Let $J_n$ be the random variable whose value is the payout of the $n$-th game. For $n=1$, we can easily calculate the expectation.
\begin{equation}\begin{split}
\mathbb{E}(J_1)&= \left(\sum_{1 \leq i \leq 20} \hphantom{.}\underbrace{\hphantom{..}2^{-i}\hphantom{..}}_{\mathbb{P}(J_1 = 2^{i-1})}\cdot 2^{i-1}\right) &+\sum_{i>20}2^{-i}\cdot 10^6\\
&=\hphantom{...........}10&+2^{-20}\cdot 10^6
\end{split}\end{equation}
Notice that if we get $21$ or more tails, we receive $10^6$ because $2^{i-1}>10^6$ for all $i\geq 21$. Also, the first summation limit is $20 = \lceil \log_2\left(10^6\right)\rceil$.
Now, precisely because of the total payout limit of $10^6$, the result of $J_{n+1}$ depends on the results of all $J_k$ with $1 \leq k \leq n$. How, then, do we calculate the expectation of $J_{n+1}$'s? For that, we use the fundamental result that $\mathbb{E}(X) = \mathbb{E}(\mathbb{E}(X|Y))$. Thus, for instance:
\begin{equation}\begin{split}
\mathbb{E}(J_2)&=\mathbb{E}(\mathbb{E}(J_2|J_1))\\
&=\left(\sum_{1\leq n \leq 20}\mathbb{E}\left(J_2|J_1=2^{n-1}\right)\cdot\mathbb{P}\left(J_1=2^{n-1}\right)\right)+\underbrace{\mathbb{E}\left(J_2|J_1=10^6\right)}_0\cdot \mathbb{P}\left(J_1=10^6\right)\\
&=\sum_{1\leq n\leq20}2^{-n} \cdot \mathbb{E}\left(J_2|J_1=2^{n-1}\right)
\end{split}\end{equation}
We first calculate $\mathbb{E}\left(J_2|J_1=2^{n-1}\right)$. We can write it as:
\begin{equation}\begin{split}
&\left( \sum_{\displaystyle 1\leq m \leq \lceil \log_2\left(10^6 -2^{n-1}\right)\rceil } 2^{-m} \cdot 2^{m-1} \right) \\
+ &\vphantom{.^{^{^{^{^{^{.}}}}}}}\hphantom{...}\sum_{\displaystyle m > \lceil \log_2 \left( 10^6 - 2^{n-1} \right) \rceil} 2^{-m} \cdot \left(10^6-2^{n-1} \right) \\
=&\hphantom{..}\frac12\cdot \lceil \log_2\left(10^6 -2^{n-1}\right)\rceil+2^{^{\displaystyle -\lceil \log_2\left(10^6 -2^{n-1}\right)\rceil \vphantom{^{^{^{^{^{.}}}}}}}}\cdot \left(10^6-2^{n-1}\right)
\end{split}\end{equation}
It follows that:
\begin{equation}\begin{split}
\mathbb{E}(J_2)&=\sum_{1\leq n \leq 20}2^{-n}\left(\frac12\cdot \lceil \log_2\left(10^6 -2^{n-1}\right)\rceil+2^{-\lceil \log_2\left(10^6 -2^{n-1}\right)\rceil }\cdot \left(10^6-2^{n-1}\right) \right)\\
&=\frac{11485739}{1048576}\vphantom{h^{h^{h^{h^{h^{h^{h^{h}}}}}}}}
\end{split}\end{equation}
We can repeat the process for the next $J_k$, and obtain that:
\begin{equation}\begin{split}
\mathbb{E}(J_3)=\sum_{\displaystyle 1 \leq i_2 \leq \lceil \log_2\left(10^6 \right) \rceil} 2^{-i_2} \cdot \left( \sum_{\displaystyle 1 \leq i_3 \leq \lceil \log_2\left(10^6 - 2^{i_2 - 1} \right) \rceil} 2^{-i_3} \cdot e(i_2,i_3) \right)
\end{split}\end{equation}
where $e(i_2,i_3)$ is
\begin{equation}\begin{split}
&\frac12 \cdot \lceil \log_2\left(10^6 - 2^{i_2 - 1} -2^{i_3-1}\right) \rceil\\
+\hphantom{..}\vphantom{^{^{^{^{^{^{^{^{^{^{^{^{.}}}}}}}}}}}}} &2^{^{\displaystyle -\lceil \log_2\left(10^6 - 2^{i_2 - 1} -2^{i_3-1}\right) \rceil}}\cdot \left(10^6 - 2^{i_2 - 1} -2^{i_3-1}\right)
\end{split}\end{equation}
from whence it follows that $\mathbb{E}(J_3)=\frac{24665391590542601}{2251799813685248}$.
