Is There a Simpler Way to Calculate the Probability of a Needle Crossing a Line?

[LCSphysicist]

TL;DR

How you would approach this problem? We need to find the accesibles position of a dropping needle

My approach to this problem is a little laborious, it involves three coordinates, probably it is right, but tiring and extensive beyond what the question wanted.

Be the origin in the rectangle middle.

It would be like: imagine a rectangle with opposite sides L and R with length l, so to find the probability the needle cross a line, i would at first find the probability the needle center fall in a position between x,x + dx; y, y + dy, so find the allowed angle to it rotate without cross a line, find the probability it falls in this range (range/2pi), ant integrate wrt dx,dy, finding Pf. But, as what we want is exactly opposite of it, we would just do P = 1 - Pf

In another words:

$$P = 1- \int\int P((x,y))P(\theta)dxdy$$

where we can express $$\theta = \theta(y)$$ (what matters to the angle is y itself, with it we can know the distance between the line and the needle center as l-y

In the end the x value will cancel $P(x,y) = \frac{dxdy}{lX}$, when we integrate dx, it will just cancel with X) and we will finish with just y.

But i believe maybe there is an easier way to do it... What would you do?

 

[fresh_42]

This experiment is known as Buffon's needle problem:
https://en.wikipedia.org/wiki/Buffon's_needle_problem

 

[LCSphysicist]

This experiment is known as Buffon's needle problem:
https://en.wikipedia.org/wiki/Buffon's_needle_problem

Wow, the approach is similar :D

 

[Office_Shredder]

Can you just do something like the angle $\theta$ between the needle and the lines is uniformly distributed between 0 and $\pi$. $\theta=0$ means the needle is parallel. Then the length covered by the needle in the direction perpendicular to the lines is $l \sin(\theta)$, and the probability it hits a line is $l \sin(\theta) / l$.

Then the probability it hits a line is
$$\frac{1}{\pi} \int_{0}^{\pi} \sin(\theta) d\theta $$
Which yields $\frac{2}{\pi}$.

This is the same as what you wrote down, but picking a slightly better perspective to avoid a double integral.

Edit to add: wikipedia has an amazing solution for this problem:
https://en.m.wikipedia.org/wiki/Buffon's_noodle

Basically the idea is:. If you curve the needle and make it a noodle (of any rigid shape), the expected number of line crossings is independent of the shape, since you can think of that shape as being broken up into small linear chunks and then observing that expectancy adds linearly. It also must be linear in the length of the noodle. If you take a circle of diameter l, it will always intersects exactly twice, so the rate of intersections is 2 per length $\pi l$. Hence if the noodle is a needle of length $l$, the expected value of the number of intersections is $l \frac{2}{\pi l} = \frac{2}{\pi}$. Since it only ever intersects 0 or 1 times, that must be the probability that it intersects.