Impact of weight of the dice

We could start off by considering the moment of inertia of a cube about it's centre of mass. That is: $I_{CM} = \frac{1}{6}ms^2$,

where $m$=mass of the cube and $s$=length of a single side.

But here, you're saying that the indents to each side due to the marking of number will have an effect on the mass distribution over the volume of the die. Thus, this will change the location of the centre of mass. So I presume that we could represent this distribution as a matrix:

$\begin{pmatrix} \frac{1}{6}m(s+x)^2 & 0 & 0 \\ 0 & \frac{1}{6}m(s+y)^2 & 0 \\ 0 & 0 & \frac{1}{6}m(s+z)^2 \\ \end{pmatrix}$

Which will give us the moment of inertia tensor, dependant on how the centre of mass is affected in relation to its middle on the $x, y, z$ plane. Where to go from here however, I do not know.

This is mostly a physics question: if you assume that the dice is thrown such that it rolls many times, then it will have a preference to finish in a position where its center of mass is the lowest. Thus, since 1 is the heaviest, you would have an ever so slightly greater chance of getting a 6.