Isotropic Scattering (Derivation of Gaussian)

Where we last left off I had shown through the conservation of momentum that if a ball coming from the left hit the left side of a peg, it would stay on the left side of the peg.  This undermines the assumption I made that there was a 50/50 probability that the ball would go left or right when hitting the pin - though that assumption was that the ball was being dropped straight down unto the peg so it would have neither left nor right leanings (in fact, conservation of momentum would have the ball be at rest atop the peg).

I want to look into this more now to see if this problem can be resolved.  On the left is a schematic of the problem (negating gravity).

Let's let b (impact parameter) represent the distance between the marble and the peg. If $$b < r_1+r_2$$ than scattering occurs, if $$b>R_1+R_2$$ than scattering does not.  We can then look at how a small change in impact parameter effects the scatter angle $$\psi$$, $$2 \pi b db = -\sigma(\psi) 2 \pi sin(\psi) d\psi$$.  From geometry we find that $$ b = (R_1 + R_2) cos(\psi/2) $$, and so we can take the derivative of that to get $$ db = (R_1 + R_2) (-1/2) sin (\psi/2) d\psi $$.  Plugging $$ \frac{db}{d\psi} $$ back into the small change in impact parameter allows us to solve for the differential scattering cross section, $$ \sigma(\psi)=1/4 (R_1 + R_2) $$, which doesn't depend on the impact parameter; i.e. it is isotropic, but all that means that the scattering cross section is the same for all impact parameters.

I wanted to show results of the DP, but I can't seem to get the DP to make sense.  I got DP to work on Pascal's Triangle, but I think Pascal's Triangle would only work with constant probability, and I want to randomly pick the probability as the ball falls down.