I have $3$ choices to make a attempt: first one costs $460$, and has a $10\%$ chance of success, and if it fails, it has a $3\%$ increase in chance for the next attempt.
The second costs $1430$, and has a $30\%$ chance, and if it fails, has a $9\%$ increase in chance for the next attempt.
The third costs $4430$ and has a $100\%$ success rate
What is the most efficient combination?
Let's calculate the expected cost of each.
The third case is the most trivial. Regardless of what you do, you'll pay a fixed cost of $\color{red}{4430}$.
In this first case, we have a $10\%$ probability of paying $460$, $90\%\cdot13\%$ chance of paying $2\cdot460=920$, etc. We can summarize this in the following sum:
$$460\cdot\sum\limits_{n=0}^{30}\bigg(n\cdot(0.1+0.03n)\cdot\prod\limits_{j=0}^{n-1}(0.9-0.03j)\bigg)$$
This is approximately equal to $\color{red}{1945.88}$.
You can use a similar process to compute the expected cost of the second case, and what you'll find is that the expected cost is approximately $\color{red}{2125.67}$.
So, the best choice is the $\textbf{first}$ choice.
P.S. When you said that there's a $9\%$ increase in probability, I presumed that when the probability went over $100\%$, the probability equals $100\%$