In a [Galton board](https://en.m.wikipedia.org/wiki/Galton_board), doesn’t the beads hitting each other affect the possibility of one falling left / right?
(i.e) will the results be same even if the beads are dropped one at a time?
If it does affect the outcome, how big of a factor do you think this could be?
In: Physics
Yes, for a big enough Galton board.
A bead dropped by itself into a Galton board can go either left or right each time it strikes a peg. The final slot it ends up in is a summary of how many times it bounces to the right. If it lands all the way to the left, it must have bounced right 0 times. The next slot over involved one rightward bounce at some point, etc.
We can model this as a binomial distribution, which counts the number of “successes” given a number of trials and a probability of success. Here, we’re saying that “success” is bouncing to the right, and the number of trials is the number of pegs a ball must strike on its way down. We’ll suppose the pegs are perfectly centered, so balls have an equal chance of bouncing in each direction. This makes the probability of success 0.5.
Any size (ideally-constructed) Galton board is an accurate model of a binomial distribution. However, we tend to think of a Galton board as a model of the n*ormal* distribution. With enough trials, the Central Limit Theorem says that the binomial distribution approaches the normal distribution. Here “enough” is about 30, which would require a Galton board with 30 layers and 465 pegs. Most Galton boards are much smaller. However, dropping the balls all at once helps to get something normal-looking even faster by adding lots of additional noise on top of the variation introduced by the pegs.
Yes it does. In fact, it might actually work *better*, I’m not sure.
The idea with a Galton board is that you’re simply showing how likely a bead is to end up in a certain space having been dropped from the middle. The balls colliding isn’t meant to be a factor. So, let’s simplify things a little.
Imagine you’ve got a checkerboard grid. There are 7 columns, and 3 rows. You have a stack of coins sitting just above the middle square in the top row of the grid. You take a coin and flip it. If it lands heads, you move it down and left. If it lands tails, you move it down and right. You repeat this process until the coin ends on the bottom row. Once this has happened, you start again with the next coin. You do this for a bunch of coins, stacking up any coins that end on a square in the last row that’s already occupied.
What you’ll find is that most of the coins end up in the middle squares. Very few end up far left or far right. To reach the far left, you need to throw heads 3 times. To reach right, you need 3 tails. However, to reach the left of middle, it could either go HHT or HTH or THH. So, you’d expect roughly 3x as many coins to go to the middle squares as the edge ones.
This is just what a Galton board is doing, but with balls and pegs instead of coin tosses. There’s no interaction needed between the coins, so there’s no interaction needed between the balls.
Yes. It does still work, as long as you drop every bead from where you want the center to be.
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