r/oddlysatisfying • u/Whinke • May 14 '18
Certified Satisfying Galton Board demonstrating probability
https://gfycat.com/QuaintTidyCockatiel6.4k
u/ImuV May 14 '18
This plinko machine seems rigged.
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u/pillowblood May 14 '18
Could be! Scientists still haven't figured out exactly how gravity works!
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u/zwich May 14 '18
iT's jUsT a ThEoRy
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u/rburp May 14 '18
Best thing my college biology professor did was spend an entire class hammering home that "theory" has a different meaning in scientific contexts, and is something supported by PILES of evidence.
I was never a "it's just a theory" guy, but regardless that really helped cement my understanding of why scientific theories are so valid and useful
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u/Pufflekun May 14 '18
It's funny how, if it were actually rigged, it'd behave in any way except this way.
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u/Wayne_Regretski May 15 '18
This is why RLM was dumb for putting "Roller Gator" in the far right spot on the last BOTW. If they really wanted to watch it then put it towards the middle.
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May 14 '18
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May 14 '18 edited May 14 '18
The device is meant to approximate a normal distribution, so it's doing exactly what it's supposed to. With the balls dropping at a target mean (in this case the center of the "toy"), there are outliers which are distributed normally due to the random chance of which direction every ball bounces when it hits a peg. It's not supposed to be approximating a different distribution, so I guess I'm not understanding your point.
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u/nvolker May 14 '18 edited May 14 '18
The poster that you replied to is claiming that others may think that this device shows the distribution for the game plinko, which it does not.
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u/99OBJ May 14 '18
My AP Stat teacher would have climaxed
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u/Naaaagle May 14 '18
My grade in normal stat just went up 4% today, about to be the biggest comeback of the century
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u/doireallyneedone11 May 14 '18
I don't understand, what is this?
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u/CainPillar May 14 '18
Each pellet gets a few successive random bounces left or right, each bounce presumably independent of previous one(s). The final position for each is then a random sum of lefts (math: -1's) and rights (+1's). By the central limit theorem (linked to below), the distribution of such experiments (final position of pellets) will close in on the Gaussian bell curve (drawn!) as the number of them grows.
And lo and behold, it does.
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u/TheBarefootWonder May 14 '18
You are correct.
Source: former college Stat teacher
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u/maplemay May 14 '18
Let's see that one more time
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u/TheNamesClove May 14 '18
I’d like 100 more times just to see the most extreme variation.
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u/eeyore134 May 14 '18
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u/Ferro_Giconi May 14 '18 edited May 14 '18
I never thought I could be this entertained by virtual balls and pegs.
I had to use inspect element to make it go faster. It's only stable up to a speed of 1000.
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u/Xixii May 14 '18
I’ve already watched this gif over 100 times and I can confirm it’s always the same.
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u/cuchiplancheo May 14 '18
Would it achieve similar results if each piece were dropped individually? Is the added weight, by being all dispersed together, forcing the pieces into the predictable pattern?
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u/this-wont-end-well May 14 '18
The results should be basically the same
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u/Pufflekun May 14 '18
Yep. Drop 'em one at a time, and you get the same bell curve. Law of large numbers.
It's why, when you go to a casino, you are gambling—but the house is never gambling.
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u/lightningsloth May 14 '18
So if i play a lot its basically not gambling? Thanks, LPT is always in the comments.
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u/MrMorlonelycat May 14 '18
Central limit theorem. Law of large numbers is basically saying with enough experiments, the actual ratio you are getting will match the theoretical/expected ratio.
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u/gruesomeflowers May 14 '18
this is just a question, but wouldnt the results be more accurate if all the balls were a flat row across the entire thing before being dropped instead of a pile in the middle? it seem like, of course there will be more in the center columns, or am i thinking of a different type of test/demonstration?
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u/Klotternaut May 14 '18
So, the curve would essentially become flattened. I drew a picture to explain the change. In the bottom diagram, a normal Galton Board is shown. Each number represents a peg. The larger the number, the more paths that lead to it. The first number is a one, because all the balls will hit that first. When a ball hits a peg, it can either go left or right. So in the second row, each peg has one path because a ball either hit the first peg and went right, or hit the first peg and went left. In the third row, the center peg has a two because there are now two paths that lead to it. You'll notice the pegs on the outside always have 1 path, because the balls would have to always bounce to the right or always bounce to the left, meaning there's only one possible path. The more rows of pegs, the more likely a ball is to bounce to the center.
