Hoops in a Field I have a simple question, but I haven't been able to find the answer I'm looking for anywhere. I'm sure there is some simple theorem or distribution that tackles this problem, but it evades me. Perhaps I haven't googled the right terms or looked in the right textbooks!
My problem is as follows. Suppose a proportion ($p$) of a field is covered in water. The water is randomly dispersed in the field. The field has an area $A$. Now we fly over the field in a helicopter and randomly drop a hoop of size $a$ (where $aN = A$) into the field. The hoop lands at a random spot within the field borders. What is the probability that there will be water inside the hoop?
My intuition about the problem is that for $a > A(1-p)$, the probability is 1 (because the area excluded by the hoop is smaller than the total area of the water), and as $a \rightarrow 0$, the probability approaches $p$. But what about for values of $a$ in the middle? Can the question be answered?
I have a sneaking suspicion that the answer is embarrassingly simple.
Thanks for your help! 
 A: This question comes up in analyses of spatial relationships, especially in studying edge effects in spatial statistics.  The answer depends intimately on the shape of the field, the shape of the water within the field, and the size of the hoop.
A hoop of radius $r$ "contains water" if and only if its center lies within distance $r$ of water, which we may view as a point set $W$ in the plane.  To solve this, construct the $r$ buffer of the water: this is the locus of all points within distance $r$ of water.  Let's call it $B_r(W).$
Assume now that the field $F$ is also modeled as a set of points in the plane, and suppose it is bounded and measurable.  Regardless of the shape of $W$ (it might not be measurable), this implies (via standard compactness arguments) that $B_r(W) \cap F$ is (Borel) measurable for all positive buffer distances $r$.  Accordingly, it is meaningful to talk about the probability that a uniformly distributed hoop center lies within $B_r(W) \cap F$.  By the very definition of a uniform distribution, that probability is precisely the buffer area divided by the area of the field.
Examples



*

*(A single puddle, at left.)  The field is a circle of diameter $R$ and the water is represented by a single interior point of distance at least $r$ from the boundary of the field.  The buffer $B_r(W)$ then is a circle of radius $r$ and its area relative to that of the field is $(r/R)^2$.

*(Multiple puddles, center.)  As before, but now the water is a finite set of $n$ points all separated by at least $2r$ from each other and all lying at least $r$ from the field's boundary.  The probability is now $n(r/R)^2.$

*(Circular pond, at right.)  Suppose the water is a circular pond of radius $w$ and, as before, lies no closer than $r$ to the field's boundary.  Its buffer is a circle of radius $w+r,$ whence the probability is $\left((w+r)/R\right)^2.$

*(Canal system.) Let the field be a rectangle and suppose the water forms a gridded irrigation system covering the field.  The spacing between the edge of one irrigation canal and the edge of the next is no greater than $2r$.  The buffer $B_r(W)$ now covers the entire field (by design!) and the probability is $1$.

This figure (taken from an answer at https://gis.stackexchange.com/a/17390) shows a small-radius buffer of an irregular set of line segments (the irrigation ditches, shown in black) within a rectangular field.  As the radius grows, eventually the buffer--the gray area--will merge into the entirety of the field.  (The pink areas are not relevant to the present discussion.)
Note that in many of these cases the water may have practically no area--it would be represented on a map as a collection of points or line segments--while the probability could end up being anything between $0$ and $1$.  When the water forms a connected, simply-connected domain that is not too tortuous and the buffer distance $r$ is very small (smaller than the width of the water at its narrowest point), then the buffer area is approximately 
$$|W| + rP + \pi r^2$$
where $|W|$ is the area of the water and $P$ is the length of the perimeter of the water (assuming it is finite).  (This formula is exact for piecewise linear convex domains: that is, convex polygons.)  To first order in $r,$ then, the probability is approximately
$$\frac{|W| + rP}{|F|}$$
where $|F|$ is the area of the field.  This shows that initially, for very small $r$, the probability grows linearly with $r$ and the slope (constant of proportionality) is the ratio of the perimeter of $W$ to the area of the field, $P/|F|.$
A: A simple answer to your question would be (but not as beautiful as thought):
If it is the case of randomly scattered infinitely small water droplets in a continuous 2-D field, then probability of water in the hoop would be $p_w \approx 1$ for all sizes of hoop. 
Why? We can imagine the hoop consisting of N number of equally sized 'pockets'. Probability of water found in hoop = $P(\text{water found any of N pockets}) = 1-(P(\text{water not found in a single pocket}))^N$
As long as size of pocket $\ne 0$, $P(\text{water not found in a single pocket}) =1-p \ne 1$ and thus $(P(\text{water not found in a single pocket}))^N \approx 0$ as $N \to \infty$.
**
If though, we have a more specific and realistic definition of water:
This can be an example:


*

*The body of water can be made up of finite number of water patches.

*Each patch of water has to be closed and bounded.


Even with this question, the answer relies very much on the form of the water patches as noted by others here. Zig-zag lines of water would mean $p_w =1$. Big circle pools of water would yield different probability(and still no easy answer). Because of the infinite number of arrangement of the water and the possible complexity of solution with each, this question is probably impossible to answer. 
Unless the we know the exact water arrangement already, I believe we wouldn't be able to quantify the chance of having water in the hoop.
