I want to propose a strategy in the limiting case $n=\infty$. Maybe this is better described as a limit of strategies, since I will allow a sequence of coin flips that are each assigned probability $\epsilon$ of success (where $\epsilon$ is infinitessimal). The total amount of probability we will "spend" before the next head appears will then be exponentially distributed, with mean 1.

I will denote by $f_k(x)$ the probability that my strategy results in success if we still need $k$ heads, and have $x$ probability remaining in our "budget." Here is how the strategy works: If $x\geq k$, we assign probability 1 to the next $k$ flips. This results in $f_k(x)=1$.

If $x\in (k-1,k)$, then we assign the next flip probability $x-(k-1)$. If this flip lands heads, we will win with probabilty 1. If the flip lands tails, we will win with probabilty $f_k(k-1)$. It follows that $$ f_k(x)=(x-(k-1))+(k-x)f_k(k-1) $$

Finally, if $x\leq k-1$, then we will assign probability $\epsilon$ to each subsequent flip, until we see a heads. This gives $$ f_k(x)=\int_0^x e^{-t}f_{k-1}(x-t)\,dt $$

We can recursively compute $f_k(x)$ for any $k$. Each $f_k$ is a continuous, piecewise-analytic function. The first few values (computed with the help of Mathematica; I hope they're correct) are: $$ f_1(x)=\begin{cases} x\text{ if }0\leq x\leq 1\newline 1\text{ if }x>1 \end{cases} $$

$$ f_2(x)=\begin{cases} -1+x+e^{-x}\text{ if }0\leq x\leq1\newline -1+\frac{2}{e}+(1-\frac{1}{e})t\text{ if }1\leq x\leq 2\newline 1\text{ if }x>2 \end{cases} $$

$$ f_3(x)=\begin{cases} -2+x+(x+2)e^{-x}\text{ if }0\leq x\leq 1\newline e^{-x}+\frac{3}{e}-2-\frac{x}{e}+x\text{ if }1\leq x\leq 2\newline -2+x+\frac{(1+e)(3-x)}{e^2}\text{ if }2\leq x\leq 3\newline 1\text{ if }x\geq 3 \end{cases} $$

While I don't have a proof this strategy is optimal, I've got a heuristic argument that assigning probability $\epsilon$ to each flip is a good idea. If our budget is $x$, then whatever our strategy, the expected number of heads we will have seen when we exhaust our budget is $x$. If the desired number of heads is much larger than $x$, we will need to make the variance in the number of heads large. If we assign probabilities $p_1,p_2,\ldots$ to the coin flips (with $p_1+p_2+\ldots=x$), then the variance in the number of heads is $\sum p_i(1-p_i)$, which is bounded above by $x$. We can make the variance arbitrarily close to $x$ by taking each $p_i$ as small as possible.

The argument is a little different if the next head that appears could cause our remaining budget to be larger than the number of additional heads we need to win.

I want to propose a strategy in the limiting case $n=\infty$. Maybe this is better described as a limit of strategies, since I will allow a sequence of coin flips that are each assigned probability $\epsilon$ of success (where $\epsilon$ is infinitessimal). The total amount of probability we will "spend" before the next head appears will then be exponentially distributed, with mean 1.

I will denote by $f_k(x)$ the probability that my strategy results in success if we still need $k$ heads, and have $x$ probability remaining in our "budget." Here is how the strategy works: If $x\geq k$, we assign probability 1 to the next $k$ flips. This results in $f_k(x)=1$.

If $x\in (k-1,k)$, then we assign the next flip probability $x-(k-1)$. If this flip lands heads, we will win with probabilty 1. If the flip lands tails, we will win with probabilty $f_k(k-1)$. It follows that $$ f_k(x)=(x-(k-1))+(k-x)f_k(k-1) $$

Finally, if $x\leq k-1$, then we will assign probability $\epsilon$ to each subsequent flip, until we see a heads. This gives $$ f_k(x)=\int_0^x e^{-t}f_{k-1}(x-t)\,dt $$

We can recursively compute $f_k(x)$ for any $k$. Each $f_k$ is a continuous, piecewise-analytic function. The first few values (computed with the help of Mathematica; I hope they're correct) are: $$ f_1(x)=\begin{cases} x&\text{ if }0\leq x\leq 1\newline 1&\text{ if }x>1 \end{cases} $$

$$ f_2(x)=\begin{cases} -1+x+e^{-x}&\text{ if }0\leq x\leq1\newline -1+\frac{2}{e}+(1-\frac{1}{e})t&\text{ if }1\leq x\leq 2\newline 1&\text{ if }x>2 \end{cases} $$

$$ f_3(x)=\begin{cases} -2+x+(x+2)e^{-x}&\text{ if }0\leq x\leq 1\newline e^{-x}+\frac{3}{e}-2-\frac{x}{e}+x&\text{ if }1\leq x\leq 2\newline -2+x+\frac{(1+e)(3-x)}{e^2}&\text{ if }2\leq x\leq 3\newline 1&\text{ if }x\geq 3 \end{cases} $$

While I don't have a proof this strategy is optimal, I've got a heuristic argument that assigning probability $\epsilon$ to each flip is a good idea. If our budget is $x$, then whatever our strategy, the expected number of heads we will have seen when we exhaust our budget is $x$. If the desired number of heads is much larger than $x$, we will need to make the variance in the number of heads large. If we assign probabilities $p_1,p_2,\ldots$ to the coin flips (with $p_1+p_2+\ldots=x$), then the variance in the number of heads is $\sum p_i(1-p_i)$, which is bounded above by $x$. We can make the variance arbitrarily close to $x$ by taking each $p_i$ as small as possible.

The argument is a little different if the next head that appears could cause our remaining budget to be larger than the number of additional heads we need to win.