Robin of Foxley has entered the FiveThirtyEight archery tournament. Her aim is excellent (relatively speaking), as she is guaranteed to hit the circular target, which has no subdivisions — it’s just one big circle. However, her arrows are equally likely to hit each location within the target.
Her true love, Marian, has issued a challenge. Robin must fire as many arrows as she can, such that each arrow is closer to the center of the target than the previous arrow. For example, if Robin fires three arrows, each closer to the center than the previous, but the fourth arrow is farther than the third, then she is done with the challenge and her score is four.
On average, what score can Robin expect to achieve in this archery challenge?
Extra credit: Marian now uses a target with 10 concentric circles, whose radii are 1, 2, 3, etc., all the way up to 10 — the radius of the entire target. This time, Robin must fire as many arrows as she can, such that each arrow falls within a smaller concentric circle than the previous arrow. On average, what score can Robin expect to achieve in this version of the archery challenge?
Let $R$ be the radius of the target.
The probability density function of an arrow can be expressed as:
$$f(r) = \dfrac{2\pi r}{\pi R^2} = \dfrac{2r}{R^2}$$An arrow landing in the target is expressed as,
$$Pr\left[0 \leq X \le R \right] = \int_{0}^{R} \left( \dfrac{2r}{R^2} \right) dr$$
$$= \dfrac{1}{R^2} \cdot {r_1^2} \big|_{0}^{R}$$
$$ = 1$$
In order to get a score of $n$, every arrow must be within the previous arrow's radius, but the $n^{th}$ arrow must be outside the $(n-1)^{st}$ arrow.
Induction will be used to show the probability of score $n$ can be expressed as:
$$P(n) = \left( \dfrac{1}{R^2} \right)^n \cdot \int_{0}^{R} \left( R^2 \frac{(2n-2)r^{2n-3}}{(n-1)!} - \frac{(2n)r^{2n-1}}{n!} \right)dr$$
$$ = \dfrac{1}{R^{2n}} \cdot \left( R^2 \frac{r^{2n-2}}{(n-1)!} - \frac{r^{2n}}{n!} \right) \bigg|_{0}^{R}$$
$$ = \dfrac{1}{(n-1)!} - \dfrac{1}{n!}$$.
To have a score of 2, the second arrow must be outside the radius of the first arrow.
$$\int_{0}^{R} \left( \dfrac{2r_1}{R^2} \right) \cdot \left(\int_{r_1}^{R} \left( \dfrac{2r_2}{R^2} \right) dr_2 \right) dr_1$$
$$= \left(\dfrac{1}{R^2}\right)^2 \cdot \int_{0}^{R} 2r_1 \cdot \left( {r_2^2} \big|_{r_1}^{R} \right) dr_1$$
$$= \dfrac{1}{R^4} \cdot \int_{0}^{R} 2r_1 \cdot \left(R^2 - r_1^2 \right) dr_1$$
$$= \dfrac{1}{R^4} \cdot \int_{0}^{R} \left( 2R^2 r_1 - 2r_1^3 \right) dr_1$$
$$= \dfrac{1}{R^4} \cdot \left( R^2 r_1^2 - \frac{r_1^4}{2} \right) \bigg|_{0}^{R}$$
$$ = 1 - \frac{1}{2}$$
$$ = \dfrac{1}{R^{2k}} \cdot \left( R^2 \frac{r^{2k-2}}{(k-1)!} - \frac{r^{2k}}{k!} \right) \bigg|_{0}^{R}$$
$$ = \dfrac{1}{(k-1)!} - \dfrac{1}{k!}$$
The upper limit $r_1$ is substituted for $R$ in the previous evaluation.
$$\int_{0}^{R} \left( \dfrac{2r_1}{R^2} \right) \cdot \left(\left( \dfrac{1}{R^2} \right)^k \int_{0}^{r_1} \left(R^2 \frac{(2k-2)r^{2k-3}}{(k-1)!} - \frac{(2k)r^{2k-1}}{k!} \right)dr \right) dr_1$$
$$= \left( \dfrac{1}{R^2} \right)^{k+1} \cdot \int_{0}^{R} 2r_1 \cdot \left( \left( R^2 \frac{r^{2k-2}}{(k-1)!} - \frac{r^{2k}}{k!} \right) \bigg|_{0}^{r_1} \right) dr_1$$
$$= \left( \dfrac{1}{R^2} \right)^{k+1} \cdot \int_{0}^{R} \left( \frac{2 R^2 r_1^{2k-1}}{(k-1)!} - \frac{2r_1^{2k+1}}{k!} \right) dr_1$$
$$= \left( \dfrac{1}{R^2} \right)^{k+1} \cdot \int_{0}^{R} \left( \frac{(k+1) - 1}{(k+1) - 1} \cdot \frac{2 R^2 r_1^{2(k+1) - 3}}{\left((k+1) - 2\right)!} - \frac{k+1}{k+1} \cdot \frac{2r_1^{2(k+1)-1}}{\left((k+1) - 1\right)!} \right) dr_1$$
$$= \left( \dfrac{1}{R^2} \right)^{k+1} \cdot \int_{0}^{R} \left( \frac{\left(2(k+1)-2 \right)R^2 r_1^{2(k+1) - 3}}{\left((k+1) - 1\right)!} - \frac{\left( 2(k+1) \right) r_1^{2(k+1)-1}}{\left((k+1) \right)!} \right) dr_1$$
$$ = \dfrac{1}{R^{2(k+1)}} \cdot \left[R^2 \frac{r^{2(k+1)-2}}{\left((k+1) - 1\right)!} - \frac{r^{2(k+1)}}{\left((k+1) \right)!} \right]_{0}^{R}$$
$$ = \dfrac{1}{\left((k+1) - 1\right)!} - \dfrac{1}{\left( k+1 \right)!}$$
Thus, $P(n)$ is true for natural $n \geq 2$.
The expected value can be expressed as:
$$\sum_{n=2}^{\infty} n \cdot \left(\frac{1}{(n-1)!} - \frac{1}{n!}\right)$$
$$ = \sum_{n=2}^{\infty} n \cdot \left(\frac{n}{n!} - \frac{1}{n!}\right)$$
$$ = \sum_{n=2}^{\infty} \frac{n(n-1)}{n!}$$
$$ = \sum_{n=2}^{\infty} \frac{1}{(n-2)!}$$
$$ = \sum_{n=0}^{\infty} \frac{1}{n!}$$
$$ = e$$