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If we choose randomly an infinite, countable set of disks in $\mathbb R^2$, what is the probability that intersection of every pair of disks from the set is an empty set?

EDIT: Because the problem in the above stated form is too hard to handle, or undecidable, or it may have many different solutions depending on the way we choose the centers and the radii I will state it with extra conditions.

First variant of the problem: Add to the original problem: "all disks have the same radius $R$".

Second variant of the problem: Add to the original problem "all disks have the same radius $R>1$ and every center has the coordinates $(x,y)$, where $x,y \in\mathbb Z$"

Question for the first and second variant of the problem: Is the problem solvable in at least one of these two variants or we need more extra conditions?

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"choose randomly" is not a well-defined prescription. It can sometimes be interpreted as a well-defined prescription, in cases where there is an obvious candidate for the most natural distribution to consider, e.g. a uniform distribution. This is not such a case. –  joriki Jan 31 '13 at 15:48
Do you want to say that this problem can`t be solved with the information I provided? –  A.P. Jan 31 '13 at 15:50
Essentially. You need to specify what "randomly" means. I.e, from what distributions are we choosing the centre coordinates and radii? –  Robert Mastragostino Jan 31 '13 at 15:58
Do you think that something changes if we set that all disks have the same radius? –  A.P. Jan 31 '13 at 16:01
That's a good start, but then you still haven't said anything about how to choose the centres. Are they supposed to be a Poisson process? If so, the answer would be zero, independent of the product of the density of the centres and the area of the disks (which would otherwise also need to be specified). –  joriki Jan 31 '13 at 16:08

1 Answer 1

Sadly, some natural formulations of this problem are extremely difficult. Take a finite square with side-lengths $n$ and then make it periodic by identifying opposite sides, i.e. making it a torus. Then you can uniformly pick the centers of each circle within the rectangle. This is called the "Pennies on a carpet problem" or "2-D hard-spheres model." The problem is extremely difficult and not too much is known about it. You can find a bit more info here

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