For the small cases, you can have F(0)=1 and F(<0)=0

I modeled the sphere using voxels, and tried to come up with the greediest steps, getting 11.
colab.research.google.com/drive/185hNX...

I reasoned that for two random points to be collinear to a given position, they must be on the same line oriented at angle theta away. The probability this happens is proportional to the length of the line segment contained within the square oriented at that theta. Then avg over all theta.

Updated plot for the simulation

For the EC, there are multiple points A from which we can reach point B in a given layer, and we must sum over the product of ways from A to B and the number of ways to reach point A. This number blows up quickly to over 1 billion.

For the regular fiddler, there is only one way to travel from points A to B within one layer, therefore the number of ways to reach each point in a given layer is identical and equal to the sum of ways to reach all points A in the current layer.

colab.research.google.com/drive/1woSjU...

colab.research.google.com/drive/1PfBxj...