With reference to question: Probability that 3 darts land in a same half of a dart board, where I know that the answer is 3/4 if $n=3$.
Q: I throw $n$ darts at a target. What is the probability that they all land in the same half of the target?
I found a rigorous solution via order statistics. Think of $n$ darts as $n$ order statistics uniformly distributed over $[0,360)$, meaning that darts are individuated by a reference angle from some axis. If I think that these order statistics are all in the same half of the target if either i) the range is less than 180°, or ii) if one of the $n-1$ gaps is larger than 180° (all these events are obviously disjoint), I could prove that the probability is equal to $n2^{1-n}$, and consistently it is equal to unity for $n=1,2$.
I wanted to attempt a solution based on a recursion argument, inspired by the question reported therein, but something is not right, and I can't track down why.
Call $\mathbb P(n)$ the probability that $n$ darts are in the same half target. Treating $n$ as an event, clearly $$\mathbb P(n)=\mathbb P(n\cap n-1)=\mathbb P(n|n-1)\mathbb P(n-1)= \mathbb P(n|n-1)\mathbb P(n-1|n-2)\mathbb P(n-2) = \dots\qquad \mathbb P(2)=1.$$ My problem is determining $\mathbb P(n|n-1)$. I try to imitate the idea in the cited question. Consider the $n-1$ darts as order statistics such that $0=X_{(0)}<X_{(1)}<..<X_{(n-1)}<X_{(n)}=180$, determining gaps with expected size $\frac {180}n$. The $n$-th dart will land in a possible half target if i) lands in $[0,180)$, or in ii) a little slice before 0 with size $\frac {180}n$, or in iii) the same thing after 180. Finally: $$\mathbb P(n|n-1)=\frac 1{360}\left(180+2\cdot \frac {180}n\right)=\frac{n+2}{2n},$$ which is wrong if I continue the reasoning. Why is not working?
