thought it didn't matter which distribution it was one would get a Gaussian, thus, it seems strange to me that this experiment shows that all distributions form Gaussians. It seems more that all distributions that are bounded in some range (given by the hole in the middle) are Gaussian, which seems too obvious and not general. Though, I don't understand this topic enough, hence my question :p

[@gorgolyt]

Brando Miranda The way it works is this: every time a ball hits a peg, that's represents a Bernouilli distribution (i.e. a coin toss). After a ball has hit n pegs, how far it's moved will therefore follow the sum of n Bernoulli distributions. If n is big, which it is, it will also look like a Gaussian distribution. This is a special case of the central limit theorem.

[@brandomiranda6703]

I see I think I understand this better. Hitting a peg means it can move left or right (the coin flip analogy/model). But, why is it that how far its moved after n pegs forms a distribution with a sum and not any other operation? i.e. if it gets hit by 1 2 3 pegs etc it seems to be a joint r.v. not sure how or why that translates to a summation of r.v's Also, sorry if this is really stubborn, but why where the balls not left to fall uniformly across the top of the box but are concentrated through the middle? I see now that its not suppose to show that its true for every distribution, but it seems to be overly restrictive to show anything meaningful. Or simply (what I assume is the case), I don't fully understand something either about the CLT or the experiment to show why its such a awesome experiment showing the CLT theorem. Thanks for the comments though! I'm understanding this better one step at a time :)

[@gorgolyt]

The sum is of -1 representing a move to the left or +1 representing a move to the right. After n steps the ball will have moved X1 + ... + Xn where each X is a Bernoulli random variable equal to +/-1. Like I say, it does indeed only show it for a special case. But it's the simplest special case, while still a good illustration of the general principle, and in fact the first historical example of the CLT -- in fact this is how the normal distribution was discovered in the first place, binomial distributions (same thing as a sum of Bernoulli distributions) for large n.

[@3D-PHASE]

Yes, ArtOfTheProblem, feel free to use it for an educational video. Good luck, petabyte99

[@Blueaspen391]

now I'm curious to know what would happen if in the third case (smallest balls) they are dropped but only one by one

[@3D-PHASE]

blueaspen propably a similar result. Have to repeat that is not physically correct. It's a simulation within cinema 4D with an old dynamics engine. But the result is pretty well I'd say.

[@AhirZamanSairi]

If there were more small balls, enough to almost fill it all, would it form a gaussian curve?

[@3D-PHASE]

I'd say yes. Think of sand. I think it would form the gaussian curve.

[@AhirZamanSairi]

@@3D-PHASE so the small balls not forming the curve was just because there weren't enough of them right?

[@michaslusarczyk6371]

what will happen if small balls would be thrown one after another? Not like in presenation together...

[@3D-PHASE]

Michał Ślusarczyk probably similar result? Would be worth to give it a try.

[@guiwald]

The last test (with small balls) would end up with a bell shaped distribution instead of a flat one.

[@balljerker377]

Can you guys really program this ?? or just pre-programmed software ??

[@chenyangli3164]

+Ball Jerker it will just be another implementation of CLT

[@fizzicist7678]

+Ball Jerker you could just use Maya and install PhysX into it.

[@eltouristoduo]

Can we explain the seeming contradiction between the distribution of odds of a set of events and the odds of each individual event? It seems to imply some calculable change in odds that occurs with each successive event. But that implies that previous events affect future ones (seemingly nonsense). The more a ball falls the same way each time, the less likely it is to fall that way the next time. But each landing on a peg and falling one way is a separate event in time, they should have no relation to each other, right? (and the odds are never zero that it will always fall the same way). The number of rows, beyond a certain few, doesn't seem to affect the approximation of the distribution. Can anyone explain this conceptually that 'easily' makes sense? (lol)

[@entoris476]

" The more a ball falls the same way each time, the less likely it is to fall that way the next time" That's called the Gambler's Paradox, and no. Just because a lot of events happen does NOT mean that it would be LESS LIKELY for it to happen again.

