So you are betting $1, 2, 4, 8,...,2^n$ dollars in each succesive round you do not win; that is, the prize of getting to the nth round is $\sum_{i=1}^{n}{2}^{n-1}={2}^n-1$.
If you win the nth bet, you get $2, 4, 8,..., 2^{n}$ dollars. To that we have to discount what you had to bet to get there, so your net win given that you win at round $n$ is $2^{n}-2^{n}+1=1$ dollar.
Now, we better calculate how much money we are expected to need in the betting process: it would not do to have to leave the table if we empty our pockets.
Let $X$ denote the money spent until we win. Then $X={2}^Y-1$, where $Y$ is the number of rounds until we win.
$Y\sim Geom(p)$, and so $P(Y=n)=p(1-p)^{n-1}\quad n\in\mathbb{N}$.
Thus:
$$
E[X]=E[{2}^Y-1]= E[{2}^Y]-1=[\sum_{i=1}^{\infty}2^iP(Y=i)] -1=
[\sum_{i=1}^{\infty}2^ip(1-p)^{i-1}] -1=p[\sum_{i=1}^{\infty}2^i(1-p)^{i-1}] -1
$$
The series only converges if $2(1-p)<1$; that is, if $p>1/2$. That means that you should be willing to bet infinite money in order to make the bet profitable, and all of that for a measly dollar.
In case you are curious, the expectancy when $p>1/2$ is:
$$
E[X]=\frac{2p}{1-2(1-p)} -1= \frac{2p}{2p-1}-1=\frac{1}{2p-1}
$$
But then again, is fairly obvious that when $p>1/2$ the game has positive expected value, so just playing with any bet is profitable.
Your optimal, risk-minimizing strategy is to bet a single chip each round
Other users have provided strong evidence of this via simulations. However, it can be mathematically proven that the one-chip strategy is superior to any other betting method, with an absurdly high probability of success.
OP believes the law of large numbers argues in favor of making big bets in the hope of reaching $10,000$ total chips bet very quickly:
What I said was, you'd want to use a strategy akin to betting as much as possible at each round. Since otherwise, from LLN, you are losing in general and just prolonging with small bets will make you lose more in the long run. I said you'd be able to maybe calculate the exact strategy using dynamic programming, with f(s,t) representing what you should bet at each combination of (current money, total cumulative so far). However I do not know if this is correct nor did I have time to fully solve it.
But it's the other way around.
Suppose in each round $k \geq 1$, I stake $C_k$ chips, where $C_k$ is a positive integer.
Since I win with probability $0.45$, but lose with probability $0.55$, my expected change in wealth at round $k$ is $$0.45 C_k - 0.55 C_k = -0.1 C_k,$$ with a variance of $$1.1^2 C_k^2 (0.45) + 0.9^2 C_k^2 (0.55) = 0.99 C_k^2.$$ This implies that the more I stake on round $k$, the more I expect to lose, and the higher the volatility/risk I assume. My loss- and volatility-minimizing strategy for each individual round is then to bet $C_k = 1$ each time, so that I lose an average of $0.1$ chip per round. Since variance of independent variables is additive, the variance per round is about $0.99$ "chips squared".
This may not sound promising in the long run, as my wealth $W_k$ at the start of round $k$ is expected to trend inexorably downwards: $$\Bbb{E}[W_k] = W_0 - 0.1 k,$$ where $W_0$ is my starting bankroll.
But remember--we're not playing to amass chips, we're playing to stay in the game as long as possible. And placing any bet $C_k > 1$ actually decreases $\Bbb{E}[W_{k+1}]$ relative to the one-chip strategy. There is no betting strategy for $C_k$ that can improve on the one-chip expectation of $\Bbb{E}[W_{k+1}] = W_0 - 0.1 (k+1)$.
Furthermore, the one-chip strategy also achieves the lowest possible volatility: placing any bet $C_k > 1$ increases $\operatorname{Var}(W_{k+1})$, relative to the one-chip strategy, which has the following variance and standard deviation of bankroll:
\begin{align*}
\operatorname{Var}(W_k) &= 0.99 k \\
\sigma_{W_k} = \sqrt{\operatorname{Var}(W_k)} &= \sqrt{0.99k}, \
\end{align*} where $\sigma_{W_k}$ is the standard deviation of my bankroll $W_k$ at the start of round $k$.
So if we only bet a single chip each round, our starting bankroll of $W_0 = 2,000$ chips puts us in an excellent position to place $10,000$ chips worth of bets before we go bust, albeit at a snail's pace.
If we bet a single chip per round, it would take us $k = 10,000$ rounds to achieve the stated winning condition. But since we started with $2,000$ chips, we have $$\Bbb{E}[W_{10,000}] = W_0 - 0.1 (10,000) = 2,000 - 1,000 = 1,000.$$ That is, we expect to amble up to round $10,000$, and win, with a very comfortable cushion of $1,000$ chips to spare!
Not only that, but we would deviate from this amount, on average, by only $$\sigma_{W_{10,000}} = \sqrt{0.99 \times 10,000} = 99.5 \text{ chips.}$$ Notice how teeny that is compared to our expected bankroll? That implies we have a strong chance of success--in fact, a very strong chance of success.
