4. In a cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ with edge length 1, points $E$, $F$, and $G$ are the midpoints of edges $A A_{1}$, $C_{1} D_{1}$, and $B C$ respectively. Then the volume of the tetrahedron $B_{1} E F G$ is $\qquad$
4. $\frac{3}{16}$.

As shown in Figure 2, draw a line through point $E$ parallel to $B_{1} G$, intersecting $A_{1} D_{1}$ at point $E^{\prime}$.
Then $V_{\text {prism } B_{1} E F C}=V_{\text {pyramid } B_{1} E^{\prime} F C}$
$$
\begin{array}{l}
=V_{\text {pyramid } G E^{\prime} F R_{1}} \\
=\frac{1}{3} S_{\triangle E^{\prime} F R_{1}} \cdot B B_{1} .
\end{array}
$$

From the problem, we have $A_{1} E^{\prime}=\frac{1}{4}$.
On the base of the cube, we have
$S_{\triangle E^{\prime} F B_{1}}$
$$
\begin{array}{l}
=\frac{1}{2}\left(1+\frac{5}{4}\right)-\frac{1}{2} \times \frac{1}{2} \times \frac{5}{4}-\frac{1}{2} \times \frac{1}{2} \times 1 \\
=\frac{9}{16} .
\end{array}
$$

Thus, $V_{\text {pyramid } B_{1} E F C}=\frac{3}{16}$.