[@Swordguyman]

Is there a reason why small balls do not look normal?

[@nachopuma13]

Lack of indepence is causing it, the smal balls behave more like a fluid, hence, it's trajectory is influenced by the balls around it. Because of that, the central limit theorem doesn't apply.

[@economicsiseverywhere1901]

The shape of the pegs have a more significant effect on the smaller balls, providing them with momentum to the left and right. You can see that the small balls become biased after that first peg.

[@1p4142136]

It works best for n sufficiently large.

[@ivancharllesalmeida2264]

O teorema central do limite na fórmula funciona ........mas não tem nada a ver essa aplicação nas bolinhas, ora, esse ensaio é bastante tendencioso, tendo em vista que o triângulo está sendo alimentado no centro, obviamente o maior número de bolinhas cairão no centro, pois a gravidade puxará as bolinhas p baixo e não para os lados..........o deslocamento da queda é maior que o deslocamento lateral ao bater nos pinos.........obviamente se a entrada das bolinhas for deslocado para direita, o gráfico mudará atingindo seu ponto máximo para direita....portanto sem fundamento.

[@insideoli]

At the last test, you NEED WAY MORE balls, as you can see, they almost cover a layer of balls at the bottom, but if you add more, you'll get the gauss mountain. You bet.

[@balljerker377]

I dare all of ya'll do this in Java

[@SinSeared]

System.out.println("Game - Bean Machine:"); Scanner input = new Scanner(System.in); System.out.print(" Enter the number of balls to drop: " ); int numBalls = input.nextInt(); System.out.print("Enter the number of slots in the bean machine: " ); int slots = input.nextInt(); int position[] = new int[numBalls]; int cubbyHoles[] = new int[slots + 1]; int poss[] = new int[slots + 1]; String destination[] = new String[numBalls]; for (int x = 0; x < cubbyHoles.length; x++){ cubbyHoles[x] = 0; } for (int x = 0; x < numBalls; x++){ position[x] = 0; } int LorR; for (int ballCount = 0; ballCount < numBalls; ballCount++){ for (int path = 0; path < slots; path++){ LorR = (int)(Math.random() * 2 + 1); if (path == 0){ if (LorR == 1){ destination[ballCount] = "L"; position[ballCount]--; } else{ destination[ballCount] = "R"; position[ballCount]++; } } else{ if (LorR == 1){ destination[ballCount] = destination[ballCount] + "L"; position[ballCount]--; } else{ destination[ballCount] = destination[ballCount] + "R"; position[ballCount]++; } } } System.out.println(destination[ballCount]); } System.out.println(); int bottom = (slots * -1); poss[0] = bottom; for (int x = 1; x < (slots + 1); x++){ poss[x] = poss[(x -1)] + 2; } for (int x = 0; x < numBalls; x++){ for (int y = 0; y < poss.length; y++){ if (position[x] == poss[y]){ cubbyHoles[y]++; } } } for (int x = numBalls; x > 0; x--){ for (int y = 0; y < cubbyHoles.length; y++){ if (x == cubbyHoles[y]){ System.out.print("| 0 |"); cubbyHoles[y] = cubbyHoles[y] - 1; } else{ System.out.print("| |"); } } System.out.print(" "); }

[@ivancharllesalmeida2264]

Essa é a coisa mais idiota que já vi...O teorema central do limite na fórmula funciona sim.........mas não tem nada a ver essa aplicação nas bolinhas, ora, esse ensaio é bastante tendencioso, tendo em vista que o triângulo está sendo alimentado no centro, obviamente o maior número de bolinhas cairão no centro, pois a gravidade puxará as bolinhas p baixo e não para os lados..........o deslocamento da queda é maior que o deslocamento lateral ao bater nos pinos.........obviamente se a entrada das bolinhas for deslocado para direita, o gráfico mudará atingindo seu ponto máximo para direita....portanto sem fundamento.

[@baynedahl7995]

haha balls

[@jameseldridge3445]

"Stop making generalizations" lol🤣