Volume of a high dimensional cone I would like to choose arbitrarily two vectors $a$ and $b$ $\in \mathbb{S}^{n-1}$ and I would like to calculate the probability they are at least some distance $\delta$ apart. This probability should be
$$\frac{|\mathbb{S}^{n-1}|-|\{ x \in \mathbb{S}^{n-1}: \|x-a\| \leq \delta \}|}{|\mathbb{S}^{n-1}|}$$
So (I think), my problem consists of finding $|\{ x \in \mathbb{S}^{n-1}: \|x-a\| \leq \delta \}|$ ($|\cdot|$ being the surface measure in $\mathbb{R}^n$). I suppose $a$ can be taken as the unit vector $(1, 0, \cdots, 0)$. By symmetry, the points $\{ x \in \mathbb{S}^{n-1}: \|x-a\| = \delta \}$ should be a "circle" with center somewhere on the line between $(0, \cdots, 0)$ and $a$. 
I need to find that center and the radius. I search for a point on the "circle" with only the first two components being non-zero: The center should have coordinates $(1-x, 0, \cdots, 0)$ and the point on the circle has coordinates $(1-x, y, 0, \cdots, 0)$. $x$ and $y$ need to verify
$$x^2 + y^2 = \delta$$ and $$(1-x)^2 + y^2 = 1$$ This gives $x = \delta / 2$ and $y = (\sqrt{3}/4) \delta$. Going back to the surface we want to calculate:
$$|\{ x \in \mathbb{S}^{n-1}: \|x-a\| \leq \delta \}| = |\{x \in \mathbb{S}^{n-1}\}\cap \{B((1-\delta / 2, 0, \cdots, 0), r = (\sqrt{3}/4) \delta)\}$$
This should just be the measure of the spherical cap of the sphere centered at $(1-\delta / 2, 0, \cdots, 0)$. To estimate this surface area, I wanted to use that surface area is the push-forward measure of a Gaussian measure on $\mathbb{R}^n$ under $x \rightarrow x/\|x\|$ (times the measure of $\mathbb{S}^{n-1}$ since the latter is a probability measure). So to estimate the above measure, we can also calculate the Gaussian measure of a cone (rays going from $0$ to points in $\{x \in \mathbb{S}^{n-1}: \|x-a\| \leq \delta\})$. Parametrizing this cone, I get $(t, \text{"disc of radius" }t*\frac{\sqrt{3}\delta}{4(1-\delta/2)}) = (t, disc)$
$$(2\pi)^{-n/2}\int_{(t, disc)}e^{-\|x\|^2/2} = (2\pi)^{-n/2}\int_0^{\infty}e^{-t^2/2}\int_{(disc)}e^{(x_2^2 + x_3^2 + \cdots + x_n^2)/2}dxdt$$
The last integral is over a radial function so
$$ = (2\pi)^{-n/2}\int_0^{\infty}e^{-t^2/2}\omega_{n-2}\int_0^{t*\frac{\sqrt{3}\delta}{4(1-\delta/2)} }e^{-s^2/2}dsdt$$
We can take $\delta$ as very small, so
$$= (2\pi)^{-n/2}\int_0^{\infty}e^{-t^2/2}\omega_{n-2}t*\frac{\sqrt{3}\delta}{4(1-\delta/2)}dt \leq (2\pi)^{-n/2}\omega_{n-2}\sqrt{3}\delta$$
EDIT: In the end, we have to multiply with $|\mathbb{S}^{n-1}|$ since the above is only the uniform probability on $\mathbb{S}^{n-1}$
Are these calculations correct? I am a bit confused about the term $(2\pi)^{-n/2}$, then $\delta$ could be quite large and we would still not occupy a large amount of the mass.
 A: I don't think this has anything to do with the Gaussian integral. 
