4. To completely fit 4 steel balls with a radius of 1 into a container shaped as a regular tetrahedron, the minimum height of this regular tetrahedron is
As shown in the figure, the centers of the spheres $O_{1}, O_{2}, O_{3}, O_{4}$ form a regular tetrahedron with a side length of 2, and the height of this tetrahedron $O_{4} M=\frac{2 \sqrt{6}}{3}$.
Assuming the height $P O$ of the regular tetrahedron $P-X Y Z$ intersects the sphere $O_{4}$ at point $N$, then the sphere $O_{4}$ is the inscribed sphere of the regular tetrahedron $P-Q R S$ with $P N$ as its height, thus $P O_{4}: O_{4} N=3: 1 \Rightarrow P O_{4}=3$. Therefore, the height of the regular tetrahedron $P-X Y Z$ is $P O=4+\frac{2 \sqrt{6}}{3}$.