Sir Francis Galton (1822-1911) was an English statistician who invented a machine called a quincunx in 1873 to demonstrate the links between some probability distributions and the normal distribution. Some other names for the quincunx include the Galton Board, Bean Machine and Plinko.

The University of Colorado at Boulder has developed an online simulation of a quincunx machine that is fun to play with. You can have a play yourself at plinko. If you want to change the probability that the balls go to either side, go to the lab setting and see the effects of changing the numbers of rows and changing the probability of balls landing on the right or left. A few screen shots from this simulation are shown below to give you the idea.

A quincunx board is an array of pegs. Balls are released from a bucket at the top, and bounce onto the top peg, and then down the rows of pegs, eventually landing in one of a row of bins at the bottom of the board. Balls can bounce to the left or the right after they hit each peg.

When balls have an equal probability of bouncing to the left or to the right from each peg, the collection of balls in the bins at the bottom of the quincunx board is shaped like a normal curve.

This is a screenshot of the standard simulation on the University of Colorado at Boulder's PhET Interactive Simulations site. The balls have an equal probability of going to the left or the right of each pin. Can you see how the final landing pattern is quite symmetric around the middle pin and looks very like a normal curve?

When the probabilities of bouncing to the left and the right are not equal, the balls in the bins at the bottom of the board form a "skewed" normal distribution.

It is possible to calculate the probability of a ball landing in a particular bin.

Suppose we want to know the probability of a ball landing in the second bin from the left-most bin after bouncing through a 6 row quincunx board.

If the probability of bouncing to the right on every pin is \(p\), the probability of bouncing to the left on every pin is \(1 - p\). Let's start by calculating the probability that our ball takes the path shown in the picture to the 2nd bin from the left-most bin.

• It bounces to the left 4 times, with probability \((1-p)^4\)
• and then to the right twice, with probability \(p^2\)

To make life simpler, I'll number the bins from the left from now on, starting with bin 0 as the leftmost bin. We need to find the number of different paths that the ball can follow to bin number 2 (it always has to bounce 4 times to the left and 2 times to the right). There are two (equivalent) ways to do this:

1. List all the paths like LLLLRR, LLLRRL, LLLRLR, etc and count them.
2. Use combinations: we need the number of ways of choosing 2 right bounces from a collection of 6 bounces, or \(\begin{pmatrix}6\\2\end{pmatrix} = \dfrac{6!}{2!(6-2)!} = 15\).

This gives a total probability of landing in bin number 2 of \(15 (1-p)^4 p^2\). If \(p = 0.5\) (i.e. there's an equal chance of bouncing to the left or to the right), then the probability of landing in bin number 2 is 0.234375.