HMMT February, 2018

Solution:
The probability of drawing $k$ marbles is the probability of drawing $k-1$ blue marbles and then the special marble, which is $p_{k} = \left(\frac{9}{20}\right)^{k-1} \times \frac{1}{20}$. The probability of drawing no ugly marbles is therefore $\sum_{k=1}^{\infty} p_{k} = \frac{1}{11}$.

Then given that no ugly marbles were drawn, the probability that $k$ marbles were drawn is $11 p_{k}$. The expected number of marbles Ryan drew is
$$\sum_{k=1}^{\infty} k\left(11 p_{k}\right) = \frac{11}{20} \sum_{k=1}^{\infty} k\left(\frac{9}{20}\right)^{k-1} = \frac{11}{20} \times \frac{400}{121} = \frac{20}{11}$$

(To compute the sum in the last step, let $S = \sum_{k=1}^{\infty} k\left(\frac{9}{20}\right)^{k-1}$ and note that $\frac{9}{20} S = S - \sum_{k=1}^{\infty}\left(\frac{9}{20}\right)^{k-1} = S - \frac{20}{11}$.)