4. Given a tetrahedron $S-ABC$ with the base being an isosceles right triangle with $AB$ as the hypotenuse, $SA=SB=SC=2, AB=$ 2, and suppose points $S, A, B, C$ all lie on a sphere with center $O$. Then the distance from point $O$ to the plane $ABC$ is $\qquad$
$\frac{1}{3}$
4. 【Analysis and Solution】As shown in the figure, let the midpoint of the hypotenuse $AB$ of the isosceles right triangle $ABC$ be $D$, then $D$ is the circumcenter of $\triangle ABC$. Given $SA=SB=SC=2$, it follows that the projection of $S$ on the base plane $ABC$ is $D$. Therefore, the center of the sphere $O$ lies on $SD$ and $OA=OB=OS$. Thus, $O$ is the center of $\triangle SAB$. Since $\triangle SAB$ is an equilateral triangle, $SD=\sqrt{2^{2}-1^{2}}=\sqrt{3}$, so the distance from $O$ to the plane $ABC$, $OD=\frac{1}{3} \cdot SD=\frac{\sqrt{3}}{3}$.