Main Puzzle
I'll assume that every archer expects the others to play optimally (this was implied in the bonus question, but I'll use it here too).
There are only two pieces of information that could influence what shot an archer decides to take:
- the probability $b$ (for "best") that was chosen by the archer currently winning or $1$ if no one has hit yet (relevant because one cannot win taking a shot with probability above $b$) and
- the number of competitors $r$ (for "remaining") to shoot after the archer (relevant because later contestants may snatch one's victory away).
So we can write the optimal probability to choose for one's shot as a function $p$ of just $b$ and $r$.
First, suppose $r=0$. Obviously, the archer should take the easiest shot available, so we have $p(b,0)=b$. For the moment, I will note that this is equivalent to $p(b,0)=\min(b,1)$.
Now take $r=1$. We need to maximize the current archer's chances of winning, $p(b,1)$, for their shot being made, times $1-p(p(b,1),0)$ for the last archer not stealing the win. The maximum of $x(1-x)$ over the interval $[0,b]$ is achieved at $x=\min(b,1/2)$, so $p(b,1)=\min(b,1/2)$.
A similar argument for $r=2$ has us maximizing $x(1-\min(x,1/2))(1-x)$ over the interval $[0,b]$. Having the $\min$ in the $1-\min(x,1/2)$ factor choose $x$ is obviously better and occurs at lower $x$ anyway, so we can equivalently maximize $x(1-x)^2$. This gives $p(b,2)=\min(b,1/3)$.
By now you can probably see the pattern: We maximize the product of one success and $r$ subsequent failures over $[0,b]$, which is equivalent—because we want all the negative mins to be small—to maximizing $x(1-x)^r$. So, in general, $p(b,r)=\min(b,1/(r+1))$.
This gives us an answer to the main question, which depends, via $b$, on how prior archers have fared:
$$d=\frac{10}{p(b,r)}=\frac{10}{\min(b,\frac{1}{n-k+1})}.$$
Or, in plain language: An archer should shoot either from a distance where they expect exactly one of the as-yet-unfired shots (hopefully their own) to hit or from the distance of the currently winning shot, whichever is further.
That makes this puzzle pretty nifty in my opinion: the result is what my intuition would like it to be, but the justification is nontrivial.
Bonus
The probability of everyone in the latter half missing is $(1-b)^{n/2}$, where $b$ is established by the first archer to hit: $1/(n-i+1)$ when the $i^{\mathrm{th}}$ archer is first. Because of the problem wording, we only care about cases where some archer in the first half scored a hit, and these cases are disjoint, so we can sum over them.
Thus, we must determine the probability of the $i^{\mathrm{th}}$ archer scoring the first hit. It is the product of the probabilities of prior archers missing, which telescopes to $(n-i+1)/n$, times the probability that the archer themself hits, $1/(n-i+1)$. In other words, just $1/n$.
(As an aside, that too is a pretty cool result: As long as these archers play optimally, we can dupe them into a fair contest by claiming the rules above but then actually awarding the victory to the first hit.)
We substitute into the sum and take the limit:
\begin{align*}\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n/2}\left(1-\frac{1}{n-i+1}\right)^{n/2}&=
\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n/2} \left(\left(1-\frac{1}{n-i+1}\right)^{n-i+1}\right)^{\frac{1}{2\left(1-\frac{i-1}{n}\right)}}\\& = \lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n/2} e^{\frac{-1}{2\left(1-\frac{i-1}{n}\right)}}\\& =
\int_0^{1/2} e^{\frac{-1}{2(1-t)}}\,dt = \frac{1}{2}\int_{1}^2 e^{-1/t}\,dt\\
&\approx 0.252396.
\end{align*}
There's slightly better than a quarter chance of the winner having taken their shot during the first half of the contest.