My winning probability if I use the one-chip strategy is insanely high
My actual bankroll $W_{10,000}$ after $10,000$ rounds, using the one-chip strategy, is given by $$W_{10,000} = 2,000 + \sum_{i = 1}^{10,000} X_i,$$ where $X_i$ are iid $\{ \pm 1 \}$-valued random variables with $\Bbb{P}(X_i = +1) = 0.45$ for all $i = 1, 2, ..., 10,000.$
$10,000$ is so large that, by the Central Limit Theorem, $W_{10,000}$ is essentially a normal random variable with mean $\mu_{10,000} = 1,000$, variance $\sigma^2_{10,000} = 10,000(0.99) = 9,900,$ and standard deviation of $\sigma_{10,000} = \sqrt{9,900} \approx 99.5$ chips.
To have run out of chips after $10,000$ bets, i.e. $W_{10,000} = 0$, would correspond to a z-score of $$z = \frac{0 - 1,000}{99.5} = -10,$$ which has an absurdly small probability, of an order of magnitude around $e^{-10^2/2} \approx 10^{-22}$, or about one sextillionth.
The probability of me being bankrupt at an earlier round than the $10,000$th is even lower than the probability of me being bankrupt at the $10,000$th round. In fact, I won't even have the opportunity to go broke until round $2,000$. So we can upper bound my losing probability by about $8,000 \times 10^{-22} \approx 10^{-18}$, or about one in a quintillion.
To put this in perspective, if I played this game once a second, it would take me a minimum of $10^{18}$ seconds $\approx 32$ billion years to see my first loss!
This explains Ionza Ieggiera's comment above:
Unfortunately, for $T = 10,000$, the script will fail by exceeding the time and memory limits of the online Magma calculator.
Even if Magma was able to simulate this game a billion times a second, it would take decades upon decades for Magma to register anything other than a win!
Accelerating my betting disproportionately increases my risk of going broke
No strategy offers a higher probability of success than the one-chip strategy, but betting more can speed up the time it takes for me to win. Since the one-chip strategy offers such a ridiculously safe bet, this is a tradeoff we're probably willing to make. But even a relatively modest gain in time saved can come with a surprisingly steep increase in risk.
Let's say I want to go $10 \times$ as fast, so I'm betting $C_k = 10$ chips each round, and it would take me $1,000$ rounds to hit the $10,000$ chip goal. I expect to lose about $0.1 C_k = 1$ chip per round with this strategy, so after $1,000$ rounds I expect to have $$\Bbb{E}[W_{1,000}] = 2,000 - 1,000(1) = 1,000$$ chips left.
(Note that using the one-chip strategy, we would expect to have far more of our starting bankroll left at this point in the game: $\Bbb{E}[W_{1,000}] = 2,000 - 1,000(0.1) = 1,900$ chips left instead. But with the one-chip strategy, we still have $9,000$ more bets to place, whereas with the ten-chip strategy, we're already finished by round $1,000$, if we make it that far.)
The resulting analysis is very similar: my actual bankroll $W_{1,000}$ after $10,000$ rounds, using the ten-chip strategy, is given by $W_{1,000} = 2,000 + 10 \sum_{i = 1}^{1,000} X_i$, where $X_i$ are iid $\{ \pm 1 \}$-valued random variables with $\Bbb{P}(X_i = +1) = 0.45$ for all $i = 1, 2, ..., 1,000.$
$W_{1,000}$ is again basically normal per CLT, with mean $\mu_{1,000} = 1,000$.
This time, though, because we're betting $10 \times$ as much per round, our variance is $100 \times$ bigger, which corresponds to a standard deviation $10 \times$ bigger. The expected value of a single bet is $-1$ chips, so the variance of a single bet is $$(10 - (-1))^2 (0.45) + (-10 - (-1))^2 (0.55) = 99 \text{ chips,}$$ and the variance of our $1,000$ bets is $\sigma^2_{1,000} = 1,000 \times 99 = 99,000$. This gives us a standard deviation of $\sigma_{1,000} = \sqrt{99,000} \approx 314.6$ chips. So the likelihood of being broke at round $W_{1,000}$ corresponds to the probability of a $z$-score of $$z = \frac{0 - 1,000}{314.6} = -3.178,$$ which is still pretty low probability (around $0.007-0.008$) but now firmly within the realm of possibility. And that's just the probability of being broke at bet $1,000$--my probability of going broke at or before bet $1,000$ is actually slightly higher. So, my impatience has multiplied my probability of losing by a thousand million million times, and deeply cut into my margin of error. At the point where you're still playing $1,000$ rounds of this game in a row, the tradeoff doesn't seem particularly worth it. Betting $2$ or $5$ chips per round, while still keeping the probability of a win essentially certain, offers an even more modest speedup. And of course, neither the five-chip strategy, nor the two-chip strategy, offers a higher win probability than the one-chip strategy.
Best Answer
If you want to maximize the expected number of rounds won, there’s no point in foregoing a win in the first round to save money hoping for a win in the second round, since that would also just count for one win. Since how much you pay in the first round depends only on whether you win, not on how much you bet in order to win, you should bet the entire sum $S$ in the first round.