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A hunter is standing in the center of an infinite 2D forest. There are point trees at all the integer lattice points. The hunter fires a gun with a bullet of zero width in a random direction. He misses (with probability one) of course, because the set of angles in which can hit the tree have measure zero, i.e. while there are an infinite number of trees, they are countably infinite.

What happens when the hunter takes many shots, whose angles make up the set $A$ when

  1. $|A|=c$, where c is finite
  2. $|A|=|\mathbb{Q}|$, a countably infinite number of shots
  3. $|A|=|\mathbb{R}|$, a uncountably infinite number of shots

When the hunter takes a finite number of shots as in 1. he still misses each shot with probability one. What about 2. and 3.? My intuition says that he still misses with probability one for all but the last one, as you can map each shot from 2. onto the rational points over $\mathbb{Q}^2$, but I'm not sure how to prove this.

Edit: At Qiaochu's suggestion, the text of question has been changed from always misses to misses with probability one.

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There is only one man who can fire infinitely many shots. But he doesn't have to shoot. He can roundhouse kick everything. – Asaf Karagila May 8 '13 at 19:40

If the hunter fires a gun once in a direction uniformly chosen at random with respect to the unique translation-invariant probability measure on angles, he misses with probability $1$. This is not the same as saying that he always misses; there's a non-empty measure-zero event corresponding to him hitting something. In probability theory the usual way to say this is that he misses almost surely, to emphasize that he doesn't miss surely. See also this math.SE question.

This remains true for a finite number of shots, but after that there are issues with defining infinite product measures that I'm not familiar with. I think things are okay for countable products of probability spaces but I don't know what happens after that.

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Thanks for the correction in terminology, I need to get in the habit being more precise - I do understand the difference between always misses and misses almost surely. I'm going to edit in the difference into the question to be more clear. To summarize the 2nd paragraph of your answer, you say 1. almost surely misses, 2. (you think) almost surely misses 3. (you have no idea). Is this correct? – Hooked May 8 '13 at 20:02
Yes. I'm uncomfortable about the third case. – Qiaochu Yuan May 8 '13 at 20:04

Restrict the shooting spree to the first quadrant. Then the tree at $(x, y)$ is at an angle with a rational tangent $y / x$. Just make the hunter shoot only at angles with irrational tangents, those are equinumerous to $\mathbb{R}$ and miss every time.

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I don’t think that this answers the question. I take the question to be: if the hunter fires $2^\omega$ independent shots, each in a direction chosen at random w.r.t. the uniform distribution on $[0,2\pi)$, does he miss with probability $1$? We know that he can miss. – Brian M. Scott May 8 '13 at 22:05
@BrianM.Scott is correct in the interpretation. The uniform distribution was implied by the fact that the hunter doesn't aim but "fires randomly". – Hooked May 9 '13 at 1:08
@Hooked, "randomly" is not "uniformly"... it just implies some distribution. – vonbrand May 9 '13 at 1:25
@vonbrand I know, I know! Like the exchange with Qiaochu, what I imply and what I say are not always the same thing. I appreciate the corrections to help get at what I'm asking. Nonetheless, Brian does indeed state the question more formally. – Hooked May 9 '13 at 1:32

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