Three uniformly distributed i.i.d points inside the unit n-ball centered at the origin are picked $p_1, p_2, p_3$ they are chosen such that $\|p_1\|>\|p_2\|,\|p_3\|$ , where $\|p\|$ is the Euclidian distance from the origin. $p_1$ is a point that lies on the surface of a hypersphere with radius $\|p_1\|$, denoted as $S_{\|p_1\|}$.
An n-hyperrectangle is then formed by taking all permutations of $p_2$ and $p_3$ as the vertices. For example in 2D $p_2= (x_2,y_2)$, $p_3 = (x_3,y_3)$, the other two vertices are then $(x_2,y_3),(x_3,y_2)$. A vertex, $v$ is then chosen at random with probability $\frac{1}{2^n}$. What is the probability that the vertex chosen lies within the hypersphere $S_{\|p_1\|}$ formed by $p_1$ i.e. $\|v\|<\|p_1\|$?
I have the form of a lower bound based on the fact that for all vertices to lie within $S_{\|p_1\|}$ then $p_2$ and $p_3$ must lie within the largest fully inscribed hypercube, $H$, inside $S_{\|p_1\|}$. Therefore the probability that all vertices lie within is $S_{\|p_1\|}$ is the ratio of the volume of the hypercube and $S_{\|p_1\|}$: $\frac{V(H)}{V(S)} = \frac{2^D\Gamma(\frac{D}{2}+1)}{D^{\frac{D}{2}}\pi^{\frac{D}{2}}} $.
The next part of the proof is where I am stuck. The approximate probability of $\Pr[\|v\|<\|p_1\|]$ takes the form of
$(1-K)\frac{V(H)}{V(S)} + K$,
where K is the ratio of number of vertices ALWAYS enclosed in $S_{\|p_1\|}$. For example in 2D it is easy to see that $K=0.75$, 3 vertices are always enclosed, $p_1 = (x_1,y_1)$, $p_2 = (x_2,y_2)$ and $v_{\min} = (\min(x_1,x_2),\min(y_1,y_2))$. Using $K=0.75$ provides extremely accurate probabilities for low dimensions $n<10$
But Monte Carlo simulations show that as $\lim_{n\rightarrow \infty} K \rightarrow 0.5$. How can I prove this and also find a functional form of $K$ with respect to $n$?
Thanks for reading!
--UPDATE--- The limit of 0.5 can be proved given that $p_2$ and $p_3$ are enclosed in $S$ then due to symmetry a max of $0.5n$ elements of either $p_2$ or $p_3$ can have a larger absolute value, in this case $2^{n/2}$ of vertices formed will lie outside of $S$.