Volume of the intersection of two simplexes Let $S_n$ be the interior of the unitary $n$-simplex, i.e 
$ S_n =\{{\bf x} \in \mathbb{R}^n \mid x_i\ge0  \wedge \sum_{i=1}^n x_i\le1\}$
Let $T_n({\bf y})$  be the reversed simplex with origin at ${\bf y}$, ie 
$T_n({\bf y}) = \{{\bf x} \in \mathbb{R}^n \mid y_i-x_i\ge 0  \wedge \sum_{i=1}^n y_i-x_i\le1\}$ 
I want to compute the intersection volume  $V({\bf y})=S_n \cap T_n({{\bf y}})$ for given ${\bf y}$ (with $y_i\ge 0$).
Clearly, $V({\bf y})=0$ if $\sum_{i=1}^n y_i>2$. Also, $V({\bf y})=\prod_{i=1}^n y_i$ if $\sum_{i=1}^n y_i\le1$ (the intersection is a rectangular parallelepiped).  But in the range $1<\sum_{i=1}^n y_i\le2$ it gets more difficult.
Given the symmetry of the problem, we could asssume WLOG $y_1 \le y_2\le  \cdots 
\le  y_n$
 A: I am restricted now to $1\leq\sum_{i=1}^n y_i<2$
First case: triangles in $\mathbb{R}^2$. Suppose that $y_i\leq 1$ for $i=1,2$. Then: 
\begin{equation}
\begin{split}
A(S_n\cap T_n(y))&=\frac{1}{2}-\frac{1}{2}(1-y_1)^2-\frac{1}{2}(1-y_2)^2-\frac{1}{2}(y_1+y_2-1)^2=\\
&=-1+y_1+y_2-y_1^2-y_2^2-y_1y_2.
\end{split}
\end{equation}
Now WLOG $y_1\leq 1$ and $y_2>1$. Then the intersection is a parallelogram and the answer is:
\begin{equation}
A(S_n\cap T_n(y))=y_1\cdot(2-y_1-y_2).
\end{equation}
The $2$-dimensional case is completely solved.
General case: symplexes in $\mathbb{R}^n$. If for any $(n-1)$-uple $\{y_{i_1},\ldots, y_{i_{n-1}}\}$ we have that $\sum_{j=1}^{n-1}y_{i_j}\leq 1$ then:
\begin{equation}
\begin{split}
Vol(S_n\cap T_n(y))&=\frac{1}{n!}-\frac{1}{n!}\sum_{i=1}^n(1-y_i)^n-\frac{1}{n!}(\sum_{i=1}^n y_i-1)^n=\\
&=\frac{1}{n!}\left(1-\sum_{i=1}^n \sum_{k=0}^n\binom{n}{k}y_i^k(-1)^{n-k}-\sum_{k=0}^n\binom{n}{k}\left(\sum_{i=1}^n y_i\right)^k(-1)^{n-k}\right)=\\
&=\frac{1}{n!}\left(1- \sum_{k=0}^n\binom{n}{k}\left(\sum_{i=1}^n y_i^k\right)(-1)^{n-k}-\sum_{k=0}^n\binom{n}{k}\left(\sum_{i=1}^n y_i\right)^k(-1)^{n-k}\right)=\\
&=\frac{1}{n!}\left(1- \sum_{k=0}^n\binom{n}{k}\left(\sum_{i=1}^n y_i^k+(\sum_{i=1}^n y_i)^k\right)(-1)^{n-k}\right).
\end{split}
\end{equation}
Suppose there exists a $(n-1)$-uple $\{y_{i_1},\ldots, y_{i_{n-1}}\}$ such that $\sum_{j=1}^{n-1}y_{i_j}>1$
Like you noticed we can WLOG assume now that $y_1\leq y_2\leq\ldots\leq y_n$.
Therefore we have two cases: either $y_1\in\{y_{i_1},\ldots, y_{i_{n-1}}\}$ or $y_1\not\in\{y_{i_1},\ldots, y_{i_{n-1}}\}$.
Thanks to the order and to the fact that we are dealing with positive numbers we can assume that $every$ $(n-1)$-uple is such that $\sum_{j=1}^n y_{i_j}>1$.
In the second case you can assume $\sum_{i=2}^n y_i>1$ and $y_1+\sum_{j=1}^{n-2} y_{i_j}>1$ for any $(n-2)$-uple $\{y_{i_1},\ldots,y_{i_{n-2}}\}$.
Note that for the limit case $\sum y_i=1$ your result and this general case coincides.
NOTE: As far as I understand at the moment there is not a smart way of computing such a volume. I am afraid one is forced to integrate. Still not sure. I neem to think more.
