I think you are correct, but the probability would go to zero if the calculation was taken that far (I stopped at at 14 for the size of set A). The probabilities are weighted, so that is why the sum equals to 1, however, on second thought, the calculation for P(A is n) is more complicated than I first thought. I think the original response I had is probably the simplest (that is, the calculations for P(B is not symbols in A | n symbols in A) in the table above. Other than that, I think I've exhausted the knowledge that can actually help you (hopefully it helped a little bit).