Are two points allowed to be identical? Otherwise the case of four points on a 2×2 grid is an obvious counterexample, as the probability of $l=3$ is 0, and of $l=4$ is 1.

Actually thinking about it, it's a counterexample in the other case, too, as there are more configurations with two points identical than all points different.

Err, actually even in the case of identical points we have $l=4$ for all configurations, as the identical points still need to be counted separately. But still a counterexample.

I think my explanation was poor. The points do not have to be in a lattice grid; they can be arranged anywhere in the box - so there are infinite arrangements.

So what is the point of the grid in the first place?

Could you define more formally in the question how the points are distributed? A continuous distribution within a fixed square would presumably lead to probability zero for any given length, as there would be uncountably with possible lengths. (This is not a proper argument, just a guess.)

I think grid is the wrong word. I meant box or square. It’s a continuous distribution so the points can be anywhere in the square (no identical points). I am concerned about the probability distribution of lengths. So, if the points are randomly distributed an arbitrarily large number of times (say, 1 billion), what proportion would have a convex hull of length 3, what proportion would have a convex hull of length 4, etc. My conjecture is that the probability distribution is uniform, but I can’t prove it.

I see, so you should simply remove the word "grid" to make your question vastly more understandable. You have a square with randomly distributed points in it. No grid that is of any relevance.

tommi‭ Well, the “length” is defined in the title as the number of the points that come to lie on the perimeter of the convex hull. It's not what conventionally would be called “length” (yes, it's another case of confusing wording, but in this case the exact intended definition is contained in the question).