Talk:Galton board

Pic
A picture of a bean machine would make this article much clearer. Michael Hardy 02:10, 16 May 2004 (UTC)
 * (Obviously, pictures have been added since Michael posted his comment. - dcljr (talk) 07:19, 20 October 2015 (UTC))

But how does it work?
Can anyone explain why it aproximates a bell curve or normal distribution? The article doesnt define why it just isnt random. —The preceding unsigned comment was added by 159.153.156.60 (talk) 10:30, 3 May 2007 (UTC).


 * Give Binomial distribution a read and see if it answers your question. Essentally, each peg in the bean machine/Galton board presents a left/right binary decision and the resulting bin that the the beans/balls eventually fall into is a result of a series of these left/right binary decisions.


 * Atlant 12:07, 3 May 2007 (UTC)


 * It IS random! A picture of such a board with the probabilities marked on it would be helpful to understanding it. For the first pin, the chances are 50% left, 50% right, right? For the second row, with 2 pins, there's a 25% chance of it falling left off the first peg, then left off the second. A 25% chance of it falling right off the first peg, then right off the second. To go right, then left, is 25%, and left, then right, is 25%. But of the latter two, left then right, or right then left, both will lead to the middle, the same place. So the odds for the middle space are 50%.


 * Then for the next row, it's 25% chance of getting there is split into two for which way it goes next, 12.5%. But then the middle channel had a 50% chance, so the two options below it are 37.5% and 37.5% (50 + 25 / 2), compared to the 12.5% of their outer siblings. Right? Notice 37.5 + 37.5 + 12.5 + 12.5 = 100. At each row, the probabilities add up to 100% since it HAS to be somewhere! Each position's odds are those of it's two parents added together. And from there, it can go two ways, so it's children inherit half the chances of it's leftmost parent, plus half the chances of it's rightmost parent. The pegs on the far left and right, though, only have one parent, so are much less likely. The pegs one row further in, could have gone, say, LLLR, or RLLL, and whichever other paths, to have arrived there.


 * If you draw that, and work out the probabilities for each position (you only need to be able to divide by 2, and add), you'll get numbers that, drawn on a graph, would form the curve.


 * I should point out, though, that it won't always be a perfect bell curve, each run can end up in any state. It's just balls after all. It's just most likely to form a bell curve, or something close to one.


 * I understand that at least but I have a question of my own. I'm not really a mathematician.


 * 84.67.73.190 (talk) 19:10, 15 January 2021 (UTC)

The statement

"According to the central limit theorem the binomial distribution approximates normal distribution provided that n, the number of rows of pins in the machine, is large."

is incorrect. The central limit theorem is freqently misapplied in this fashion. It has to do with the distribution of the SUM of the random variables. This does not mean that if you "sum" enough graphs of the binomial distribution that it will bcome a normal distribution. —Preceding unsigned comment added by 192.25.240.225 (talk) 15:13, 31 March 2008 (UTC)


 * does that statement really wrong? I do not think so. As I knew, when a ball is falling down from the top of the machine, it will bounce left and right as they hit each pin. And each hit is an binomial experiment. The final position a ball get depends on the SUM of the results of the experiments(hits). Then, the height curve of the balls is the Probability density function curve of binomial distribution. according to CLT, it will be a bell curve.Chaosconst (talk) 06:35, 16 September 2008 (UTC)


 * It's not wrong. Technically, the de Moivre–Laplace theorem gives the result, but since that theorem is a special case of the CLT, saying the CLT gives the result is just as correct. - dcljr (talk) 07:19, 20 October 2015 (UTC)

Clarity of machine description
I had not come across Galton's box before. It would have been helpful to point out that balls are fed to the machine centrally at the top.

Tony.payton (talk) 10:41, 15 September 2008 (UTC)


 * Actually, given the wide "bowl" (or "twin ramps") shape at the top of Galton's diagram and then the narrower "bowl" below, you should be able to feed in the bean anywhere at the top, not just in the very center. (Unless you just meant that it's not fed in from the side of the box. In which case, yeah.) But come to think of it, he probably did that so he could pour in lots of beans at once (in which case the beans start interfering with each other's movements, and the motions are no longer independent). Perhaps I'm overthinking this… - dcljr (talk) 07:53, 20 October 2015 (UTC)

Distribution of the balls
The article states "If a ball bounces to the right k times on its way down (and to the left on the remaining pins) it ends up in the kth bin counting from the left." Surely this cannot be correct? If the ball bounces right k times and left l times it will end up (k - l) bins to the right of the central bin?

