Four spheres of radius $1$ are placed so that each is tangent to the other three. What is the radius of the smallest sphere that contains all four of them?

1. The centers form a regular tetrahedron of edge $2$.
2. The center of the smallest enclosing sphere is the tetrahedron's center, at distance $\dfrac{\sqrt6}4\cdot2=\dfrac{\sqrt6}2$ from each vertex (the circumradius of a regular tetrahedron of edge $a$ is $\frac{\sqrt6}4a$).
3. The enclosing sphere must reach the far side of each small sphere, so $R=\dfrac{\sqrt6}2+1=\boxed{1+\dfrac{\sqrt6}2}$.