, 2024

Solution:

The probability that the rank $N$ player passes round $i$ (where $i=0$ is implied as 1) is
$$\frac{\binom{1024-2^{i}}{N-1}}{\binom{1023}{N-1}}$$

Summing this from $i=0$ to $9$ (each represents the expectation of advancing one round), we must find the minimum $N$ for which the sum goes below $2$. Let $r$ indicate $N-1$.

The first term is $1$, the second simplifies to $\frac{1023-r}{1023}$, and round $i$'s contribution is
$$\prod_{j=0}^{2^{i}-2} \frac{1023-r-j}{1023-j}$$

At this point, a program may output $N=350$.

However, one may notice that for larger $i$, this term vanishes; in fact, estimating this product with $k^{2^{i}-1}$ where $k=\frac{1023-r}{1023}$.

The trailing terms in the product will be a bit smaller than $k$, but vanishes. If we do find the estimate for $k$ where $\sum_{i=0}^{9} k^{2^{i}-1}=2$, $k$ will be a bit too small, the true $k$ being something slightly greater.

The main concern is how many great of an upper limit for $i$ we choose to approximate with for $k$. If we stop at $i=2$, for instance, we have $1+k+k^{3}=2$, where we might find $k=\frac{2}{3}$ a just estimate, $\frac{2}{3}+\frac{8}{27}$ being just shy of $1$. We may produce/cursory check $0.67$ or $0.68$ as an estimate for $k$ (from this equation, around $0.683$). This produces $N=339$ or $N=328$.

Estimating the real solution to $1+k+k^{3}+k^{7}=2$ produces very close to the maximum number of points. Given our estimate for $k_{1}$, just above $\frac{2}{3}$, we can estimate that $k^{7} \approx \frac{128}{2187} \approx 0.06$.

Then, we estimate the solution to $k+k^{3}=1-0.06$, for which we might notice that if $k=k_{1}-\delta$, then by Newton's method, we can approximate $k$ by
$$\delta \cdot\left(1+3 k^{2}\right) \approx \delta \cdot\left(1+3 \cdot \frac{4}{9}\right)=0.06$$

making $\delta$ about $\frac{0.06 \cdot 3}{7} \approx 0.026$. If this is subtracted from $k_{1}=0.683$, then $k=0.657$, producing $N=351$.