If the stars are distributed randomly within the universe, what is the probability for a star to be the nearest neighbor of a star that is its nearest neighbor? What if the number of spatial dimensions is higher than 3 or even grows without limit?
PS. I am interested in the asymptotic behavior with the number of stars growing without bound.
PPS. It's a well-known problem (the "birds on a wire problem" in higher dimensions) and I know the answers. However, no idea how to solve this analytically.
Let's assume an infinite $n$-dimensional universe with stars distributed according to a Poisson process with density $\lambda$. The probability that there is no nearest neighbour within a hypherspherical volume $V=\alpha_nr^n$ of some star $S$ is $\exp(-\lambda V)=\exp(-\lambda\alpha_nr^n)$. Thus the density for finding the nearest neighbour at distance $r$ is $\lambda\alpha_nnr^{n-1}\exp(-\lambda\alpha_nr^n)$.
Now conditional on the nearest neighbour of $S$ being at distance $r$, there is no star in the sphere around $S$ with radius $r$, and the sphere around $S$'s nearest neighbour $T$ with radius $r$ overlaps this sphere in a symmetrical lune, whose volume is proportional to $r^n$, say, $\beta_nr^n$, with $\beta_1=1$ and $\beta_2=2\pi/3-\sqrt3/2\approx1.2284$ (see MathWorld). The probability that there is no other star within distance $r$ of $T$, and hence $S$ is the nearest neighbour of $T$, is $\exp(-\lambda(\alpha_n-\beta_n)r^n)$. Thus the total probability of $S$ being the nearest neighbour of $T$ is
$$ \int_0^\infty\lambda\alpha_nnr^{n-1}\exp(-\lambda\alpha_nr^n)\exp(-\lambda(\alpha_n-\beta_n)r^n)\mathrm dr=\frac{\alpha_n}{2\alpha_n-\beta_n}\;. $$
For $n=1$, we have $\alpha_1=2$ and $\beta_1=1$, so the probability is $p_1=2/3$. For $n=2$, we have $\alpha_2=\pi$ and $\beta_2=2\pi/3-\sqrt3/2$, so the probability is
$$ p_2=\frac\pi{2\pi-2\pi/3+\sqrt3/2}=\frac1{4/3+\sqrt3/(2\pi)}\approx0.6215\;. $$
We can write the probability for $n$ dimensions as
$$ p_n=\frac{\displaystyle\int_{-\pi/2}^{\pi/2}\cos^nx\,\mathrm dx}{2\displaystyle\int_{-\pi/2}^{\pi/6}\cos^nx\,\mathrm dx}\;, $$
the ratio of the volume of a sphere to the volume of the union of two spheres with centres on each other's surfaces. Here's a Sage worksheet that calculates these probabilities for $n$ up to $10$:
sage: t=var('t');
sage: def f(n,x): return integral (cos (t)^n,t).substitute (t=x);
sage: def g(n): return (f (n,pi/2) - f (n,-pi/2)) / (2 * (f (n,pi/6) - f(n,-pi/2)));
sage: [g(n) for n in range (1,11)]
[2/3, 6*pi/(8*pi + 3*sqrt(3)), 16/27, 12*pi/(16*pi + 9*sqrt(3)), 256/459, 30*pi/(40*pi + 27*sqrt(3)), 2048/3807, 840*pi/(1120*pi + 837*sqrt(3)), 65536/124659, 840*pi/(1120*pi + 891*sqrt(3))]
sage: [g(n).n () for n in range (1,11)]
[0.666666666666667, 0.621504896887431, 0.592592592592593, 0.572465550279709, 0.557734204793028, 0.546588665492621, 0.537956396112425, 0.531153988127765, 0.525722170079978, 0.521339529895594]
The probability for odd $n$ is rational, and the probability for even $n$ is of the form $(4/3+q_n\sqrt3/\pi)^{-1}$ with $q_n$ rational. (You can show this analytically by deriving a recurrence relation for $\int_0^x\cos^nx\,\mathrm dx$ by integrating by parts.) Here are tables for odd $n$
$$ \begin{array}{r|c|c} n&p_n&p_n\\\hline 1&\frac23&0.66667\\\hline 3&\frac{16}{27}&0.59259\\\hline 5&\frac{256}{459}&0.55773\\\hline 7&\frac{2048}{3807}&0.53796\\\hline 9&\frac{65536}{124659}&0.52572\\ \end{array} $$
and even $n$:
$$ \begin{array}{r|c|c} n&q_n&p_n\\\hline 2&\frac12&0.62150\\\hline 4&\frac34&0.57247\\\hline 6&\frac9{10}&0.54659\\\hline 8&\frac{279}{280}&0.53115\\\hline 10&\frac{297}{280}&0.52134\\ \end{array} $$
Thus, for stars in our three-dimensional universe the probability is $16/27\approx0.59259$. Our star, the sun, is not its nearest neighbour's nearest neighbour: Its nearest neighbour is Proxima Centauri at a distance of about $4.2$ light-years, whose nearest neighbour is the binary star system Alpha Centauri.
The ratio of the volume of the lune to the volume of the sphere goes to zero with $n\to\infty$, so the probability for a star to be the nearest neighbour of its nearest neighbour goes to $1/2$.