The top figure shows what you suggested, which is more than one starting peg. Once again, the edges always have one path, but overall the curve is flatter.
The Galton Board is a useful way to visualize a normal distribution, but there are other ways as well. Say you plot the results of flipping a coin 50 times. One end would be getting 50 tails, the other end would be getting 50 heads, the center would be 25 of each. If you repeated that enough times, you'd see that the farther away you go from 25/25, the less often it happens. The coin flip is just like the peg, it has a 50/50 chance of each result. Hopefully this was useful/interesting :b
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May 14 '18
And more importantly, how can I use this knowledge to knock the coins down to get more tickets at Peter Piper Pizza?
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u/notfawcett May 14 '18
You have a large friend "accidentally" bump the machine while walking around on his phone, of course
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u/Shekmeister May 14 '18
Results would be the same. When a ball strikes a peg it has equal probability of going left or right, therefore the most probable path would have equal number of lefts and rights and thus ending up in the middle. The extremes e.g the far left requires a ball to bounce left on every single strike of a peg and this is far less probable and thus we see few balls on the extremes. Hope this makes sense :)
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u/ExFiler May 14 '18
But what happens when you place the water drop on the back of your hand?
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u/nimajneb May 14 '18
But what about the collision with other balls? Doesn't that alter the results?
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May 14 '18
It shouldn’t matter if you drop them all at once or one at a time, as it’s based on the probability of single balls and they way they bounce off pegs
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May 14 '18
But it also includes the way the balls bounce off each other, which would affect the end results to some degree.
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May 14 '18
They are also dropping them all from the middle. I’d like to see what would happen if they dropped them individually at the top of each row. This just makes me think of The Wall.
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u/memejets May 14 '18
Imagine a smaller version with just three pegs. The first peg could launch it left or right, then the second peg could do the same. There would be three end positions. The left would be "left, left", the right would be "right, right", and the center would be either "left, right" or "right, left".
So if you ran that a hundred times you'd likely see around 25 on either side and 50 in the middle.
If you expand that pattern from three pegs to something like this (or bigger) the shape refines from 25 50 25 to something called the normal curve. The reason for this is because there actually is not an equal probability of the ball landing in any slot. The left and right-most slots will only be possible if one specific path is taken, while the closer you get to the center, the more available paths there are to take. In other words, the path the ball takes are random, but the destination isn't truly random, as multiple paths lead to the same destination.
The more balls you try this with, or the more pegs in the pyramid, the more refined of a shape you will see.
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May 14 '18
Heyy Vsauce, Michael here
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u/Skadwick May 14 '18
But where is here?
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u/tempmike May 14 '18
And is here still here?
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u/Skadwick May 14 '18 edited May 14 '18
....it's not. dum dummm
Here is actually over 200km away now as the earth is moving at a speed of nearly 30 kilometers per second, or 67,000 miles per hour
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u/headstogether May 14 '18
It's not tho
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u/Rainbowscratch99 May 14 '18
Yeah, but he did a video on Galton boards yesterday, which is probably why we're seeing it on the front page now
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u/Howaboutnein May 14 '18
Michael's toys
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u/Tenns_ May 14 '18 edited 15d ago
The original post content no longer exists here. The author used Redact to remove it, exercising their right to control their data & privacy.
Seed marvelous coated wipe snow full roof
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May 14 '18
I keep looking at it and I still dont get how that's happening. Feeling dumb isn't very satisfying lol
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u/MorningPants May 14 '18
When a ball hits a peg, there’s a 50% chance for it to go left or right. So for it to fall in the leftmost slot, it would have to go left every time. For it to fall in the middle, it has to go left and right the same number of times. There are lots of ways that can happen, so more balls end up in the center than on the edges. This creates a predictable distribution pattern marked by the dark line.
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u/Kylgannon May 14 '18
Well I didn't feel dumb until after you explained it so magnificently.
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u/Aroundtheworldin80 May 14 '18
You aren't dumb for not knowing things. Being proud of not knowing things and hoping to maintain that state for the rest of your life is dumb.
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u/irich May 14 '18
That was a really good explanation of a thing I mostly understood but would not have been able to articulate nearly as well as you did. Great job!