Similarly,
\begin{equation}\begin{split}
\mathbb{E}(J_4)=\sum_{\displaystyle 1 \leq i_2 \leq \lceil \log_2\left(10^6 \right) \rceil} 2^{-i_2} \cdot \left( \sum_{\displaystyle 1 \leq i_3 \leq \lceil \log_2\left(10^6 - 2^{i_2 - 1} \right) \rceil} 2^{-i_3} \cdot \\
\left( \sum_{\displaystyle 1 \leq i_4 \leq \lceil \log_2\left(10^6 - 2^{i_2 - 1} - 2^{i_3 - 1} \right) \rceil} 2^{-i_4} \cdot e(i_2,i_3,i_4) \right) \right)
\end{split}\end{equation}
where $e(i_2,i_3,i_4)$ is
\begin{equation}\begin{split}
&\frac12 \cdot \lceil \log_2\left(10^6 - 2^{i_2 - 1} -2^{i_3-1} - 2^{i_4-1}\right) \rceil\\
+\hphantom{..}\vphantom{^{^{^{^{^{^{^{^{^{^{^{^{.}}}}}}}}}}}}} &2^{^{\displaystyle -\lceil \log_2\left(10^6 - 2^{i_2 - 1} -2^{i_3-1} - 2^{i_4 - 1}\right) \rceil}}\cdot \left(10^6 - 2^{i_2 - 1} -2^{i_3-1} - 2^{i_4 - 1}\right)
\end{split}\end{equation}
It follows that $\mathbb{E}(J_4) = \frac{808234073895490893065}{73786976294838206464}$.
I believe at this point the pattern should be clear on how to set up the recursion that yields $\mathbb{E}(J_{n+1})$ from previous $J_k$'s. That said, I'm not sure we're any wiser in doing so, and calculations become quickly quite troublesome. Here are the first few approximate values for $\mathbb{E}(J_{k})$:
- $\mathbb{E}(J_{1}) = 10.9536743164062$
- $\mathbb{E}(J_{2}) = 10.9536542892456$
- $\mathbb{E}(J_{3}) = 10.9536342620864$
- $\mathbb{E}(J_{4}) = 10.9536142349288$
- $\mathbb{E}(J_{5}) = 10.9535942077726$
- $\mathbb{E}(J_{6}) = 10.9535741806178$
Unsurprisingly, with each game the expected payout is very slightly smaller than that of the previous game (because the limit payout is reduced from our previous winnings). This also implies that the difference between them is shrinking:
- $\mathbb{E}(J_{1}) - \mathbb{E}(J_{2}) = 0.0000200271606$
- $\mathbb{E}(J_{2}) - \mathbb{E}(J_{3}) = 0.0000200271592$
- $\mathbb{E}(J_{3}) - \mathbb{E}(J_{4}) = 0.0000200271576$
- $\mathbb{E}(J_{4}) - \mathbb{E}(J_{5}) = 0.0000200271562$
- $\mathbb{E}(J_{5}) - \mathbb{E}(J_{6}) = 0.0000200271548$
So we could write
$$\mathbb{E}(J_k) = \mathbb{E}(J_1) - \left( \sum_{1 \leq i \leq k-1}\mathbb{E}(J_i) - \mathbb{E}(J_{i+1}) \right) > \mathbb{E}(J_1) - (k-1) \cdot \big( \mathbb{E}(J_1) - \mathbb{E}(J_2) \big)$$
We thus have the following rough estimate for all $k \geq 2$:
$$\mathbb{E}(J_1) - (k-1) \cdot \left( \mathbb{E}(J_1) - \mathbb{E}(J_2) \right) < \mathbb{E}(J_k) < \mathbb{E}(J_1)$$
Accordingly, for instance, $10.9336671829668 < \mathbb{E}(J_{1000}) < 10.9536743164062$. It is clear, I believe, that even over the course of the one thousand games, the effect is minuscule.
For the total expected payout $\mathbb{E}(\sum_{1 \leq k \leq 1000} J_k) = \sum_{1 \leq k \leq 1000} \mathbb{E}(J_k)$, we thus have the following estimates:
$$10943.6707496865 < \mathbb{E}\left(\sum J_k\right) < 10953.6743164062$$
Hence, $\mathbb{E}\left(\sum J_k\right) \simeq 10948.6725330464$, give or take $5$ dollars (an error of about $0.0457$%).
Therefore, if we're paying $10.94$ dollars or less per game, we should expect profit over the course of a thousand games. However, if we're paying $10.96$ or more per game, we should expect loss. For $10.95$ we're not quite sure, but since our upped bound is so lazy and yet 'barely' above $10950$, I'd hazard you're slightly in the loss here.
So yeah, up to $10.94$ dollars.