The following formula was recursively derived by Rob Corlett. The arrangement is nice and regular, with the replacement pegs down the left and right sides.Īs for the bonus question, this arrangement generalizes to other cases for which the number of bins is a power of 2. Here is the solution showing how the pegs are arranged and how they route 128 marbles. Following the convention of Lionel Lincoln and Rob Corlett, we can call the equal-probability pegs standard (S) pegs. The replacement pegs are simple: They merely route any marble that strikes them to their left (L pegs) or to their right (R pegs). It turns out that just 10 pegs need to be replaced out of the 28. Rob Corlett correctly answered this problem, and Thana Somsirivattana submitted a formal proof.

What is the smallest number of pegs you will need to replace, and with what left-right ratios, in order to achieve the goal of complete equality for all bins?Īs a bonus question, can you derive and justify a formula that generalizes the above result to a Galton board of any size? You can choose pegs that direct a marble entirely to their left, entirely to their right, or in any proportion between these two extremes. You know that you will have to replace some of the traditional pegs with new ones that direct the marbles unequally to their left or right side.

You wish to modify it so that each of the bins at the bottom will collect an equal number of marbles. This one has bins in place of the eighth row and possesses traditional equal-probability pegs. Imagine that you have a Galton board of the kind shown in the picture above.