The following problem is motivated by a certain aggravating side-quest in an old video game. The side-quest is itself a game which follows a sequence of rounds; each round can be a success or a failure, and the game ends after $N$ successes. In its simplest form, we take the probability of failure after $k$ successes to be $k/N$, in which case we recover the coupon collector's problem. (In the original setting the failure of probability is rounded up to the nearest whole percentage less than 100%, e.g., 84.5% rounds to 85% but 99.5% rounds to 99%; I chose not to include this detail for simplicity.)
Thus far there is nothing for the player to do besides keep drawing until success. To add a strategic element, we include a base cost for each round (one "token"). Furthermore, we allow the player to bet multiple tokens: each token past the first reduces the probability of failure by 1%, to a minimum of 0%. Here the analogy with the coupon collector's problem breaks down: drawing multiple coupons can lead to more than one new coupon being drawn, and the probability to get any new coupons would moreover not be a linear function of the number already found. So this is a coupon collector's problem where the probabilities have been 'linearized'.
The question is then what strategy to pursue, and this will very much depend on what goal we pursue. If what we want to minimize is the number of rounds, then we should always bet enough tokens to eliminate the chance of failure and thus finish in $N$ rounds. But these bets grow linearly and thus the total number of tokens required scales as $N^2$. On the other hand, we can simply bet the one-token minimum each round. In that case the number of tokens is the same as the number of rounds, which scales as $N\log N$ per the coupon collector's problem. This strategy therefore minimizes the token count at the cost of more rounds, especially near the end when the probability of failure is quite high.
This last point suggests that a reasonable tradeoff between rounds and tokens can be achieved by a strategy which interpolates between the two outlined above: Bet the minimum initially, but transition towards higher bets as the success probability shrinks. But it's not at all clear to me what an 'optimal' strategy is, or even how to properly formulate the tradeoff, and I would appreciate some perspective on how this might best be framed.
You have not said what $k$ is, but presumably it is the number of successes so far.
Let's suppose your strategy is to bet $b_k$ in such a situation so (if I have understood correctly) the probability of failure becomes $\frac{k}{N}-\frac{b_k-1}{100}$. So to get the next success, you expect
to need an average $\dfrac1{1-\frac{k}{N}+\frac{b_k-1}{100}}$ attempts
at an expected cost of $\dfrac{b_k}{1-\frac{k}{N}+\frac{b_k-1}{100}}$ tokens
The derivative of that last expression with respect to $b_k$ seems to be $\frac{100 N \left( 99 N-100 k\right) }{{{\left( b_k N+99 N-100 k\right) }^{2}}}$ which is positive if $k < 0.99 N$ and negative if $k > 0.99 N$ and zero if $k=0.99N$. This suggests that if your aim is to minimise the number of tokens then
If $N>100$, this may be slightly better than always betting the minimum but still $O(N \log(N))$ in terms of the expected total number of tokens needed.