Probabilities of choosing a red ball:Case 1) 10 distinct balls,8 red and 2 black Case 2) 10balls ,8 identical red and 2 identical black

Question

I know the probability in both cases evaluate out to be 8/10. I want to know the intuition behind solving the probability of choosing a red ball when there are 8 identical red balls and 2 identical black balls. Had all balls been distinct, the result would have been obvious that there are 10 possible outcomes( since balls are distinct, each ball drawn forms a distinct outcome). Using the classical definition of probability since each outcome is equally likely the probability of choosing red ball would turn out to be number of favorable outcomes/ total number of outcomes ie 8 /10. However if 8 red balls are identical and 2 black balls are identical, the possible outcomes are only 2, either it would be a red ball or a black ball and these outcomes aren't equally likely, so we cannot apply classical definition of probability here ie total number of favorable outcomes/total number of outcomes. I want to know the intuition in such cases, about how we can relate it to the case if balls weren't identical and in what type of situations we can do so?