4. Given a tetrahedron $S-ABC$ with the base being an isosceles right triangle with hypotenuse $AB$, $SA=SB=SC=2, AB=2$, and points $S$, $A$, $B$, $C$ all lie on a sphere with center $O$. Then the distance from point $O$ to the plane $ABC$ is $\qquad$ (Proposed by the Problem Group)
4. $\frac{\sqrt{3}}{3}$.

In the isosceles right triangle $ABC$ with hypotenuse $AB$ and midpoint $D$, $D$ is the circumcenter of $\triangle ABC$. Given $SA=SB=SC=2$, it follows that the projection of $S$ onto the base $ABC$ 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 $ABC$, $OD=\frac{1}{3} \cdot SD=\frac{\sqrt{3}}{3}$.