Buffons Needle Problem Revisited 5

Probability Level pending

Suppose you drop a needle onto a square grid. What is the probability that the needle lies across two different grid lines? (Find a closed form for this probability.)

Details and Assumptions:

(0) The grid is infinite in the x- and y- directions:

Source: grid image

(1) The length of the needle equals the length of each grid square.

(2) The center of the needle lands anywhere on the grid with equal probability.

(3) The needle can be rotated (around its center) at any angle with equal probability.

(i.e., the position and rotation of the needle are each distributed uniformly, as in Buffon's needle problem)

Inspired by Buffon's needle problem ..


The answer is 0.318.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Arjen Vreugdenhil
Dec 22, 2017

The probability that the needle lies across both a horizontal and a vertical gridline is 1 π 0.318 . \frac{1}{\pi} \approx \boxed{0.318}. The analysis is similar to my work for problem 1 in this series. Now we work with P ( θ ) = sin θ cos θ , \mathbb P(\theta) = \sin \theta \cos \theta, and 1 π / 2 0 π / 2 P ( θ ) d θ = 1 π 0 π / 2 sin 2 θ d θ = 1 2 π ( cos π cos 0 ) = 1 π . \frac{1}{\pi/2} \int_0^{\pi/2} \mathbb P(\theta)\:d\theta = \frac{1}{\pi} \int_0^{\pi/2} \sin 2\theta\:d\theta = -\frac{1}{2\pi} (\cos \pi - \cos 0) = \frac{1}{\pi}.

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...