# Shift a Galton board's distribution by manipulating pegs

My question is related to this one but is simpler and more specific. It's also related to this this question which doesn't answer my question.

In a standard Galton board, balls are dropped onto a board in which apparently identical pegs or other obstructions, usually round, are arranged in rows, with the pegs in each row staggered to be halfway between those in the preceding row. Slots at the bottom of the peg array catch the balls so that one can see where each ball dropped. Dropping a sufficient number of balls into a Galton board with a sufficient number of rows should approximate a Gaussian distribution with high probability.

My question is whether I can shift the distribution to the right or the left by manipulating the pegs.

(1) Suppose I changed the staggering scheme, so that pegs in one row were, say, shifted to the right by 0.6 rather than 0.5 relative to the spacing in the previous row?

(2) Suppose that I use the normal staggering scheme, but instead of round pegs, I use oval pegs. If the ovals are all vertical, I should get the same behavior as with round pegs, I would assume. However, if I slant all of the pegs to so that the upper end points left, let's say, would that make it more likely that balls would bounce to the right? Maybe this would depend on the precise kind of curve we had at the top of the oval, but the idea is to make an oval in which more of the upper surface causes balls to bounce in one direction rather than the other.

(3) If neither of these schemes would work to shift the distribution, what would you suggest?

(If you have any relevant textual citations that come to mind, they'd be much appreciated.)

EDIT: I thought I had figured out a partial answer for option (1), but it's not at all clear to me that it works. I would welcome feedback from those with more physical insight than I have.

About (1), shifting the pegs each row so that e.g. each peg is to the left of the horizontal midpoint of the two pegs nearest to it in the previous row:

Call the two pegs above A1 and A2, and the lower peg between them in the row below, B. The usual Galton board arrangement looks something like this:

         A1     A2
o     o     o     o
o     o     o
B


Assume that a ball will hit B if and only if it bounces off of A1 or A2. The following assumptions seem reasonable (to me, at least):

(i) The points at which balls hit B's upper surface are symmetrically distributed around the top center point of B.

(ii) The angles of velocity vectors for balls hitting B are symmetrically distributed around a vector that points directly down.

(iii) The lengths of velocity vectors of balls are distributed in the same way whether a ball has bounced off of A1 or A2.

These assumptions imply, I believe, that balls bouncing off of B are evenly distributed between those going right and those going left, as is usually assumed for Galton boards.

But now what if we shift B and all of the pegs in its row a little bit to the left?

         A1     A2
o     o     o     o
o     o     o
B


By assumption (i), it seems that now on average more balls will hit B to the right of its top center point than to the left of it. This implies that on average more balls will bounce to B's right than to B's left.

So far this sounds like we can predict that the distribution of balls at the bottom of the board will be shifted to the right (and perhaps distorted in other ways). The idea is that we could the same staggered pattern in every row, like this:

row A   ...  o     o     o     o    ...
row B   ...    o     o     o     o  ...
row C   ...  o     o     o     o    ...
row D   ...    o     o     o     o  ...


But now this reveals a possible flaw in the reasoning. If more balls bounce of the right side of pegs in row B, doesn't that mean that more balls will bounce off the left side of pegs in row C? Would that counteract the effect of the bias in how balls bounce off of row B? Is the end result that it doesn't matter how the pegs are staggered? (As long as there is enough space for balls coming from the previous row to easily bounce off both sides of a peg.)

• A pity no one answered (yet). If I had the time I'd simulate it to see what happens (there are some simulations/applets available online, but I don't know if they're good enough and easy to modify). – stafusa Dec 19 '17 at 22:59
• Thanks @stafusa. I don't know enough physics to simulate it. However, I now think the question is a bit simplistic. I've discovered there are quite a few published papers investigating more subtle and sophisticated manipulations of Galton boards. I don't know whether it's possible to simply shift the distribution to the right or to the left; that might be difficult. However, it is possible to change the overall distribution quite a lot by manipulating physical properties of the pegs and their arrangement. I should probably add links. – Mars Dec 20 '17 at 18:27
• I wouldn't call it simplistic - it's only one of many possibilities, true, but a valid one. I was also a bit surprised at first by the richness of the system, but it becomes natural when we see the link to billiard systems. – stafusa Dec 20 '17 at 18:57
• @stafusa, OK, right, but I thought it might have a simple answer! Now I think that all answers to similar questions, including mine, are complicated . :-) Among other things, I wasn't thinking much about spacing, and wasn't thinking at all about size and elasticity of the pegs (or whatever they might be) and the balls (or whatever they might be). – Mars Dec 20 '17 at 20:43