Four balls of radius $1$ are placed in space so that each of them touches the other three. What is the radius of the smallest sphere containing all of them?
1. **Understanding the Problem:**
   We have four balls of radius 1 placed in space such that each ball touches the other three. We need to find the radius of the smallest sphere that can contain all four balls.

2. **Geometric Configuration:**
   The centers of the four balls form a regular tetrahedron with side length equal to twice the radius of the balls, i.e., 2 units. This is because each ball touches the other three, and the distance between the centers of any two balls is the sum of their radii (1 + 1 = 2).

3. **Finding the Center of the Tetrahedron:**
   The center of the smallest sphere containing all four balls will be the centroid of the tetrahedron. The centroid of a regular tetrahedron is equidistant from all its vertices.

4. **Distance from a Vertex to the Centroid:**
   To find the radius of the sphere, we need the distance from a vertex of the tetrahedron to its centroid. For a regular tetrahedron with side length \( a \), the distance from a vertex to the centroid is given by:
   \[
   d = \frac{a \sqrt{3}}{3}
   \]
   Here, \( a = 2 \), so:
   \[
   d = \frac{2 \sqrt{3}}{3}
   \]

5. **Height of the Tetrahedron:**
   The height \( h \) of the tetrahedron (distance from a vertex to the opposite face) can be found using the Pythagorean theorem in the triangle formed by the height, the centroid, and the midpoint of a side:
   \[
   h = \sqrt{a^2 - \left(\frac{a \sqrt{3}}{3}\right)^2} = \sqrt{4 - \frac{4}{3}} = \sqrt{\frac{8}{3}}
   \]

6. **Distance from the Centroid to the Center of the Sphere:**
   The distance from the centroid of the tetrahedron to the center of the sphere is half the height of the tetrahedron:
   \[
   OP = \frac{\sqrt{\frac{8}{3}}}{2} = \frac{\sqrt{6}}{2}
   \]

7. **Radius of the Sphere:**
   The radius of the sphere containing all four balls is the distance from the centroid to a vertex plus the radius of one ball:
   \[
   R = \frac{\sqrt{6}}{2} + 1
   \]

8. **Simplifying the Expression:**
   \[
   R = \frac{\sqrt{6}}{2} + \frac{2}{2} = \frac{\sqrt{6} + 2}{2}
   \]

Therefore, the radius of the smallest sphere containing all four balls is:
\[
\boxed{\frac{\sqrt{6} + 2}{2}}
\]