I came across this question involving calculating probabilities of hitting dartboard on youtube: Link for the video
The problem discussed there is:
There is a dart game, where you start by taking a shot at the dart board. Let's say you hit the point C. Now draw the chord BD through C perpendicular to its radial direction. Now the board shrinks to the diameter being equal to the chord BD. You take a shot again, hitting J. So the board shrinks again to its diameter being equal to IH. You go on until you miss (i.e. do not hit the board). Your score is the number of shots you take (So you get 1 point for taking the first shot even if you miss it....watch the video for a clearer description).
The question is that under the assumption that the initial size of the board is 1 unit (radius) and the player has a uniform probability to hit any point in the square formed by perpendicular tangents to it, what is the expected score?
The way it was done in the video is let's say it hits a point $C(x_o,y_o)$.
So the probability to score at least 1 is the probability that the distance OC is less than 1
i.e. $\sqrt{x_o^2+y_o^2} \leq 1$
$$x_o^2+y_o^2 \leq 1$$
Now since we are also bound by $|x| \leq 1$ and $|y| \leq 1$, the probability: $P(S > 1)=\frac{Ar(circle \ of \ radius \ 1)}{Ar(square)} = \frac{\pi (1)^2}{4}$
For the next shot, the new radius ($r_1$) can be found by connecting OA and applying the Pythagoras Thm.
$$r_o^2 = x_o^2+y_o^2 + r_1^2$$
For S>2, the distance of the new point $(x_1,y_1)$ from O must be less than $r_1$
$$x_1^2+y_1^2 \leq r_1^2$$
$$x_1^2 +y_1^2 \leq 1 - x_o^2 -y_o^2$$
$$x_o^2+y_o^2+x_1^2+y_1^2 \leq 1$$
Under the same bounds, the probability that this inequality is true is :$P(S > 2)=\frac{Volume \ equivalent \ of \ unit \ ball \ in \ \mathbb{R}^4 }{Volume \ equivalent \ of \ cube \ in \ \mathbb{R}^4 }=\frac{\pi^2}{2(2)^4}$
Similarly you can check for $P(S > 3)$ you will move analogously to $\mathbb{R}^6$, for $P(S > 4)$ move to $\mathbb{R}^8.....$
So now the expected score is :
$$E(S)= 1.P(S=1) +2.P(S=2)+3.P(S=3)+....$$
$$E(S)=1(P(S>0)-P(S>1))+2(P(S>1)-P(S>2))+....$$
$$E(S)=P(S>0) + P(S>1) +P(S>2)+P(S>3)+P(S>4)+...$$
$$=1+\frac{\pi}{4}+ (\frac{\pi}{4})^2 \frac{1}{2!}+ (\frac{\pi}{4})^3 \frac{1}{3!}+...$$
$$=e^{\frac{\pi}{4}}$$
Now my question is finding out alternate approaches to solve this problem.