There are 800 marbles in a bag. Each marble is colored with one of 100 colors, and there are eight marbles of each color. Anna draws one marble at a time from the bag, without replacement, until she gets eight marbles of the same color, and then she immediately stops. Suppose Anna has not stopped after drawing 699 marbles. Compute the probability that she stops immediately after drawing the 700th marble.
It is equivalent to looking at the problem backwards, where the probability of 99 balls are each of a color and the 100 ball could be any color and are picked from the bag without replacement and thus the probability is
Invoking hypergeometric distribution, we get
Prob=$\frac{{792\choose 1}.{786\choose 1}\cdots{8\choose 1}{700\choose 1}}{{800\choose 100}} = \frac{8.99\times8.98\times\cdot 8.1 \times 7.100}{800\choose 100}=\frac{7.8^{99}.100!}{800\choose 100}$
I don’t understand all of what you did (in particular where $7.8^{99}$ came from), but in any case this is huge overkill.
If Anna hasn’t stopped after drawing $699$ marbles, then there is at least one marble of each colour left, and thus exactly one colour of which $2$ marbles are left. The probability that she draws one of the $99$ marbles that would end the process is just $\frac{99}{101}$.