Galton board
Each ball meets one peg in every row and bounces right with probability p or left otherwise, each bounce independent of the others. The bin it lands in counts its bounces to the right, so the bins fill up to the binomial distribution: the dots show how many balls the binomial expects in each bin.
Goes with chapter 5. The binomial distribution
- Balls in each bin
- What the binomial expects
Where this run ends: 500 balls with an average bin of 5, against np = 5.
The bins in numbers
| Bin (bounces to the right) | Balls | The binomial expects |
|---|---|---|
| 0 | 0 | 0.5 |
| 1 | 7 | 4.9 |
| 2 | 23 | 22.0 |
| 3 | 58 | 58.6 |
| 4 | 86 | 102.5 |
| 5 | 142 | 123.0 |
| 6 | 101 | 102.5 |
| 7 | 57 | 58.6 |
| 8 | 20 | 22.0 |
| 9 | 6 | 4.9 |
| 10 | 0 | 0.5 |
What to look for
A ball ends in bin k when k of its bounces go right, and there are many more paths to the middle bins than to the edges: one path reaches bin 0, but 252 reach bin 5 of a board with 10 rows. That count of paths is the binomial coefficient, and it is why the pile is highest in the middle.
Move p away from 0.5 and the pile leans: most balls bounce the likelier way, and the long tail points to the other side. Add rows and the pile widens and looks more and more like a bell, which is where the normal distribution comes from.