The start has been substantially changed as a result of Matt's comments, affecting the result.
Let $f(N,H,D)$ be the number of ways of putting $D$ balls into $N$ buckets where there are strictly fewer than $H$ balls in each bucket, and counting different orders for the balls as distinct (i.e. labelled balls and labelled buckets). We can calculate this with the recurrence
$$f(N,H,D) = \sum_{i=0}^{H-1} {D \choose i} f(N-1,H,D-i)$$
starting with $f(0,H,D)=0$ except $f(0,H,0)=1$ and remembering ${x \choose y} = 0$ if $x<y$. This stems from adding an extra bucket and putting from $0$ to $H-1$ balls in it, in an order mixed with the balls in the other buckets. For example $f(4,3,4)=204$.
How many of these have $H-1$ balls in the first bin? That is like removing $H-1$ balls and one bin, so is $f(N-1,H,D-H+1) {D \choose H-1}$ which in the example is $9 \times {4 \choose 2} = 54$.
So the probability that (a) you put the next ball in the first bin and (b) fill the first bin by doing so is $\frac{1}{D} \cdot \frac{f(N-1,H,D-H+1) {D \choose H-1}}{f(N,H,D)}$, and so the probability of filling any of the bins with the next ball is $D$ times that, i.e.
$$\frac{f(N-1,H,D-H+1) {D \choose H-1}}{f(N,H,D)}$$
or in this example $\dfrac{54}{204} = \dfrac{9}{34}$ confirming Matt's result.
One possible statistic that you could use to check this: if the probability of success for event $i$ is reported as $p_i$, let $Y_i = 1 - p_i$ in case of success and $-p_i$ in case of failure, and let $S = \sum_{i=1}^n Y_i$ (where there are $n$ events). Assuming the probability is accurate and the events are independent, $S$ has mean $0$ and standard deviation
$\sigma = \sqrt{\sum_{i=1}^n p_i (1-p_i)}$; if $n$ is large and at least some positive fraction of the $p_i$ are not too close to $0$ or $1$, the distribution should be close to a normal distribution with that mean and standard deviation. The null hypothesis (that the probabilities are accurate) would be rejected at the 5% confidence level if
$|S| > 1.96 \sigma$.
For your data I get $S = -5.46$ and $\sigma = 3.122627099$, so $S/\sigma = -1.748527707$.
That's not enough to reject the null hypothesis.
Best Answer
Let $P(i)$ be the probability of having caught the Pokémon after the $i$th attempt. We have:
$$P(1) = 0.01$$ $$P(2) = P(1) + 0.02 (1 - P(1)) = 0.02 + 0.98 P(1) = 0.0298$$ $$P(3) = P(2) + 0.03 (1 - P(2)) = 0.03 + 0.97 P(2) \approx 0.0589$$ $$\ldots$$ $$P(100) = 1$$
Using this approach, we find that $P(11) \approx 0.4968$ and $P(12) \approx 0.5572.$ For the expected number of attempts, we can start calculating from the back. When starting the $n$th attempt, the expected value for the remaining turns $E[X_{n}]$ equals:
$$E[X_{n}] = \frac{n}{100} \cdot 1 + \left(1 - \frac{n}{100}\right) (E[X_{n+1}] + 1)$$
Working recursively, we find:
$$E[X_{100}] = 1$$ $$E[X_{99}] = 0.99 \cdot 1 + 0.01 (E[X_{100}] + 1) = 1.01$$ $$E[X_{98}] = 0.98 \cdot 1 + 0.02 (E[X_{99}] + 1) = 1.0202$$ $$\ldots$$ $$E[X_{2}] = 0.02 \cdot 1 + 0.98 (E[X_{3}] + 1) \approx 11.32$$ $$E[X_{1}] = 0.01 \cdot 1 + 0.99 (E[X_{2}] + 1) \approx 12.21$$
Alternatively, you could use the formula:
$$E[X] = \sum_{i=1}^{100}iP(i),$$
where $P(i)$ is the probability of catching the Pokémon on the $i$th attempt. Since the first $i-1$ attempts must fail, and the $i$th attempt must succeed, we find:
$$P(i) = \prod_{j=1}^{i-1} (1-P(j)) \frac{i}{100}$$
This ultimately results in:
$$E[X] \approx 12.21$$