Ambient occlusion of a sole object in a scene So, the german wikipedia states about ambient occlusion:The ambient occlusion method is based on the assumption that an object, a set of objects or an entire scene is surrounded by a unit sphere of light.

Now, let's suppose we've got a scene in which the only object is a pipe, closed at one end, at the center of the scene. If I got this right, then to get the ambient occlusion this object would now be evenly light from every direction.

But wouldn't lighting the object evenly from everywhere mean that for every point in the pipe there would also be a ray of light that would directly reach it, leading to no ambient occlusion at all (no shadows inside the pipe)?

What did I get wrong here?
When you compute ambient occlusion (AO) for a point on an object's surface, you sample multiple directions and measure what fraction of them escape. That is, you're trying to approximate what fraction of the unit hemisphere is visible to that point on the surface. (Note that it's a unit hemisphere because half of the unit sphere is on the wrong side of the surface to start with.)

So in your pipe example, the outside of the pipe is not at all occluded. From every point on the outside, any ray fired in that hemisphere will escape without intersecting the object. All of the outside would be lit with 100% of the ambient light.

The inside is another matter. The inside of the end cap can only see a small circle of that hemisphere, through the other end of the pipe. The exact fraction depends on the radius and length of the pipe, but call it 10% for the example. 90% of rays fired from a point on the end cap will hit the inside of the cap, and the other 10% will escape along the length of the pipe and out of the other end. These points are 90% occluded, so they'll be lit with 10% of the ambient light.

Consider a point at the open end of the pipe, but on the inside. This point is 50% occluded. Half of the rays fired from it go into the pipe and intersect with it at some point. The other half go straight out and escape to the unit hemisphere. This point will be lit with 50% of the ambient light, and it's easy to see that the 50% will smoothly decrease to the 10% (or however much is the minimum) as you travel down the inside of the pipe towards the end cap, as less of that hole is visible.

The important point I think you misunderstood is that you measure or estimate what fraction of the unit hemisphere is visible from the point you're shading, not whether it's visible at all. Casting many rays and counting what fraction escape is one way of estimating this.