4. The vertices of a cube with edge length 1 are alternately colored red and blue, so that the two vertices on each edge are of different colors. The volume of the common part of the tetrahedron formed by the red vertices and the tetrahedron formed by the blue vertices is $\qquad$
4. $\frac{1}{6}$.

From the problem, we know
$S_{\triangle A_{1} B D}=\frac{\sqrt{3}}{4}(\sqrt{2})^{2}=\frac{\sqrt{3}}{2}$.
Let $A C_{1}$ intersect the plane of $\triangle A_{1} B D$ at point $M$. Then $A M=h$.

According to the volume relationship of the tetrahedron $A_{1} A B D$,
$$
\frac{1}{3} \times \frac{1}{2}=\frac{h}{3} \times \frac{\sqrt{3}}{2} \Rightarrow h=\frac{\sqrt{3}}{3} \text {. }
$$

Since $A C_{1}=\sqrt{3}$, then $C_{1} M=\frac{2 \sqrt{3}}{3}$.
In the regular tetrahedron $A B_{1} C D_{1}$ and the regular tetrahedron $A_{1} B C_{1} D$, the plane $A_{1} B D \parallel$ plane $C B_{1} D_{1}$, and the other three pairs of bases are also correspondingly parallel. Therefore, the part of the regular tetrahedron $A B_{1} C D_{1}$ outside the regular tetrahedron $A_{1} B C_{1} D$ consists of four congruent regular tetrahedra, and let the volume of one of them be $V_{A}$.
$$
\begin{array}{l}
\text { Then } V_{\text {regular tetrahedron } A_{1} B C_{1} D}=\frac{1}{3} C_{1} M \cdot S_{\triangle A_{1} B D} \\
=\frac{1}{3} \times \frac{2 \sqrt{3}}{3} \times \frac{\sqrt{3}}{2}=\frac{1}{3}, \\
\frac{V_{A}}{V_{\text {regular tetrahedron } A_{1} B C_{1} D}}=\left(\frac{1}{2}\right)^{3}=\frac{1}{8} . \\
\text { Hence } V_{\text {common }} \\
=V_{\text {regular tetrahedron } A_{1} B C_{1} D}-4 \times \frac{1}{8} V_{\text {regular tetrahedron } A_{1} B C_{1} D} \\
=\frac{1}{2} V_{\text {regular tetrahedron } A_{1} B C_{1} D}=\frac{1}{6} .
\end{array}
$$