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.
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