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 $\qquad$ .
4. As shown in the figure, in the isosceles right triangle \( \triangle ABC \) with hypotenuse \( AB \), the midpoint of \( AB \) is \( D \), so \( D \) 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 \( ABC \) is \( OD = \frac{1}{3} \). \( SD = \frac{\sqrt{3}}{3} \).