What is the sample space of a dice labelled with $1,2,2,3,3,3$ for the standard dice? When we roll a dice labeled with $1,2,2,3,3,3$ for the standard dice. What is the sample space of this activity?
If someone argues the probability of getting $1$ is $\frac{1}{3}$. Because the person argues the sample space $S=\{1,2,3\}$ and the event of getting $1$ is $E=\{1\}$. So $$P(1)=\frac{n(E)}{n(S)}=\frac{1}{3}.$$ How can we correct this arguing?
Similar question:
If we draw a ball in a bag which contains $99$ black balls and $1$ white ball. If someone says the sample space $S$ should be $\{black, white\}$ and the event of getting a black ball $E=\{black\}$. So the probability of getting a black ball is $$P(black)=\frac{n(E)}{n(S)}=\frac{1}{2},$$ what's is the wrong with this reasoning?
 A: To correct your first argument, although the sample space is $\{1, 2, 3\}$, the key observation is that the probability masses assigned to each of these points are no longer equal. So the "classical probability" model no longer applies to this case and it's now not legitimate to use the formula $P(E) = n(E)/n(S)$. 
Specifically, we have $P(\{1\}) = 1/6, P(\{2\}) = 1/3$ and $P(\{3\}) = 1/2$. 
A: Possibly a return to the definitions is needed. The sample space is simply the set of possible events - absolutely nothing about the probability masses is implied by a sample space, other than if the set is complete, the sum of the probabilities is one. 
The formula you state refers to a property of a fair die - by definition each side is equally likely to appear. There is nothing 'classical' about this, it's a set of words used to establish a situation that we can reason about .
There is no reason to not choose any the three probabilities that sum to one as the three probability masses assigned to the sample space.
