TITLE: Probability that a self-avoiding walk on $\mathbb{Z}^3$ closes to a polygon
QUESTION [6 upvotes]: The probability that a random walk on $\mathbb{Z}^3$ returns to the origin is about 34%.
This is (part of)
Pólya's theorem.
I have been looking for an analogous (numerical) result for the probability
that a random "myopic" self-avoiding walk on $\mathbb{Z}^3$ returns to the origin to form a self-avoiding polygon.
By a "myopic self-avoiding walk," I mean
a walker who takes a sequence of random (unit) steps, each restricted to avoid previously visited sites (to quote Yoav Kallus's and Vincent Beffara's comments).
Any other related information—e.g., probability of the walk
ending in a cul-de-sac before $n$ steps (see below)—would be welcomed,
as would either small-$n$ results or asymptotics.

 
 
 


 
 
 
Origin: green. Cul-de-sac: red, reached after 335 steps.

A 2D version of this question was posed earlier, but not entirely answered:
"Probability that a “closable” self-avoiding random walk forms a polygon"

REPLY [3 votes]: A little data from a limited Monte Carlo simulation (as suggested by Ben Crowell).
Letting paths run up to $2000$ steps, I find:

56% are still free after $2000$ steps. 
33% have ended in a cul-de-sac before reaching $2000$ steps. 
11% close to a self-avoiding polygon.

The 11% doesn't contradict Ben's $\frac{1}{6}$ 
because I did not count the doubly covered segment EW as a self-avoiding
polygon. I wanted to restrict attention to simple polygons;
for my count, a $1 \times 1$ square is the smallest polygon.
Here is one self-avoiding polygon, of length $433$ steps:

 
 
 


Next question: What is the knot complexity (say, expected
crossing number) of these self-avoiding polygons?
Difficult to tell in the example above, but it may well be the unknot.