Linking Probability and tensor calculus. A special dice has the property that the probability of throwing 1 is 50 %, whereas all the
other numbers appear with the probability of 10 % each. My question is: can we make any statements about
the properties of the dice, such as center of mass, or tensor of inertia?
Assumption-the dice is cubic.
 A: Suppose you have a perfect cube in the same shape as some ideal die, and assume rigid body mechanics, and that its inertia tensor is a multiple of the identity, and its COM (center of mass) is at its center; then its motion must be exactly the same as that of the fair die, in which case (by symmetry) all faces are equally likely: if some toss comes up "3", then by rotating the die before tossing, you can make any other face occupy the place where "3" was, and thus make that face come up. Why? Since nothing else changes -- the geometry, inertia tensor, and COM are all identical in both cases, and those are the only things involved in the equations of rigid body mechanics, the equations of motion for the re-oriented die are the same as those of the originally-oriented die. 
Thus, by a contrapositive argument, if you have a die that's perfectly cubical, and it's NOT a fair die, then it must in fact either have a messed up COM or inertia tensor. 
Post-comment addition: Consider a "die" made from a thin square piece of lead that forms one side of the die, with the rest made from say, a very light hollow shell of carbon fiber, with the "dots" painted in lightweight paint. Then (assuming the cube has side 2, and is centered at the origin, and the "heavy" face is at $x = +1$, we get (approximately):
$$
COM = (1,0,0)\\
I = \begin{bmatrix}
0 & 0 & 0 \\
0 & K & 0 \\
0 & 0 & K
\end{bmatrix}
$$
for some positive value $K$. This die is surely biased, in the sense that it's very likely to land with the lead side down. And that means that any die whose COM and $I$ are similar enough to these is also biased.
