Krishna v/s Robin Hood!

Let the distances of Krishna's three arrows from the centre of the target be the random variables $R_1,R_2,R_3$ . Then $R_1,R_2$ have the same distribution, with PDF $f(r)$ and CDF $F(r)$ given by $f(r) \; = \; 2r \hspace{1cm} F(r) \; = \; r^2 \hspace{2cm} 0 \le r \le 1$ while $R_3$ has PDF $g(r)$ and CDF $G(r)$ given by $g(r) \; =\; 4r - 4r^3 \hspace{1cm} G(r) \; = \; 1 - (1 - r^2)^2 \; = \; 2r^2 - r^4 \hspace{2cm} 0 \le r \le 1$ The distance of the middle of Krishna's arrows from the centre is the random variable $M$ which has PDF (we need to deal with whether the third arrow is the nearest, middle or furthest from the centre) $\begin{array}{rcl} f_M(r) & = & 2G(r) \times f(r) \times (1 - F(r)) + 2 F(r) \times g(r) \times (1 - F(r)) + 2 F(r) \times f(r) \times (1 - G(r)) \\ & = & 4r^3(1-r^2)(5 - 4r^2) \end{array}$ for $0 \le r \le 1$ . Krishna's point score is $W \; = \; \lfloor 100(1 - M^2) \rfloor$ and so we need to calculate $\mathbb{E}[W] \; = \; \int_0^1 \lfloor 100(1 - r^2) \rfloor \, f_M(r)\,dr$ A simplistic calculation shows that $\int_0^1 100(1-r^2)f_M(r)\,dr \; = \; \tfrac{170}{3} \; = \; 56\tfrac23$ This is, of course, greater than $\mathbb{E}[W]$ , which is in fact equal to $\sum_{j=-0}^{99} j \int_{\sqrt{1 -\frac{j+1}{100}}}^{\sqrt{1 - \frac{j}{100}}}\,f_M(r)\,dr \; = \; \frac{280833333}{5000000} \; = \; 56.1667$ Both of these values lie between $56$ and $57$ .

Thus the largest integer score that Robin Hood can obtain, and still have Krishna expect to win, is $\boxed{56}$ . Of course, with a graphic like the one with the question (and according to the stories about Robin Hood's splitting arrows), Robin would be able to score a perfect, or near perfect, score, so Krishna would need to practise a lot more!