When you can look through a forest?

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Yesterday I was coming back home by train through various forests and I realised, that through some of them one can see to the other side and through some of them not. Can this be formalised? That is, what is the chance of seeing what is on the other side of the forest depending on the size of the forest, average number of trees per unit area and the diameter of their trunk?

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A partial answer, showing how someone who does this kind of thing regularly gets started on the process

I hope that this "here's how I'm thinking through this" answer is of some value to you.

Here's a sketch of an answer to a very specific question that's at least closely related to yours.

Consider an long strip in the $xy$-plane, oriented along the $x$-axis, whose bottom edge extends from $(-L, 1)$ to $(L, 1)$, and whose top edge extends from $(-L, 1+a)$ to $(L, 1+a)$, where $L$ is some large number (compared to the diameter of a tree-trunk). This is my model of an overhead view of a forest of "depth" $a$ and extent $2L$. We're going to scatter trees of some fixed diameter $w < 1$ in this strip (with the tree-centers uniformly distributed in the $2L \times a$ rectangle, so the tree edges might extend a little outside of it), and imagine that the sun is just rising at $(0, \infty)$, casting rays towards the $x$-axis. The question then becomes, "What is the probability that the origin is not in the shadow of any tree-trunk?"

This amounts to asking "If the train stops and I look straight out at the forest, what is the probability that I can see through it?"

The nice thing about this model is that the shadow cast by a tree of diameter $w$ on the $x$-axis is just an interval of width $w$ (because the sun is effectively "infinitely far away"). Let's suppose that the tree "density" is $c$, so that within a forest region of area $A$, the expected number of tree-centers is $cA$.

If we're considering the origin as our "test point", the only trees than can possibly shadow us are those whose centers have $x$-coordinates between $-\frac{w}{2}$ and $\frac{w}{2}$, and $y$-coordinates between $1$ and $1 + a$ (because those are the $y$-limits of the forest). That's a region of area $aw$, so we expect there to be $caw$ trees in it.

To do this right, I should consider the number of trees in that region as a random variable, and then take an average over all possible values of that random variable, but I'm going to keep it simple (for now) and say that I've got $caw$ trees.

So my simplified problem is this: given $caw$ intervals of length $w$, with centers in $[-w/2, w/2]$, what's the probability that $0$ is not in any of them? Well...the probability is zero, because all those possible trees (assuming we use closed intervals) actually overlap the origin. The only subtelty here is that $caw$ might be much less than $1$, so when I say I "have $caw$ intervals", I might be saying to "consider $0.024$ intervals of length $w$ ...", and so on.

But wait: suppose that $caw$ is actually $1/7$ (just to pick a round number). That means that $1/7$ of the time (on average) there's a tree whose projection covers the origin, so $6/7$ of the time I can see through the forest! Aha! In this simple version where I say that "caw" is the number of trees, I get that $1 - caw$ is the probability I can see through the forest.

That's actually a pretty good approximation when $caw$ is small (say, less than $0.5$). But what if $caw = 1$? Then the average count of trees (in all imaginable forests) shadowing the origin is $1$. But in some of those, two or three trees might be shadowing the origin, and in others, none might be shadowing it. So we can't really say that the probability that the origin is unshadowed is zero. Hmmm.

In fact, at this point, you're asking a question analogous to this one: suppose I flip a fair coin $N>>1$ times. I expect the number $S$ of "heads minus tails" to be zero, which it is, on average. But what's the probability (as a function of $N$) that $S > 1$? or that $S > 2$? I.e., what fraction of the time is some random variable far from its average?

Feller's introduction to probability has a whole chapter on questions like this, but it's upstairs, and I need to go run errands, so I'm gonna stop here.