Draw cards repeatedly, until we find the ace of spades. Probability that we draw between 20 and 30 cards?

The probability that the $i-th$ card is the ace of spades is $\frac{1}{52}$for every $i=1,...,52$. So, the desired probability is $\frac{11}{52}$ because $[20,30]$ contains $11$ numbers.

When you shuffle the deck, the ace of spades is as likely to be in any spot as any other. You win if it is in $11$ of the $52$ spots. You can do the calculation your way, but you have to start with the chance that the ace is not drawn in the first $19$ cards. The chance of that is $\frac {51}{52} \cdot \frac {50}{51} \cdot \dots \cdot \frac {33}{34}=\frac {33}{52}$ Then the chance that it is in the next $11$ cards is $\frac {11}{33}$. The product of these is $\frac {11}{52}$ as it should be. If you want to keep going your way, the chance it is the $20$th card given it has not been found is $\frac 1{33}$. The chance it is the $21$st card, given that it is not found by card $19$, is $\frac {32}{33} \cdot \frac 1{32}=\frac 1{33}$ You can keep going this way, but it is a much harder approach and will still come out $\frac {11}{52}$ in the end.

This is a pretty straightforward application of the Negative Hypergeometric Distribution which is used to model the number of trials to achieve a success when sampling without replacement. If you have a population of N (52 cards) with m successes (1 Ace of Spades), and you wish to draw till you achieve j successes (draw the Ace of Spades once), then the probability that you need to draw k times is

$$ P(x=k) = \frac{{k-1 \choose j-1}{N-k \choose m-j}}{N \choose m} $$

Thus, what you are interested in is (as each event k=1, k=2, etc is mutually exclusive)

$$ \sum_{k=20}^{30} P(x=k) = \sum_{k=20}^{30} \frac{{k-1 \choose 0}{52-k \choose 0}}{52 \choose 1} = \sum_{k=20}^{30} \frac{1*1}{52} = \frac{11}{52} $$