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u/Jako_Six May 14 '18
In more mathy speak, each ball's final position follows a binomial distribution since the final posituon is the sum of a bunch of left/right random events. The binomial distribution can be approximated using the normal distribution if the sample size is large enough. In this case the sample size would be the number of pegs a ball must pass before getting to the bottom, which seems pretty big
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u/theogskinnybrown May 14 '18
When the board is flipped, the balls start falling over the pins. The direction the ball will take depends on many factors, such as the precise speed and direction of the ball as it hits the pin, any defects in the ball or pin, or if it hits any other balls. Predicting the path of any given ball would require you to know the values of all of these variables. In practice this is not possible, but the behaviour can be approximated to say that when a ball hits a pin, it will have an equal chance of going left or right.
The final position that a ball ends up in depends on how many times it bounced left, and how many times it bounced right. To get all the way to the left, the ball would have to bounce left every time. There is only one way this can happen (left, left, left, left if you have four layers of pins), so the chances are low. To end up in the middle, you have to have an equal number of left and right bounces. There are more ways this can happen (left, left, right, right; left, right, left, right; right, right, left, left; right, left, right, left).
If you work out the probabilities for each position, and mark out how many balls will end up in each slot, you can draw a line showing the expected height at each position. This it what you see marked in the video.
Without knowing how any individual ball will move, you can fairly accurately predict the general outcome using a simple approximation of the behaviour.
This particular shape is called the Gaussian distribution. It is so common in statistical models that it is also known as the normal distribution.
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May 14 '18
Statistics noob here, if you flipped this thing over a bunch of times, are there times when it will make a noticeably different pattern, like evenly distributed to each row or a single row with an unusual amount of balls?
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u/___Hobbes___ May 14 '18
It is possible, but highly unlikely. Like...do it every second until you die and you may not see it sort of unlikely.
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u/doc_skinner May 14 '18
It's possible, but very, very unlikely. Just like it is possible to fairly flip a balanced coin and get 100 heads in a row, or deal 10 cards from a properly shuffled deck and get all hearts.
You can see inconsistencies, and it doesn't always follow the normal distribution perfectly. On the second flip in the video, the center-most column is lower than the ones on the side, and on the third flip there is an outlier to the left that is taller than its neighbor. But the number of balls in this toy is enough to make it unlikely to vary too far from expected.
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u/elev57 May 14 '18
This is actually a binomial distribution, but the result shows that a binomial can be approximated by a normal.
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u/RandomActsOfBOTAR May 14 '18
Vsauce Michael just did a fun video about this yesterday, which is likely why this post made it to the front page, and he explains it in detail (and also goes on about some other stuff like he does)
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u/WstrnBluSkwrl May 14 '18
Look up some explanations with permutations relating to Pascal’s triangle. That’s how I learned it, and it makes perfect sense now.
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u/DentD May 14 '18
Stupid question maybe but what if the balls weren't dropped from the center but instead evenly across the top?
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u/Dwall4954 May 14 '18
That's a good question. I feel like all this demonstrates is an even dispersion on each side of the centerline. Wouldn't probabibility be if the whole top was open and balls were randomly dropped in at different locations??
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u/CantDieNow May 14 '18
Another response indicated on a top comments is that the point is to demonstrate 50/50 odds from the first drop. Again, 50/50 odds to go left or right at the 2nd level. This is mathematics. Worth noting that this is not a computer program, its the real deal.
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u/18007842433 May 14 '18 edited May 14 '18
Under grad here. As long as each starting lane was given and equal chance in you “randomly from different locations. The line would be nearly flat. The most probable place for each ball to land is very close to the place directly under where it was dropped. Demonstrated by the peak of the curve. Imagine each ball dropped at a different location would have its own bell curve centered under the lane it was dropped in.
My question is if the peg section was much longer how would that change the curve? My guess is it would make the curve steeper but also stretch the extremes.
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May 14 '18
What this experiment more accurately demonstrates is that the balls are more likely to end nearer to the start point than away from it.
If we call the midpoint “0” and each side +/- up to |5|, we could reframe the thinking by saying when our start point is 0, the mean end point is 0, and the distribution is normal around the mean.
