The premise is basically a 2D plane with a single point, the starting point.

Now a landmark sought by a hiker is a certain distance from that point.

If the hiker can only see 1 mile in any direction, the chance of him finding the landmark by walking straight in a random direction from the starting point seems like it will go down the further away the thing is.

Practically speaking if something is 20 yards from the starting point then any direction I choose to walk I will see it right away. But if something is 20 miles from the starting point then my chance of finding it by walking in a straight line should reduce drastically.

I am imagining a circle that has a radius equal to the distance of the landmark (the center of the circle being the starting point). The further the landmark is, the bigger the circle is, which means if I divide by 360 the 'chunks' will be bigger.

What am I talking about? What's the math for this?

Also, please evaluate my tags, because I am not sure which category this would fall under but I will adjust if you leave a comment (or edit if you have the power).



If your hiker can see 1 mile in any direction, then we can define success as the probability of choosing a direction vector that will intersect a circle of radius 1 mile around the target.

Now, if we let $\theta$ be the randomly chosen angle with $\theta \sim U(0,2\pi)$ (in radians), and our target is $R$ miles away from you, then the probability of finding the target is equal to the length of the subtended arc $\theta_s$ generated with a radius $R$ that passes through the target point and has endpoints at the the points where the arc intersects the 1-mile radius circle around the target point:

$\theta_s=2\sin^{-1}\left(\frac{1}{R}\right)\implies P(Success|R)=\frac{\sin^{-1}(\frac{1}{R})}{\pi}$

enter image description here

  • $\begingroup$ A diagram of your geometry would be useful; I'm not quite persuaded as yet. $\endgroup$ – Semiclassical Jul 24 '14 at 13:18
  • 1
    $\begingroup$ I realized that I meant $sin^{-1}$ not $tan^{-1}$...ah, pictures are helpful, aren't they ;-) $\endgroup$ – user76844 Jul 24 '14 at 13:38
  • $\begingroup$ One thing you can also see from that picture is that the question only makes sense if $R\geq 1$ (which is obvious but important for checking normalization.) On that note: how is the probability normalized? $\endgroup$ – Semiclassical Jul 24 '14 at 13:46
  • $\begingroup$ @Semiclassical well, I was taking the OP's statement of "random direction" to mean the direction is uniform on $[0,2\pi]$, hence the normalizing constant is $\frac{1}{2\pi}$ $\endgroup$ – user76844 Jul 24 '14 at 13:55
  • $\begingroup$ Can you check that with your result? That'd be a nice addition to your answer. $\endgroup$ – Semiclassical Jul 24 '14 at 14:02

Draw a circle of radius $r=1$ mile around landmark. From our starting point at a distance $d$ from the landmark draw a straight line L to the landmark. We want our path to intersect the circle. To do this, the direction chosen must be at an angle less than $\theta=\arcsin(r/d)$ from L (on either side). We choose a direction with a uniform distribution on $[0,2\pi].$ If $d\le r$ then we start inside the circle and the answer is 1. If $d\gt r $ then some trig gives the probability as:

$$ \frac{\arcsin(r/d)}{\pi} $$

For example, if $r=1,d=20$ the probability of success is $ 0.016 $ and if $r=1,d=2$ then the probability is 0.167 and is $0.456$ at $d=1.01$ .


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