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Two people, A and B, starts from two different points and move in a perfectly straight line in an infinite plane. When they move they leave a visible trace after them.

Question: What is the probability that their path (of traces) will intersect at some point regardless of where they start?

What I've tried so far is to draw two circles and split it in quadrants. That helped a little bit but didn't really solve the problem, just got an overview.

Here's some examples of interesecting paths (first row) and non intersecting paths (second row) which gives you an idea of the criterias for when they intersect and don't intersect:

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How would you approach and solve this problem?

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    $\begingroup$ Questions with random choices of paths are often confusing and can be ill-posed (like the famous "random chord in a circle" problem). So I very tentatively offer as a comment not an answer: consider the half-planes each side of $AB$. If paths occupy different halves (probability $\frac12$) they will not intersect. If they occupy the same half (probability $\frac12$) then they will intersect iff (referring to the upper half in your diagrams) $\angle A<\angle B$ where angle is measured from the direction $\vec{AB}$, prob $\frac12$ again. So the probability of NO intersection is $\frac34$. $\endgroup$
    – David
    May 29 at 4:42
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    $\begingroup$ If you draw the two full lines through the starting positions of $A,B$, the probability that the lines intersect is $1$. The "paths" (or rays, or half-lines) will intersect iff both $A,B$ walk towards the intersection point. That's assuming a plausible definition of "random direction" for the walks, though not necessarily the only conceivable one. $\endgroup$
    – dxiv
    May 29 at 4:44
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    $\begingroup$ @dxiv Same answer by a different argument, so maybe we are right :) $\endgroup$
    – David
    May 29 at 4:46
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    $\begingroup$ @David Right ;-) though, as you wrote, there is room for ambiguity in the statement of the problem. $\endgroup$
    – dxiv
    May 29 at 4:48
  • $\begingroup$ It makes sense to try and answer this question fixing the two points first. There is no problem defining the angles to each be uniformly chosen from $[0,2\pi)$, leading to a random ray being defined which starts from these fixed points, so let's tackle that first (it's not too difficult). Then allow the points to be randomly chosen. $\endgroup$
    – Sarvesh Ravichandran Iyer
    May 29 at 5:11

2 Answers 2

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The answer is $1/4$.

A line in the plane goes to infinity in both directions. The chances that two random lines are parallel is negligible. If we have two lines, they will meet with probability $1$. Label the intersection point $X$.

Does the trace starting at $A_0$ (person A's initial position) pass through $X$? The chance is $1/2$. Similarly and independently for the trace starting at $B_0$, B's initial position. So the chance that the traces intersect is $1/4$.

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    $\begingroup$ Very nice and simple explanation! $\endgroup$
    – Prim3numbah
    May 30 at 11:44
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By translation, rotation and scaling we can assume $A$ is at the origin and $B = (1, 0)$ and that $A$'s direction is the $(\cos a, \sin a)$ for some $a\in [0, 2\pi]$ and $B$'s direction is $(\cos b, \sin b)$ for some $b\in [0, 2\pi]$.

So we're working on the probability space $[0, 2\pi]^2$ with uniform distribution. Denote the event $I = $ "the rays intersect"

To solve the intersection, we need to find a solution with $t,s>0$ in

$$\begin{cases} t\cos a = 1+s\cos b \\ t\sin a = s \sin b \end{cases}$$

We get inequalities (use angle sum formulas there)

$$ t = \frac{\sin b}{\sin(b-a)} > 0 \\ s = \frac{\sin a}{\sin(b-a)} > 0 $$

We can see the good region is two triangles with base and height $\pi$. So the answer is

$$ \mathbb{P}(I) = \frac{\pi^2}{(2\pi)^2} = \frac{1}{4}. $$

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    $\begingroup$ NIce! But I don't really follow eveything of this approach and found the other answer more easy to follow since it uses a very simple concept. $\endgroup$
    – Prim3numbah
    May 30 at 11:44

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