Consider the case $n=3$ first. We want to compute the area of the spherical cap $C\subset S^2$ comprising all points $x\in S^2$ having distance $|x-e_3|\leq d$ from the point $e_3\in S^2$. To this end we transform $d$ into the corresponding spherical distance $\alpha$ related to $d$ by the formula $d=2\sin{\alpha\over2}$. The cap in question, call it $C_\alpha$,  now consists of all points $x\in S^2$ satisfying $0\leq\theta\leq\alpha$, whereby $\theta$ is the complement of the geographical latitude on $S^2$, i.e., $\theta=0$ at the north pole $e_3$. Planes $z={\rm const.}$ partition $C_\alpha$ into "infinitesimal lampshades" of width ${\rm d}\theta$ and circumference $\omega_1\sin\theta$, whereby $\omega_1=2\pi$ is the one-dimensional measure of $S^1$. The area of such an infinitesimal lampshade is then given by $2\pi\sin\theta\>d\theta$, and the area of $C_\alpha$ computes to
$$2\pi\int_0^\alpha\sin\theta\>d\theta=2\pi(1-\cos\alpha)=4\pi\,\sin^2{\alpha\over2}\ ,\tag{1}$$
hence is $\approx \pi d^2$ when $d\ll1$, as expected.
After this exercise we are ready for the general case. We want to compute the area of the spherical cap $C_\alpha\subset S^{n-1}$ comprising all points $x\in S^{n-1}$ having spherical distance $\leq \alpha$ from the point $e_n\in S^{n-1}$. Denote the $r$-dimensional area of the complete $r$-dimensional unit sphere by $\omega_r$ $(r\geq0)$. Then we obtain instead of $(1)$ the formula
$${\rm area}_{n-1}(C_\alpha)=\omega_{n-2}\int_0^\alpha\sin^{n-2}\theta\>d\theta\ .$$ The probability you are interested in is then given by
$$p(n-1,\alpha)={\omega_{n-2}\over\omega_{n-1}}\int_0^\alpha\sin^{n-2}\theta\>d\theta\ .$$
Depending on your needs you can compute the integral in elementary terms or in case $\alpha\ll1$ obtain good approximations.
A: 
The surface area of the unit sphere in $\mathbb{R}^n$ is
$$
\omega_{n-1}=\frac{2\pi^{n/2}}{\Gamma(n/2)}\tag1
$$
Thus, the surface area on the unit sphere of a spherical cap with an angular radius of $\theta$ is
$$\newcommand{\Beta}{\operatorname{B}}
\begin{align}
\int_0^\theta\omega_{n-2}\sin^{n-2}(t)\,\mathrm{d}t
&=\frac{\omega_{n-2}}2\int_0^{\sin^2(\theta)}u^{\frac{n-3}2}\left(1-u\right)^{-\frac12}\,\mathrm{d}u\\
&=\frac{\pi^{\frac{n-1}2}}{\Gamma\!\left(\frac{n-1}2\right)}\,\Beta\left(\sin^2(\theta);\frac{n-1}2,\frac12\right)\tag2
\end{align}
$$
where $\Beta(x;a,b)$ is the Incomplete Beta Function:
$$
\Beta(x;a,b)=\int_0^xt^{a-1}(1-t)^{b-1}\,\mathrm{d}t\tag3
$$
Dividing $(2)$ by $\omega_{n-1}$ and subtracting from $1$ (since you ask for the probability of being off the cap), we get a probability of
$$
\left\{\begin{array}{}
1-\frac{\Beta\left(\sin^2(\theta);\frac{n-1}2,\frac12\right)}{2\Beta\left(\frac{n-1}2,\frac12\right)}&\text{if }0\le\theta\le\frac\pi2\\
\frac{\Beta\left(\sin^2(\theta);\frac{n-1}2,\frac12\right)}{2\Beta\left(\frac{n-1}2,\frac12\right)}&\text{if }\frac\pi2\le\theta\le\pi
\end{array}\right.\tag4
$$
where $\theta=2\sin^{-1}\left(\frac\delta2\right)$ is the angular distance between the points on the sphere, and $\delta$ is from the question. Using these, we get the probability
$$
\left\{\begin{array}{}
1-\frac{\Beta\left(\delta^2\left(1-\frac{\delta^2}4\right);\frac{n-1}2,\frac12\right)}{2\Beta\left(\frac{n-1}2,\frac12\right)}&\text{if }0\le\delta\le\sqrt2\\
\frac{\Beta\left(\delta^2\left(1-\frac{\delta^2}4\right);\frac{n-1}2,\frac12\right)}{2\Beta\left(\frac{n-1}2,\frac12\right)}&\text{if }\sqrt2\le\delta\le2
\end{array}\right.\tag5
$$
When $n=3$, the plot of the probability is

When $n=100$, the plot of the probability is

