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 $O$. Then the distance from point $O$ to the plane $ABC$ is $\qquad$.
4. $\frac{\sqrt{3}}{3}$.

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