The Galton board: order from random bounces
Starter
A board has 3 rows. Draw every left/right path a ball can take. How many are there?
2 × 2 × 2 = 8 paths.
How many of the 8 paths end in each of the 4 bins?
1, 3, 3, 1: one path to each edge, three to each middle bin.
What is the chance that a ball lands in one of the two middle bins?
6 ÷ 8 = 75 %.
Core
With 10 rows, how many different paths can one ball take?
2¹⁰ = 1,024.
How many of them end in the middle bin (exactly 5 bounces to the right)?
C(10, 5) = 252.
What is the chance of the middle bin?
252 ÷ 1,024 ≈ 24.6 %.
What is the chance of one of the two outer bins?
2 ÷ 1,024 ≈ 0.2 %.
Out of 1,000 balls, about how many land in bins 4, 5 or 6?
(210 + 252 + 210) ÷ 1,024 ≈ 65.6 %, so about 656 balls.
Challenge
Every peg now sends 60 % of balls to the right. Which bin of the 10 is most likely, and how likely?
Bin 6: C(10, 6) · 0.6⁶ · 0.4⁴ ≈ 25.1 %. The centre of the bell moves to 10 × 0.6 = 6.
With an even number of rows n, the middle bin gets C(n, n/2) ÷ 2ⁿ of the balls. From how many rows is that below 10 %?
From 64 rows: C(64, 32) ÷ 2⁶⁴ ≈ 9.9 %. The bell spreads, so the middle bin gets a smaller share.