Let's pick the point of the center of a sphere on a 3-d axis the $x$ coordinate, $y$ coordinate, and $z$ coordinate of the sphere's center, $(x, y, z)$ are each separately from a uniform distribution $[0, 1] $
From these coordinates, we make a sphere with radius $1$.
What is the percent of the surface area (on average) that lies within the cube region with vertices at:
$(0,0,1), (1,0,1), (0,1,1), (1,1,1),$
$(0,0,2), (1,0,2), (0,1,2), (1,1,2)$
So my first thought was to just assume that $x,y,z$ are each $0.5$ since that is the expected value of each (or the average) and then calculate it from there, but I know that that logic is flawed because of monte carlo simulations- although I can't figure out why.
Also, I know that the absolute upper bound would be $1/6$ as the percent of the surface area that is in the cube region above it's center on average should be the same percent of the surface area that is in the cube region below and likewise the the $4$ around it as well by symmetry. It cannot equal $1/6$ because some of the non orthagonally adjacent cube regions will have some of the surface area in them.