If you throw a 6-sided die there is a probability of 1/6 to throw any specific value.
However, this assumes that dice are exactly symmetrical, and we all know that they are not, in reality. Let's assume that the sides with more eyes weigh less, because the eyes are carved in the surface of the dice.
Would that mean that the probability of throwing a 6 becomes bigger ? At first sight, that seems to make sense, because the heavier side of the dice is impacted more by gravity.
On the other hand, while the die makes a circular movement through the air, the heavier side is more likely to hit ground first. Assuming that a dice always roles after hitting the ground, that again makes it just a little less likely to actually finish in the position that it landed at.
So, do you think that weight increases/decreases the probability ?
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.