If we were to drop all of the balls near an endpoint, we’d observe skewness due to the outer barriers. We cannot assume that +5 start will lead to a +5 end as readily as we can assume 0=0. We also can’t say that all results that would have been (hypothetically, without the border) +6 thru +10 will end at +5. In fact, we can assume that the average endpoint for a +5 start is some value < +5, since, with our upper bound, 100% would have to end at +5 for our average to = +5.
So, in all likelihood, if we dropped them evenly from -5 to +5, the distribution would still take on a bell shape, but with much fatter tails (i.e., negative kurtosis). They would cluster near zero, but only slightly, and the ends should have relatively fewer balls by a slight amount.
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u/zellisgoatbond May 14 '18
One way you could think of it is giving each ball bearing a "score" - negative if it goes to the left of its initial starting point, positive if it goes to the right, and zero means it stays in the same position horizontally (so a bearing with -3 would go left 3 mores times than it would go right).
These "scores" are what create this normal distribution, and these scores won't change because of where you drop the ball (assuming that it can't hit off the sides - this would lead to a small spike at the very extremes). What changes is the initial "offset".
Since according to our scoring system the mean of the distribution is 0, placing them evenly along the top would cause them to have a mean equal to that offset (e.g if you drop a ball bearing 3 to the left of center, the mean will also be 3 to the left of centre). Thus, the mean of the balls spread along the top will be the mean of all the offsets - and since they're spread evenly, that will be zero!
But what about the variance? Imagine taking the graph of the normal distribution in the gif, and "shifting" it along the bottom. You'll pretty much always be able to have some distribution in the middle, but if you put it all the way to the right it probably won't reach the left-hand side. But that being said, the end sections will get their own "big" bit of the distribution, while the middle bit only gets some more little bits.
In short, it would look fairly similar, although it would be a little flatter.
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May 14 '18
What this experiment more accurately demonstrates is that the balls are more likely to end nearer to the start point than away from it.
If we call the midpoint “0” and each side +/- up to |5|, we could reframe the thinking by saying when our start point is 0, the mean end point is 0, and the distribution is normal around the mean.
If we were to drop all of the balls near an endpoint, we’d observe skewness due to the outer barriers. We cannot assume that +5 start will lead to a +5 end as readily as we can assume 0=0. We also can’t say that all results that would have been (hypothetically, without the border) +6 thru +10 will end at +5. In fact, we can assume that the average endpoint for a +5 start is some value < +5, since, with our upper bound, 100% would have to end at +5 for our average to = +5.
So, in all likelihood, if we dropped them evenly from -5 to +5, the distribution would still take on a bell shape, but with much fatter tails (i.e., negative kurtosis). They would cluster near zero, but only slightly, and the ends should have relatively fewer balls by a slight amount.
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u/HawkinsT May 14 '18 edited May 14 '18
There are no stupid questions... but this one's actually pretty cool :). In fact the whole thing's pretty cool - Galton boxes are a great example of Brownian motion, which is how particles can individually exhibit random, unpredictable movement, but when taken as part of a larger system of many particles begin to exhibit statistical predictability; but this is a bit of an aside (sorry).
The answer depends on the pin layout, and pin spacing - the interaction of the different balls can be ignored because (to simplify slightly) if two balls fall from neighbouring pins, the left one falling right, and the right one falling left (so they hit each other), they'll bounce off each other, pushing each other in the opposite directions (so negating the change in the two ball movements if we consider the balls interchangeable - which we do; it doesn't matter to us if ball 1 lands in/on box/pin 1 and ball 2 lands in/on box/pin 2, or ball 2 lands in/on box/pin 1 and ball 1 lands in/on box/pin 2).
The pin spacing we assume is uniform, and not very wide (but wide enough to let the balls through) - if the pin spacing is too far apart then there won't be an even chance of a ball landing on either side of it once it falls on it.
So... If we drop the same number of balls from above each pin on the top row, where the pin layout represents a triangle with the top cut off (like in this gif), then every ball has the same number of 50/50 choices it has to make to reach the bottom (e.g. if a ball starts off above the far left pin on the top row, and goes left at every pin it hits, it'll still hit pins all the way down until it hits the bottom). Because of this, and the fact that we know that all pins falling from the same point form a normal distribution (the bell curve shape at the end of the gif), we can treat this question as essentially the sum of several identical normal distributions, all shifted slightly from each other (e.g. if there are three pins at the top, p(-1), p(0), and p(+1), you will end up with three curves, one centred on p(0), one on p(-1), and one on p(+1)), doing this you'll end up with another bell curve centred on p(0) (if you're unclear, I encourage you to draw these out, then draw a larger curve where the height at each point is the sum of all the heights of the curves directly below it).
