Two distinct regular tetrahedra have all their vertices among the vertices of the same unit cube. What is the volume of the region formed by the intersection of the tetrahedra?

$ \textbf{(A)}\ \frac{1}{12}\qquad\textbf{(B)}\ \frac{\sqrt{2}}{12}\qquad\textbf{(C)}\ \frac{\sqrt{3}}{12}\qquad\textbf{(D)}\ \frac{1}{6}\qquad\textbf{(E)}\ \frac{\sqrt{2}}{6} $
1. **Understanding the Problem:**
   We need to find the volume of the region formed by the intersection of two distinct regular tetrahedra whose vertices are among the vertices of the same unit cube.

2. **Volume of a Regular Tetrahedron:**
   The volume \( V \) of a regular tetrahedron with side length \( a \) is given by:
   \[
   V = \frac{a^3 \sqrt{2}}{12}
   \]
   For a unit tetrahedron (side length \( a = 1 \)):
   \[
   V = \frac{1^3 \sqrt{2}}{12} = \frac{\sqrt{2}}{12}
   \]

3. **Volume of a Tetrahedron with Side Length \( \sqrt{2} \):**
   The tetrahedra in the problem have side length \( \sqrt{2} \). Using the volume formula:
   \[
   V = \frac{(\sqrt{2})^3 \sqrt{2}}{12} = \frac{2\sqrt{2} \cdot \sqrt{2}}{12} = \frac{2 \cdot 2}{12} = \frac{4}{12} = \frac{1}{3}
   \]

4. **Intersection of Two Tetrahedra:**
   Each tetrahedron can be divided into 8 smaller tetrahedra with side length \( \frac{1}{2} \). The volume of each smaller tetrahedron is:
   \[
   V_{\text{small}} = \frac{\left(\frac{1}{2}\right)^3 \sqrt{2}}{12} = \frac{\frac{1}{8} \sqrt{2}}{12} = \frac{\sqrt{2}}{96}
   \]
   However, we need to consider the volume of the intersection of the two larger tetrahedra. The four tetrahedra at the corners of each large tetrahedron do not intersect with the other large tetrahedron. Thus, half of the volume of one tetrahedron will not be inside the other tetrahedron.

5. **Calculating the Intersection Volume:**
   Since half of the volume of one tetrahedron is not inside the other, the volume of the intersection is:
   \[
   V_{\text{intersection}} = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6}
   \]

Conclusion:
\[
\boxed{\frac{1}{6}}
\]