2. 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. Then, the volume of the common part of the tetrahedron formed by the red vertices and the tetrahedron formed by the blue vertices is $\qquad$
2. $\frac{1}{6}$.

As shown in Figure 5, note that
$$
S_{\triangle A_{1} B D}
$$
$$
=\frac{\sqrt{3}}{4}(\sqrt{2})^{2}=\frac{\sqrt{3}}{2} \text {. }
$$

Let $A C_{1}$ intersect the plane of $\triangle A_{1} B D$ at point $M$, and let $A M=h$.
According to the volume relationship of the tetrahedron $A_{1} A B D$, we have
$$
\frac{1}{3} \times \frac{1}{2}=\frac{h}{3} \times \frac{\sqrt{3}}{2} \Rightarrow h=\frac{\sqrt{3}}{3} \text {. }
$$

Since $A C_{1}=\sqrt{3}$, then $C_{1} M=\frac{2 \sqrt{3}}{3}$.
In the regular tetrahedra $A B_{1} C D_{1}$ and $A_{1} B C_{1} D$, the plane $A_{1} B D \parallel$ plane $C B_{1} D_{1}$, and the other three pairs of bases are also 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 $V_{A} 、 V_{C} 、 V_{B_{1}} 、 V_{D_{1}}$.
$$
\begin{array}{l}
\text { Also, } V_{A_{1} B C_{1} D}=\frac{1}{3} C_{1} M \cdot S_{\triangle A_{1} B D} \\
=\frac{1}{3} \times \frac{2 \sqrt{3}}{3} \times \frac{\sqrt{3}}{2}=\frac{1}{3}, \\
\frac{V_{A}}{V_{A_{1} B C_{1} D}}=\left(\frac{1}{2}\right)^{3}=\frac{1}{8},
\end{array}
$$

Thus,
$$
\begin{array}{l}
V_{\text {common }}=V_{A_{1} B C_{1} D}-4 \times \frac{1}{8} V_{A_{1} B C_{1} D} \\
=\frac{1}{2} V_{A_{1} B C_{1} D}=\frac{1}{6} .
\end{array}
$$