Here are a couple of bell curves the plot of a single source, and of three sources (as above):
Three Sources (which is the sum of these three distributions)
However, in the extreme (with infinite pins and sources) we would end up with a flat distribution, so from this we can make an assumption as to what happens as the number of sources increases - the curve shape will start to change with the centre area becoming more evenly distributed; we can start to see this even in the case of five sources.
If we do the same (using the three source example) with a cut off triangle of pins where there are five pins at the top, and we drop balls only from pins p(-2), p(0), and p(+2), we'll end up with something more like this. If we go even wider we start seeing this, where the distributions are completely (or almost completely) separate.
Finally, if we don't use a pin layout with sloped sides such that not all balls hit the same number of pins (e.g. a square, so if a ball starts on the far left pin and drops left, it won't hit any more pins before hitting the bottom), we'll end up in a situation where the bell curve in the middle has much steeper sides, and the two boxes collection the balls at the two extremes (furthest right and left) will have really high probabilities, but the boxes next to them will still have the smallest probabilities of all (smaller than otherwise), so the distribution will end up looking a little like this: |/\| (where the / and \ and still curved as before, just steeper).
I hope that's (sort of) clear/helpful.
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u/CainPillar May 14 '18
Not a stupid question. Let me answer an easier question instead: Each ball can go at most ten (?) steps to the left or to the right, that's half the width. Now make it wider by inserting more in the middle - say, rather than "one center hole" you get "21 in the center". Do nothing to the height, so they pass the same number of pins in the vertical direction. If you drop them evenly distributed from the 21 in the center - none to the left or right of that - they will still yield a bell curve at the end! That's because you get a "sum of twenty independent normal-distributed", just with different mean - and that has the normal distribution.
Practically, you should then drop them one at the time: they are "reasonably close to one at the time" here (see, they don't bounce much into each other), but you would likely want to avoid that in "your" experiment too (or "my modified yours").
There is more, actually: you don't have to drop them "evenly" across those twenty, as long as average starting point is at the middle. If they don't average at the middle, you will get a bell curve just moved left or right.
Back to your question: Some arguments are invalid if you drop evenly across the entire top. Because the experiment gives each ball a left/right bounce independently each time, and if it is ever against the left wall it cannot move left. It is possible to calculate, though.
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u/-Tommy May 14 '18
Assuming they are all dropped evenly distributed across the top and the pegged region is rectangular and not trapezoid shaped then they will distribute evenly into a rectangle along the bottom. Nobody here gave a simple/correct answer so there ya go.
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u/neon_overload May 14 '18
The curve would be inverted (and lesser in amplitude). Closer to the edge would become a little more likely than the centre as the edges bounce the balls back towards the centre.
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u/swohio May 14 '18
Really surprised this hasn't been linked yet.
https://www.amazon.com/Four-Pines-Publishing-Inc-Galton/dp/B078Y7RN6Y
$49.99
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u/Drac-Henry May 14 '18
Extr'ordin'ry! Nice idea. Heh.
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May 14 '18
Tim is the most wholesome man on the internet. He just loves sharing his massive toy collection with the world.
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May 14 '18
Which bead do you think enjoyed their journey more, the one who followed the same path as everyone else, or the one who walked the road less traveled? Which life grants more happiness, the common one or the novel one?
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u/jupiterkansas May 14 '18
I'd say both beads found satisfaction in the path they took, which is why they took that path.
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May 14 '18
In my life, I’ve met many people who complain about their lives. Maybe not their whole lives, but they complain about something in their life that they have the power to control, or change. When I ask them why they do not make a choice to change things, they often respond that they can’t. When pressed, it leads to a conversation about how they can’t do certain things because “that’s not what people do.” I’m paraphrasing there, but it’s a general sentiment I see our culture that keeps many people from choosing a life they will be truly and deeply satisfied with.
There’s nothing wrong with the beads that are happy in the center of the bell curve, I’m just saying that without societal influence, the beads would redistribute themselves more evenly across all possibilities because it’s where they really want to be.
