Take a smooth closed curve in the plane. At each self-intersection, randomly choose one of the two pieces and lift it up just out of the plane. (Perturb the curve so there are no triple intersections.) I don't really know anything about knot theory, so I don't even know if I'm asking the right questions here, but I'm wondering: What is the probability that this is the trivial knot? What can we say about how knotted this knot might be, and with what probabilities? (Measure "knottedness" in whatever way you like.) More generally, can we say anything about the probability of the various possible values in the usual invariants that people use to study knots?


I only have an idea of how to approach the first question, and even then it's only by brute force. I was just playing around with the easiest cases, and I think that with 0, 1, or 2 intersections, all knots are trivial, and with 3 intersections the knot is trivial with probability 75%.


A general analysis should presumably involve calculating the probability that we can simplify using various Reidemeister moves, but I don't know how to incorporate this. I'd imagine a computer could brute-force the first few cases pretty easily (I'm not so bold as to venture an order-of-magnitude guess on whether it's the first few hundred or the first few million)...



Probabilistic knot theory
One possible route to a model of random knots would be through the braid group. Every knot can be expressed (non-uniquely) as the closure of a braid. So, for example, you could apply the braid generators uniformly $n$ times across $k$ strands, close the braid using your favorite closure, and then ask this question sensibly. I don't think you can directly ask about the $n \to \infty$ limit for the braid group, though, because I don't think there is a notion of uniform measure for that group. Actually, perhaps I will post this as a separate question, but is the braid group amenable? I would wager that in this model, the probability of having the unknot decreases very quickly with $n$ and $k$.


To test if you have the unknot, it is conjectured that you just have to check the Jones polynomial. But even this is *still* hard in general, unless *even if* you happen to have a quantum computer. :) 


(Edit: Thanks Greg Kuperberg, below, for the correction.)


