Four disks with disjoint interiors are mutually tangent. Three of them are equal in size and the fourth one is smaller. Find the ratio of the radius of the smaller disk to one of the larger disks.
1. Consider the arrangement of the four disks. Three of the disks are large and equal in size, and the fourth disk is smaller. The large disks are mutually tangent to each other and also tangent to the smaller disk.

2. Let the radius of each of the large disks be \( r \) and the radius of the smaller disk be \( r_s \).

3. Connecting the centers of the three large disks forms an equilateral triangle with side length \( 2r \) (since the disks are tangent to each other).

4. The center of the smaller disk is equidistant from the centers of the three large disks. This point is the incenter of the equilateral triangle formed by the centers of the large disks.

5. The distance from the incenter to any vertex (center of a large disk) of the equilateral triangle is given by the formula for the inradius of an equilateral triangle:
   \[
   \text{Inradius} = \frac{\sqrt{3}}{3} \times \text{side length} = \frac{\sqrt{3}}{3} \times 2r = \frac{2r\sqrt{3}}{3}
   \]

6. This distance is the sum of the radius of the smaller disk \( r_s \) and the radius of a large disk \( r \):
   \[
   r_s + r = \frac{2r\sqrt{3}}{3}
   \]

7. Solving for \( r_s \):
   \[
   r_s = \frac{2r\sqrt{3}}{3} - r
   \]

8. Simplify the expression:
   \[
   r_s = r \left( \frac{2\sqrt{3}}{3} - 1 \right)
   \]

9. The ratio of the radius of the smaller disk to the radius of a large disk is:
   \[
   \frac{r_s}{r} = \frac{r \left( \frac{2\sqrt{3}}{3} - 1 \right)}{r} = \frac{2\sqrt{3}}{3} - 1
   \]

10. Simplify the final expression:
    \[
    \frac{2\sqrt{3}}{3} - 1 = \frac{2\sqrt{3} - 3}{3}
    \]

The final answer is \(\boxed{\frac{2\sqrt{3} - 3}{3}}\).