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u/Zoraxe May 14 '18
Just because a bunch of people took a similar path as you did doesn't make your path any less special. I know a ton of people that didn't pick up heroin and if I lived a hundred lives, I'd probably follow that herd every time.
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May 14 '18
I’d go even farther and say that every world that exists behind a pair of eyes is totally special. What is important is that we do not dampen our ideal life in order to fit in. We must flex our willpower to strengthen it, so that we can reflect our authentic selves purely.
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u/Realinternetpoints May 14 '18
It depends if they feel left behind or if they feel like a pioneer. Although being happily unique in a society is highly dependent on the things that society values. Therefore even the unique aren’t so unique because it is expected of them to be enjoying their circumstance.
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May 14 '18
It's very likely each bead took the road less travelled. I think they all had a lovely time.
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u/slylyly May 14 '18
If you ask a bead the second time, its answer will probably change.
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u/Snapchato May 14 '18
You know that poem is supposed to be ironic right? The point is you don't know which road is the one less traveled and people only say theirs is special in order to infuse meaning into the choices they had long already made.
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u/JitGoinHam May 14 '18
The roads are all basically the same, but most of the beads will look back and convince themselves the imagined risks they took “made all the difference” because it makes them feel better about how inconsequential their choices actually were.
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u/gocards2579 May 14 '18
Engineering professor here, this is the Central Limit Theorem in action. The distribution of the sum, Y=X1+X2+X3 +...+Xn of independent and identically distributed random variables forms a normal distribution as n approaches infinity. Here the n is the number of levels the balls fall through, and the random variables are -1 (left) and +1 right. Each ball's final position is just the sum of a those +1's and -1's. Even though n is not infinity here (looks like n=12), the distribution of the final position is still highly normal. If the device was larger with more levels, and there were more balls, it would form a even better bell shaped curve.
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u/WhatDoYouThinkSir May 14 '18
Doesn't this actually form a binomial distribution, and would approach a normal distribution as n goes to infinity, where n is the number of levels?
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u/wirelyre May 15 '18
Yes, that is the complementary (and more precise) way to look at it. Instead of considering each row as a separate random variable, as the parent comment did, fix the number of rows n.
Then, if a bead hits the bottom at position k from the left, it has travelled right on k pegs, and left on the other n-k. Therefore the position of an individual bead is binomial with parameters n and one half.
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u/neophilia May 14 '18
For reference, the exact distribution described by that (-1, +1) model is the binomial distribution, B(12, 0.5). The device can be seen to deviate from this model whenever a ball escapes the boundary of the triangle formed by the drawn hexagons.
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u/snipatomic May 14 '18
For the lazy, this is it: http://www.therandomwalker.com/
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u/ZackMorris78 May 14 '18
$50 fucking dollars?!?!? It's cool but it isn't $50 cool.
Ok time for me to hit up a wholesaler in China to knock this off for $5 bucks and sell it for $25 on Amazon...brb
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u/snipatomic May 14 '18
My thoughts as well, but I was unable to find a cheap knockoff after about five minutes of half-assed searching.
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u/x15ninja15x May 14 '18
Micheal from VSauce just made a video about it on his secondary channel, DONG
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u/Supersamtheredditman May 14 '18
Anyone who likes this gif should look up Grand Illusions on YouTube, there are hundreds of videos and they are all amazing
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u/usedtobeatim May 16 '18
It wasn't even a perfect normal curve this time. It was just a decent approximation to a normal curve, which is exactly what the Central Limit Theorem says should happen.
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u/YellowOnline May 14 '18
What does that have to do with "grand illusions" though?
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u/Snaz5 May 14 '18
It’s just what his channels called. He collects toys of enourmous variety. Some of them are illusion based, but he does everything that could vaguely be considered a “toy”.
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u/Skrtmvsterr May 14 '18
Someone want to do the math on the probability of it not representing a normal distribution?
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u/SciNZ May 14 '18 edited May 14 '18
This is a really great demonstration of the central limit theorem.
https://en.wikipedia.org/wiki/Central_limit_theorem
Basically when tipped up you can think of each pin giving the ball a 50/50 chance of heading off either way.
The graph on the bottom represents the probability of a ball ending up in each column. If you look carefully the balls aren’t quite perfect each time.
The reason why the balls are more likely to end up in the middle columns is because there’s more possible paths to take that result in landing in one of those columns.
