Volume of $n$-dimensional spherical orthant in upper diagonal halfspace Consider an $n$-dimensional Euclidean Space. Consider orthants in that space.  Each orthant occupies  $\frac{1}{2^n}$ of the volume of an $n$-dimensional unit sphere. Let's call that a spherical orthant.
Let's index spherical orthants by a binary $n$-tuple, e.g. for $n=4$, $(+,+,-,+)$ is the spherical orthant given by 
$$x_1 > 0, \qquad x_2 > 0, \qquad x_3 < 0, \qquad x_4 > 0.$$
Now consider the "upper diagonal halfspace" given by 
$$x_1 + x_2 + ... + x_n > 0.$$
What is the volume of the intersection of this "upper diagonal halfspace"
with a spherical orthant, given as a fraction of the volume
of that  spherical orthant?
$\newcommand{\Vol}{\mathrm{Vol}}$
Let's see some examples. $D_n$ will be the upper diagonal half-space in $\mathbb{R}^n$. $\Vol(O_n)$ will be the volume of a spherical orthant.
$$(n= 2)$$
We have:
\begin{align*}
\frac{\Vol((+,+)\cap D_2)}{\Vol(O_2)}=1 && \frac{\Vol((+,-)\cap D_2)}{\Vol(O_2)}=\frac12 \\
\frac{\Vol((-,+)\cap D_2)}{\Vol(O_2)}=\frac12 && \frac{\Vol((-,-)\cap D_2)}{\Vol(O_2)}=0.
\end{align*}
$$(n=3)$$
We have:
\begin{align*}
\frac{\Vol((+,+,+)\cap D_3)}{\Vol(O_3)}={}&1 & \frac{\Vol((+,+,-)\cap D_3)}{\Vol(O_3)}={}&y \simeq 0.78 \\
\frac{\Vol((+,-,+)\cap D_3)}{\Vol(O_3)}={}&y\simeq 0.78  & \frac{\Vol((-,+,+)\cap D_3)}{\Vol(O_3)}={}&y \simeq 0.78 \\
\frac{\Vol((+,-,-)\cap D_3)}{\Vol(O_3)}={}&1-y \simeq 0.22 & \frac{\Vol((-,+,-)\cap D_3)}{\Vol(O_3)}={}&1-y \simeq 0.22 \\
\frac{\Vol((-,-,+)\cap D_3)}{\Vol(O_3)}={}&1 - y \simeq 0.22 & \frac{\Vol((-,-,-)\cap D_3)}{\Vol(O_3)}={}&0,
\end{align*}
with $y = 2 - \frac4\pi\arccos(\frac{1}{\sqrt3}) \simeq  0.78$.
$$(n= 4)$$
We have:
\begin{align*}
\frac{\Vol((+,+,+,+)\cap D_4)}{\Vol(O_4)}={}&1 & \frac{\Vol((+,+,+,-)\cap D_4)}{\Vol(O_4)}={}&\frac{11}{12} \simeq 0.92\,\, \text{(3 pluses)} \\
\frac{\Vol((+,+,-,-)\cap D_4)}{\Vol(O_4)}={}&\frac12\,\, \text{(2 pluses)} & \frac{\Vol((+,-,-,-)\cap D_4)}{\Vol(O_4)}={}& \frac{1}{12} \simeq 0.08\,\, \text{(1 plus)}\\
\end{align*}
$$\frac{\Vol((-,-,-,-)\cap D_4)}{\Vol(O_4)}=0.$$
Can you give the result  for general $n$, and spherical orthants with $k$ times "+" ($0\leq k\leq n$) ?
Thank you,
Andreas
 A: This can be solved with the probabilistic method. The probability we want is, if you sample a random point from a particular orthant, what is the probability that the sum of the coordinates is at least 0? In fact, it suffices to work with the surface of the orthant instead: the sign of the sum doesn't change if you scale the point to the surface.
Why should we use the surface of the orthant? Because you can sample a random point from the orthant by picking from standard normal distributions for each coordinate, using the correct signs, then normalizing. Call a random point generated in this way $(x_1, x_2, \dots, x_n)$.
The quantity you have been computing for various $k$ and $n$ is
$$\Pr[x_1 + x_2 + \cdots + x_n \geq 0]$$
Following your notation and letting $k$ be the number of plus signs for this orthant, there are $k$ positive terms here and $n-k$ negative terms. Again, actually we can ignore the normalizing factor and just assume $x_i$ are either folded normal distributions or their negations (depending on which orthant you want to be in). 
Let $F_m$ represent a random variable that is the sum of $m$ i.i.d folded normal distributions. We need to compute the CDF by computing $\Pr[F_m \leq t]$ for fixed $t$. Geometrically, thinking of one dimension for each of the $m$ summands as normal variables, for $m=2$ this is a rotated square and for $m=3$ the region is an octahedron. I don't know this integral off the top of my head so I made a question for it here.
All that remains is to compute
$$\Pr[F_k \geq F_{n-k}]$$
Once we have the CDF and PDF of $F_m$, call them $F_m(t)$ and $f_m(t)$, the answer is
$$\int_{0}^\infty f_{k}(t) F_{n-k}(t) dt$$
