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 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.
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