Why do objects sometimes fall on their own?

This will be a conceptual answer essentially without math although basically everything I say here can be made mathematically rigorous by way of the potential energy function and the concept of a local minimum of such a function.

There can be a large number of reasons for the phenomenon you describe, but all of the reasons I can think of boil down to the following:

When you set the object down, it stays there, at least initially, because it is in a position of stable mechanical equilibrium.

This basically means that if you consider all of the configurations of the object that are sufficiently close to its initial one, the net force on the object is such that if it is in any of these nearby configurations (which form what I'll call the "stability neighborhood"), then it will be pushed back to the initial configuration. However, if you perturb the object too much, then it will no longer be pushed back or stay where it is, it will be pushed away instead. If his happens, then it might reach another point of stable mechanical equilibrium, or it may "run away" completely.

The main point here that answers your question is that mechanical equilibrium is a local notion. This is captured by the fact that if you displace the object too much, then it will "fall over" to another local equilibrium point or otherwise completely leave the vicinity of it's initial configuration.

How does this answer your question? Well, if an object is in one of these configurations of stable equilibrium for long enough, then chances are that one of the following will happen

Case 1. At some point some external agent exerts a large enough perturbation on the object that its configuration leaves the neighborhood in which it will be restored to its initial configuration.

Case 2. Small perturbations build up over time and kick the object from one stable configuration to another (or to an unstable configuration) so that at some late enough time, the stability neighborhood of the object is right next to a position which would cause the object to move far away from its initial, stable configuration, (or the object would just run far away from the original stability neighborhood).

Here are examples to illustrate what I mean:

Case 1 - Real Example. This actually happened to me today! I balanced a monster stuffed animal I have on top of my computer screen, and I noticed that I could even give him a little kick, and he wouldn't fall over, so I thought "all right, I think he'll sit there for a while." About an hour later, I noticed that Darren (yes my monster stuffed animal's name is Darren) had fallen from his perch. What had happened? Well, in between the time when I first sat him down, and the time at which he had fallen over, I had opened the sliding door to my patio, and I speculate that a small, gust of wind blew him off. In more technical terms, the gust of wind large enough so that Darren's configuration changed sufficiently that he no longer was in the stability neighborhood around his initial configuration, and the force of gravity did the rest of the work to push him over.

Case 2 - Real Example. Yesterday I was listening to music with my subwoofer on blast. I had a quarter on my desk. The quarter was an inch or two from the edge, so you could have moved the quarter slightly from its initial configuration with its having fallen over. However, over time, the vibration from the woofer caused the quarter to inch closer and closer to the edge. At some point, the quarter was so close to halfway over the edge (though still in a position of stability by perhaps a fraction of a millimeter) that one more vibrational kick from the woofer caused it to fall over the edge of the desk.

Consider the potential energy of an object:

$V({\vec r})$. It is well known that the force on that object due to whatever interaction generating the potential is given by ${\vec F} = -{\vec \nabla}V$. Therefore, if I want to keep the object stationary, I merely need to place it in some place where the gradient of the potential vanishes.

But wait! What happens if I nudge my stationary thing a little bit? My potential might not have a zero gradient wherever it ends up. If this is the case, then I have a few possibilities:

1) The new force on my object pushes the object back to where it came from. This is great, because then, our nudges restore the object to equilibrium

2) the new force on the object pushes the object further away from its rest point. If this happens, it's sure to end up somewhere crazy

3) it pushes it in some other direction. If this is the case, the results will be somewhere between cases 1) and 2).

From this argument, it should be clear that if I want my object to permanently stay in the position where it started, it's necessary that the potential be such that, if I displace in any direction away from my equilibrium point, then any force on the particle will push me back toward equilibrium. It turns out that, mathematically, what this means is that $\frac{\partial^{2}V}{\partial x^{1}x^{2}} > 0$ for any choice of spatial variables $x^{1}$ and $x^{2}$, or, in more precise mathematical terms, the Hessian of the potential has components that are all greater than zero.