My favourite dartboard is a disc of unit radius and centre \(O\). I never miss the board, and the probability of my hitting any given area of the dartboard is proportional to the area. Each throw is independent of any other throw. I throw a dart \(n\) times (where \(n>1\)). Find the expected area of the smallest circle, with centre \(O\), that encloses all the \(n\) holes made by my dart.
Find also the expected area of the smallest circle, with centre \(O\), that encloses all the \((n-1)\) holes nearest to \(O\).
My other dartboard is a square of side 2 units, with centre \(Q\). I never miss the board, and the probability of my hitting any given area of the dartboard is proportional to the area. Each throw is independent of any other throw. I throw a dart \(n\) times (where \(n>1\)). Find the expected area of the
smallest square, with centre \(Q\), that encloses all the \(n\) holes made by my dart.
Determine, without detailed calculations, whether the expected area of the smallest circle, with centre \(Q\), on my square dartboard that encloses all
the \(n\) holes made by my darts is larger or smaller than that for my circular dartboard.
Solution:
Firstly, we consider the probability that all darts lie within a distance \(s\) from the centre, ie
\begin{align*}
\mathbb{P}(\text{all darts within }s) &= \prod_{k=1}^s \mathbb{P}(\text{dart within }s) \\
&= \left ( \frac{\pi s^2}{\pi} \right)^n \\
&= s^{2n}
\end{align*}
Therefore the pdf is \(2ns^{2n-1}\), and the expected area is \(\int_{s=0}^1 \pi s^2 \cdot 2n s^{2n-1} \d s = 2n \pi \frac{1}{2n+2} = \frac{n}{n+1} \pi\).
\begin{align*}
\mathbb{P}(\text{n-1 within }s) &= \underbrace{s^{2n}}_{\text{all within }s} + \underbrace{ns^{2n-2}(1-s^2)}_{\text{all but 1 within }s}\\
&= ns^{2n-2}-(n-1)s^{2n}
\end{align*}
Therefore the pdf is \(n(2n-2)s^{2n-3} - 2n(n-1)s^{2n-1} = 2n(n-1)(s^{2n-3}-s^{2n-1})\) and the expected area is:
\begin{align*}
\int \pi s^2 \cdot2n(n-1)(s^{2n-3}-s^{2n-1})\d s &= 2n(n-1) \pi \left ( \frac{1}{2n} - \frac{1}{2n+2} \right) \\
&= n(n-1)\pi \frac{2}{n(n+1)} \\
&= \frac{n-1}{n+1} \pi
\end{align*}
Now consider a square of side-length \(s\), we must have \(\mathbb{P}(\text{all darts within square}) = \left ( \frac{s^2}{4} \right)^n\) and therefore the pdf is \(n \frac{s^{n-1}}{4^n}\). Therefore the expected area is \(\displaystyle \int_0^2 s^2 \cdot n \frac{s^{n-1}}{4^n} \d s = \frac{n}{n+1} \frac{2^{2n+1}}{2^{2n}} = \frac{4n}{n+1}\)
It is clearly larger as the square dartboard contains all of the circular dartboard, and there will be some probability that the darts land outside the circular dartboard, making the circle much larger.