4. In the unit cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, $E, F, G$ are the midpoints of edges $A A_{1}, C_{1} D_{1}, D_{1} A_{1}$, respectively. Then the distance from point $B_{1}$ to the plane containing $E F G$ is $\qquad$ .
4. $\frac{\sqrt{3}}{2}$

Solution 2: Equal Volume Method, it is easy to know that $S_{B_{1} F G}=1-S_{B_{1} A_{1} G}-S_{B_{1} C_{1} F}-S_{D_{1} F G}=1-\frac{1}{4}-\frac{1}{4}-\frac{1}{8}=\frac{3}{8}$, and the distance from point $E$ to plane $B_{1} F G$ is $h_{0}=\frac{1}{2}$, so $V_{E B_{1} F G}=\frac{1}{3} h_{0} S_{B_{1} F G}=\frac{1}{16}$.
Also, $E F^{2}=E A_{1}^{2}+A_{1} F^{2}=E A_{1}^{2}+\left(A_{1} D_{1}^{2}+D_{1} F^{2}\right)=\frac{1}{4}+1+\frac{1}{4}=\frac{3}{2}$,
thus $E F=\frac{1}{2} \sqrt{6}$, $G F=G E=\frac{\sqrt{2}}{2}$,
$\cos \angle E G F=\frac{G E^{2}+G F^{2}-E F^{2}}{2 G E \cdot G F}=-\frac{1}{2}, \angle E G F=120^{\circ}$, then $S_{\triangle E G F}=\frac{1}{2} G E \cdot G F \sin 120^{\circ}=\frac{1}{8} \sqrt{3}$,
if the distance from $B_{1}$ to plane $E F G$ is $h$, then $\frac{1}{16}=V_{E B_{1} F G}=\frac{1}{3} h \cdot S_{\triangle E G F}=\frac{\sqrt{3}}{24} h$, so $h=\frac{\sqrt{3}}{2}$.