Keep calm and practice archery!

Let $R$ be the distance from the centre of the $N$ th best shot out of $Nk$ shots. Then $R$ has PDF $g(r) \; = \; Nk \times {Nk-1 \choose N-1} (r^x)^{N-1} \times xr^{x-1} \times (1-r^x)^{N(k-1)} \; = \; \frac{x r^{x-1} \times r^{x(N-1)} (1-r^x)^{N(k-1)}}{B(N,N(k-1)+1)} \hspace{1cm} 0 \le r \le 1$ and so $f(x,k)$ is the limit of the integral $\pi \int_0^1 r^2 g(r)\,dr \; = \; \frac{B(N+\frac{2}{x},N(k-1)+1)}{B(N,N(k-1)+1)}\pi$ as $N \to \infty$ . When $x = \tfrac{2}{m}$ , where $m$ is a positive integer, we deduce that $f\big(\tfrac{2}{m},k\big) \; =\; \frac{\pi}{k^m}$ and since $\sum_{m=1}^\infty \frac{\pi}{k^{2m}} \; = \; \frac{\pi}{k^2(1 - k^{-2})} \; = \; \frac{\pi}{k^2-1} \; = \; \tfrac12\pi\left(\frac{1}{k-1} - \frac{1}{k+1}\right)$ we deduce that $\sum_{m \ge 1 \atop m\, \mathrm{even}} \sum_{k \ge 2} f\big(\tfrac{2}{m},k\big) \; = \; \tfrac12\pi\sum_{k=2}^\infty\left(\frac{1}{k-1} - \frac{1}{k+1}\right) \; = \; \tfrac34\pi$ making the answer $\boxed{7}$ .

Let $r$ be the minimum radius containing his shots including $t = \left \lfloor\frac{n}{k}\right \rfloor$ best shots.

Our aim will be to find the probability that the worst among the best $t$ shots lie within $r+dr$ and $r$ .

We select which $t$ shots are best and then which among those best shot is the one lying between $r$ and $r + dr$ .

$\displaystyle \text{P(radius containing best t shots is r)} = \binom{n}{t} \binom{t}{1} ({(r+dr)}^x - r^x) r^{x(t -1)} {(1 - r^x)}^{n-t}$

$\displaystyle \text{P(radius containing best t shots is r)} = \binom{n}{t} t x r^{xt -1} {(1 - r^x)}^{n-t}$

$\displaystyle \text{Expected Area} = \frac{\pi \int_0^1 {r^{tx +1} {(1 - r^x)}^{n-t} \ dt}}{ \int_0^1 {r^{tx -1} {(1 - r^x)}^{n-t} \ dt}}$

$\displaystyle = \frac{\pi \int_0^1 {u^{t-1 +2/x} {(1 - u)}^{n-t} \ dt}}{\int_0^1 {u^{t -1} {(1 - u)}^{n-t} \ dt}}$

$\displaystyle = \frac{\pi \Gamma\left(t + \frac{2}{x}\right)\Gamma(n-t+1)}{\Gamma\left(n +\frac{2}{x}+1\right)} \frac{\Gamma(n+1)}{\Gamma(t) \Gamma(n-t+1)}$

$\displaystyle = \frac{\pi \Gamma\left(\left \lfloor\frac{n}{k}\right \rfloor + \frac{2}{x}\right) \Gamma(n+1)}{\Gamma\left(n +\frac{2}{x}+1\right)\Gamma(\left \lfloor\frac{n}{k}\right \rfloor)}$

We have to find this expected area as $n \to \infty$ .

Now we invoke this (something I recently discovered)-

$\displaystyle \lim_{n \to \infty}{\dfrac{\Gamma\left(\frac{p}{m}+n+1\right) \Gamma\left(\frac{q}{m} + \frac{n}{t} + 1\right)}{\Gamma\left(\frac{q}{m}+n+1\right) \Gamma\left(\frac{p}{m} + \frac{n}{t} + 1\right)}} = \dfrac{1}{t^{|p-q|/m}}$

Therefore,

$\displaystyle \lim_{n \to \infty}{\left(\text{Expected Area}\right)} = f(x,k) = \frac{\pi}{k^{2/x}}$

$\displaystyle f\left(\frac{2}{m}, x\right) = \frac{\pi}{k^m}$

$\displaystyle \sum_{k\geq 2}{ \sum_{m=1}^{\infty}{\frac{\pi}{k^{2m}}}} = \sum_{k \geq 2}{\frac{\pi}{k^2 -1}}$

which telescopes to

$\frac{3\pi}{4}$