A question about a metric space I have a question regarding the following problem. 

Let $A,B,C$ be the three independently selected, uniformly distributed points on the unit sphere $S^3$ in $\Bbb R^4$. What's the probability that the unique chordal triangle $ABC$ is acute?

I just started analysis, and I have little knowledge about metric space.(I just learned the definition of the metric space.) 
I'm wondering if it is okay to change "the unit sphere $S^3$ in $\Bbb R^4$" to "the unit sphere $S^3$ in $\Bbb R^3$". 
Thanks for your help.
 A: I'll assume that by "the unique chordal triangle $ABC$" you mean simply the triangle $ABC$.
A triangle can have at most one obtuse angle. Thus the probability for it to have an obtuse angle is three times the probability of a given one of the three angles being obtuse. Thus we just have to calculate that probability.
Imagine $A$ and $B$ having been chosen, and place a pole at the midpoint between them. Then the angle $ACB$ will be obtuse if and only if $C$ is chosen closer to the pole than $A$ and $B$. This yields a double integration for the desired probability:
$$
\begin{align}
P(\angle ACB\text{ obtuse})
&=
\frac1{S_{d+1}^2}\int_0^\pi\mathrm d\theta\,S_d\,\sin^d\theta\int_0^{\theta/2}\mathrm d\phi\,S_d\,\sin^d\phi
\\
&=\left(\frac{S_d}{S_{d+1}}\right)^2\int_0^\pi\mathrm d\theta\,\sin^d\theta\int_0^{\theta/2}\mathrm d\phi\,\sin^d\phi\;,
\end{align}
$$
where $d$ is one less than the dimension of the sphere (in your case $d=2$), $\theta$ is the angular distance between $A$ and $B$, $S_d$ is the surface area of the unit $d$-sphere, $S_d\,\sin^d\theta$ is the angular density of points $B$ at angular distance $\theta$ from $A$, $\theta/2$ is the angular distance from the pole to $A$ and $B$, $\phi$ is the angular distance from the pole to $C$ and $S_d\,\sin^d\phi$ is the angular density of points $C$ at angular distance $\phi$ from the pole.
To warm up, we can recover the known result $1/4$ for $d=0$ (a circle):
$$
\left(\frac{S_0}{S_1}\right)^2\int_0^\pi\mathrm d\theta\int_0^{\theta/2}\mathrm d\phi=\left(\frac{2}{2\pi}\right)^2\int_0^\pi\mathrm d\theta\,\frac{\theta}2=\frac14\;.
$$
Next, with $d=1$, the result for a sphere:
$$
\left(\frac{S_1}{S_2}\right)^2\int_0^\pi\mathrm d\theta\,\sin\theta\int_0^{\theta/2}\mathrm d\phi\,\sin\phi
=
\left(\frac{2\pi}{4\pi}\right)^2\int_0^\pi\mathrm d\theta\,\sin\theta\left(1-\cos\frac\theta2\right)=\frac16\;.
$$
And now, with only slightly more involved integrals, your case $d=2$:
$$
\begin{align}
\left(\frac{S_2}{S_3}\right)^2\int_0^\pi\mathrm d\theta\,\sin^2\theta\int_0^{\theta/2}\mathrm d\phi\,\sin^2\phi
&=
\left(\frac{4\pi}{2\pi^2}\right)^2\int_0^\pi\mathrm d\theta\,\sin^2\theta\left[\frac12(\phi-\sin\phi\cos\phi)\right]_0^{\theta/2}
\\
&=
\frac2{\pi^2}\int_0^\pi\mathrm d\theta\,\sin^2\theta\left(\frac\theta2-\sin\frac\theta2\cos\frac\theta2\right)
\\
&=
\frac14-\frac4{3\pi^2}\;.
\end{align}
$$
This is the probability for a given angle to be obtuse; the probability for any angle to be obtuse is three times that, and so the probability for the triangle to be acute is
$$
1-3\left(\frac14-\frac4{3\pi^2}\right)=\frac14+\frac4{\pi^2}\approx0.6553\;.
$$
(I checked these results with numerical simulations.)
