4. Given a tetrahedron $S-ABC$ with the base being an isosceles right triangle with hypotenuse $AB$, $SA=SB=SC=2, AB=2$. Suppose points $S, A, B, C$ all lie on a sphere with center $O$. Find the distance from point $O$ to the plane $ABC$.
(Let the midpoint of $AB$ be $D$, then $D$ is the circumcenter of $\triangle ABC$. From $SA=SB=SC=2$, we know that the projection of $S$ on the base $ABC$ is $D$. Therefore, the center of the sphere $O$ is on $SD$. Also, since $OA=OB=OS$, point $O$ is the center of the equilateral $\triangle SAB$, so the distance from point $O$ to the plane $ABC$ is $OD=\frac{1}{3} SD=\frac{\sqrt{3}}{3}$.)