To win the doll in exactly two throws means that the first throw was a failure (90% likely) and the second throw a success (10% likely). Assuming each throw is independent of the others, you can just multiply the probabilities together to get
$$
0.9 \cdot 0.1 = 0.09.
$$
More generally, if exactly $x$ throws are required, then the first $x-1$ throws were failures and the final throw a success. Multiplying it all out gives
$$
0.9^{x-1}\cdot 0.1.
$$
The probability of requiring more than three throws to win can be seen as the complement of requiring three or fewer throws to win. That is,
$$
\Pr(\text{more than three throws}) = 1 - \Pr(\text{three or fewer throws}).
$$
As is common with problems using "at most" or "at least", it is useful to break the event into several instances of "exactly":
$$
\begin{align*}
& \Pr(\text{3 or fewer throws})\\
= & \Pr(\text{exactly 1 throw}) + \Pr(\text{exactly 2 throws}) + \Pr(\text{exactly 3 throws})\\
= & 0.9^0 \cdot 0.1 + 0.9^1 \cdot 0.1 + 0.9^2 \cdot 0.1\\
= & 0.271.
\end{align*}
$$
Finally,
$$
\begin{align*}
\Pr(\text{more than three throws}) &= 1 - \Pr(\text{three or fewer throws})\\
&= 1 - 0.271\\
&= 0.729.
\end{align*}
$$
For this, you would want to use multiple binomial distributions; for this, you would have to sum N elements, each of the following form -
$\binom{n}{k} \times (P)^k \times (1 - P)^{n - k}$
Where k is the number of trials - in this case, you would sum this little formula with k as eight, then k as nine, then k as ten, all the way to 100. A mathematical translation of this is as follows -
$\sum_{i = 8}^{100} \binom{100}{i} \times (0.01)^i \times (1 - 0.01)^{100 - i}$
The reason this works is that, for each iteration, you are finding the chance that a set amount of hardwares do not fail (which is the second term in the product), the chance that a set amount of hardwares fail (which is the third term), and then the number of combinations of hardwares this is possible with (the first term).
A shorter way to calculate may be to sum from zero to seven, and then subtract that from one, as P(Less than 8 hardwares failing) = 1 - P(At least 8 hardwares failing) - this would be written as follows -
$1 - \sum_{i = 1}^{7} \binom{100}{i} \times (0.01)^i \times (1 - 0.01)^{100 - i}$
Either way, the answer will be about $8.22020473856651522786926141017182175041798 \times 10^{-6}$
Best Answer
After $N$ trials, you can either have 9 failures in a row or not, and the sequence of trials can end in anywhere between $0$ and $8$ failures in a row if 9 failures have not occurred in a row so far. Write down recurrence relations that give you the probabilities for these cases for $N+1$ trials in terms of the probabilities for these cases for $N$ trials. Then a computer program will very very quickly give you the answer if you build up the cases in terms of increasing $N$ and remember the previous values for smaller $N$. Or, you can solve the system analytically but that will likely involve finding the roots of a 9th degree polynomial and then computing with powers of those roots.