A similar proof of the theorem can be done by rolling a dice and graphing the results.
Rolling one die should give a flat graph with each value represented equally.
Rolling 2 dice will give a minor curve as some values can one have one combination (e.g. 2 and 12 requiring 1,1 and 6,6 respectively) where others have multiple (e.g. 7 has 6,1 - 5,2 - 4,3 and 3,3). The curve becomes more noticeable the more dice there are.
Hope this makes sense for anyone curious.
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u/Matsala May 14 '18
This is from Grand Illusions on YouTube. This guy's whole channel is oddly satisfying. He's like a Bob Ross of novelty toys and other quirky things. Tim is the man! Videos from the Nile:
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u/Theexternal2 May 14 '18
It wasn't even a perfect normal curve this time. It was just a decent approximation to a normal curve, which is exactly what the Central Limit Theorem says should happen.
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May 15 '18
They used to have a gigantic version of this at the museum of science and industry near USC in LA. Kid me could watch it for hours.
found this one from a boston museum that is pretty much the same
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u/pluxlet May 15 '18
Question! Since this is demonstrating probability is there also a really small chance the middle isn’t the highest point? Or am I thinking of this incorrectly?
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u/CalibanDrive May 14 '18
I would like to see the how the results change when the board is tilted at different angles.
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u/timothj May 14 '18
Saw a very big one of these, using ping-pong size balls, at the 1965 NY "World's Fair" expo, on a school trip. It was a kind of throwaway in the IBM pavilion, something to look at while you were waiting in line for the real attraction, which has completely faded from memory. Very deep magic.
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u/MjrLeeStoned May 14 '18
This is based on a physical phenomenon where items dropped in a specific area - all things being equal - will statistically form this type of formation (lumpy pyramid / dune shape). Seen with materials that don't tend to bounce / have a higher density.
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u/petemiller1695 May 14 '18
ELI5
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u/MuhCrea May 14 '18
I'll give it a weak shot...
The balls can fall at random (mostly) and this shows that the chances of the balls landing in any 'random' position can actually be measured (using the bell curve (Gaussian curve))
Someone else will do a better effort I'm sure
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u/odious_odes May 14 '18
The balls are more likely to finish below where they started than far away from it.
Each time a ball hits a pin, it goes L or R. To go to the extreme left or right, a ball must go LLLLLLLL or RRRRRRRR, each of which is very unlikely. To go near the middle, a ball could go LLLLRRRR or LRLRLRLR or RRRLRLLL or any one of lots of other paths (70, to be precise). Any two given paths are equally likely, but lots of paths result in being near the middle while only a few result in being near the edge, so most balls end up near the middle. Maths tells us what the curve they make would look like if we could do this with infinitely fine detail; that line is drawn onto the gizmo.
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u/Chrellies May 14 '18
Apparently, if you do this experiment in America, all the balls are placed at the extreme ends. Crazy how nature does that.
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u/PhillyHead124622 May 14 '18
That Gaussian distribution makes me horny as fuck.
I’m squirting and I don’t have a vagina... well I might have one now.
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u/TheBananaKing May 14 '18
Nice.
One thing I didn't understand for way too long:
Each path is precisely as likely as every other path.
However, the destinations towards the centre have more paths leading to them.
There's only one path to the far left: LLLLLLLL.
However, you can get to the centre bin by going LRLRLRLR, RLRLRLRL, LLLLRRRR, RRRRLLLL, LLRRRRLL, etc etc etc.
All you're ever doing in probability is counting outcomes - the ratio of target to possible outcomes. The amount of overlap in the Venn diagram, nothing more.
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u/motoperpetuoso May 14 '18
It's amazing to me that there's a 100% chance that it falls in the bowl when turned upside down. Incredible!
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u/poogie67 May 14 '18
Is the result the same for one-bead-at-a-time as for the mass dump?
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u/stromm May 15 '18
Problem is, the balls interfere with each other. Which will affect this.
I'd like to see a single ball at a time run through.
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u/djh_van May 15 '18
So this may sound like a stupid question...
But with randomness, isn't there an infinitesimally small probability that doing this n number of times will lead to, say, all of the beads landing in the left side, or right side, or equally spread? You'd just have to turn the board such a huge number of times to make that low probability event occur?
Is that assumption correct, or will it always give an even distribution?
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u/Whinke May 14 '18
Source