This is more of a statistics/probability question than algebraic....assuming that each "turn" consists of three throws then there are three chances to score a 3 or an 8 score...
you have two possible outcomes, and it is possible for an outcome to repeat (you can score three threes in a turn, unlike drawing cards from a deck, where once the card is drawn it is impossible to draw it again)..so this equation tells you the # of possible outcomes in a turn:
n^r = total combinations of outcomes per round
n=2 (number of different outcomes per throw)
r= 3 (number of turns)
so you have 2^3 = 8 possible combinations of scores in a given turn. If the greatest score you can earn is 24 (8+8+8) and the lowest is 9 (3+3+3), then the difference between your greatest and smallest outcomes is 15, your scoring range.
From the earlier equation you can show that there are only 8 possible results in this range, so 15-8 = 7 outcomes that are impossible to score in the range of scoring...
As one other poster mentioned, there are an infinite number of impossible outcomes, if you are considering all numbers and not just those you would consider in the range of scoring. My answer assumes that you are only considering the range of scores you would expect per turn.
If you provide additional information about the parameters I might be able to give you a more specific answer!