Probabilistic paradox: Making a scratch in a dice changes the probability? For dices that we cannot distinguish we have learned in class, that the correct sample space is $\Omega _1 = \{ \{a,b\}|a,b\in \{1,\ldots,6\} \}$, whereas for dices that we can distinguish we have  $\Omega _2 = \{ (a,b)|a,b\in \{1,\ldots,6\} \}$.
Now here's the apparent paradox: Suppose we have initially two identical dices. We want to evaluate the event that the sum of the faces of the two dices is $4$. Since $ 4=1+3=2+2$, we have $P_1(\mbox{Faces}=4)=\frac{2}{|\Omega_1|}=\frac{2}{21}$. So far so good. But if we now make a scratch in one dice, we can distinguish them, so suddenly the probability changes and we get $P_2(\mbox{Faces}=4)=\frac{3}{|\Omega_2|}=\frac{3}{36}=\frac{1}{12}$ (we get $3$ in the numerator since $(3,1) \neq (1,3$)).
Why does a single scratch change the probability of the sum of the faces being $4$ ? 
(My guess would be that either these mathematical models, $\Omega _1,\Omega _2$, don't describe the reality - meaning rolling two dices - or they do, but in the first case, although the dices are identical we can still distinguish them, if we, say, always distinguish between the left dice and the right, so applying the first model was actually wrong. But then what about closing the eyes during the experiment ?)

The correct probability distribution for dice treats them as distinguishable.  If you insist on using the sample space for indistinguishable dice, the outcomes are not equally likely.
However, if you are doing quantum mechanics and the "numbers" become individual quantum states, indistinguishable dice must be treated using either Fermi or Bose statistics, depending on whether they have half-integer or integer spin.