Tony.payton (talk) 10:47, 15 September 2008 (UTC)


 * In the second picture (photograph), if a ball never goes to the right, you can see it would end up in the "zeroth" (leftmost) bin; if it goes to the right every time, it would end up in the last (rightmost) bin. You don't need to keep track of both left and right movements. The top picture (drawing) suffers from the fact that a ball can actually reach the left or right side of the box and get bounced back towards the center (thus every once in a while, the sequence of random movements is interrupted by a non-random movement towards the center). If the box were made wider, the balls could be prevented from hitting the sides of the box, avoiding that problem. - dcljr (talk) 07:19, 20 October 2015 (UTC)

Three dimensional bean machine
I am preparing a paper in which I use the bivariate binomial distribution proposed by Aitken and Gonin (1935), which is based on a fourfold sampling procedure in contrast to the twofold sampling procedure in the ordinary bean machine. For a demonstration one actually needs a three dimensional bean machine. In the traditional bean machine, after the ball has fallen on the first pin, the ball can fall on one of two pins with probabilities p1 and p2 with p1+p2 = 1. In the case of a fourfold sampling procedure according to Aitken and Gonin after the ball has fallen on the first pin, the ball can fall on one of four pins, arranged as the corner points of a square, with probabilities p1, p2, p3, p4 with p1+p2+p3+p4 = 1 and so on. Does one know of such three dimensional bean machine.

Reference

Aitken, A.C. and Gonin, H.T. On fourfold sampling with and without replacement, Proc. Roy. Soc. Edinburgh 55, 114–125, 1935. — Preceding unsigned comment added by Ad van der Ven (talk • contribs) 18:51, 30 January 2012 (UTC)

Bean machine, bean made wrong?
I'm somewhat disappointed that the media clip 'ending' has the middle 3 columns all the same size (when the image above shows different) Was there a reason for this? Thanks 92.0.244.195 (talk) 09:00, 10 September 2018 (UTC)


 * The middle column is _slightly_ higher. And it's not an infinitely large machine, nor an ideal one. It came pretty close! It's a real video, not a simulation, plastic beads on a wooden board. So the results are always gonna be random. Might be that he filmed 10 goes, and this is the best one he got. Maybe he tried it 30 times. How much difference would that make, towards getting a more ideal result? How many tries would it take to get a perfect bell curve, with 1000 beads?


 * Nah, I'm not here to set you homework! So yeah this is the difference between stochastic randomness and real, actual, random random.


 * 90.243.234.191 (talk) 21:58, 2 February 2021 (UTC)

Why a rectangle of pegs?
It's unclear why the pegs fill a complete rectangular area instead of a triangular, maybe slightly "blown up" to a semi-circular shape. I understand well that the beans "flow" in a high density instead of (more or less) one by one (why?!?), which makes that many (or even most) beans are influenced more by their "neighbors" than by the pegs (obviously a bug in the design, unless intended to be a feature? :frown:), and they are pushed to either of the sides much more and further than they would by the pegs only. Nonetheless, the upper left and upper right pegs are never touched (especially on the design by the inventor which has a "guide" for the beans to focus them to the center of the uppermost row) and so these pegs could be omitted. &mdash; MFH:Talk 13:11, 5 February 2019 (UTC)


 * I don't believe there is any need for it to be rectangular other than it looks nice and is simpler to construct. You can see at least 3 of the photographed examples in the article already diverge from the rectangle design into more triangular shapes, and other ones do as well (like the log-normal one). --Gwern (contribs) 17:53 24 July 2019 (GMT)


 * I suppose it looks "fairer" to the layman. Just a big field of pegs, no suspicious triangles, because a bell curve is sort-of triangular, to cheat the balls into any sneaky formations. It gives a suggestion of an infinite number of identical pegs, which would give a perfect result with an infinite number of balls and time. Each actual run is just a sample of that.


 * It's not like the extra pegs cost a whole lot more.


 * 84.67.73.190 (talk) 18:10, 15 January 2021 (UTC)

Probabilities of 1000 runs of the machine
Imagine you ran a machine like this 1,000 times. Each run, you note how close the distribution is to an actual proper binomial one that the odds predict. Measure this by how many balls are in the "wrong" location compared to a nice graph that's printed on the front of the machine. How many runs of the 1,000, would be out by only 1 ball? How many would be out by 2? Graph that. I've got a funny suspicion what the graph would look like, but I'm not smart enough to figure out how to calculate it, short of writing the code to actually simulate it. But surely there's an equation that could do it. Ta for help, I really don't know the answer to this.

84.67.73.190 (talk) 18:15, 15 January 2021 (UTC)

WP:COMMONNAME
What's the WP:COMMONNAME for this, "bean machine" or "Galton Board?" Firestar464 (talk) 05:19, 24 March 2021 (UTC)

Bean machine
Is this usually called a bean machine? And does bean machine usually mean this? Online searches suggest this is more commonly called a Galton board or Galton box or Galton quincunx (though quincunx usually means five points: four in a square and one in the middle). Bean machine usually seems to involve making coffee, or perhaps somebody who eats a large amount of beans.2A00:23C7:7B0E:7400:BC5F:152F:8ACC:7C (talk) 15:16, 28 October 2021 (UTC)
 * Moved to Galton board Rumping (talk) 15:27, 28 October 2021 (UTC)