From Buffon's Needle to Buffon's Noodle(mbmccoy.dev) |
From Buffon's Needle to Buffon's Noodle(mbmccoy.dev) |
So taking the limit of a large number of segments converging to a circle of diameter W leads to the result that the average number of intersections must be 2L/\pi.
It's interesting that the number of crossings is independent of whether L/W is less than or greater than 1, but the probability of crossings is equal to 2pi * L/W only in the short case. This makes sense since in the short case the noodle can at most cross a single line.
The point is that the "right" quantity to be considering for the problem is the average number crossings, since that naturally extends to curved noodles, lines of any length, and even circles. The number of crossings is also known as the Euler characteristic of the intersection, and there's a rather deep and beautiful theory of geometric probability that takes this as the jumping off point.
I think it's because only the closed polygon is totationally symmetric, so you don't get errors from the edge case at the edge of the finite sample space. But I'm not sure.
The illustration is missing the more interesting visualization of how linearity of expectation applies to all possible rotations and translations of all segments of the needle/noodle. Each noodle is equivalent to a curve of discs, like a string of pearls. And those pearls do not even need to be connected!
But for a perfect circle of diameter W, it will alway hit exactly two vertical lines.
Yes, under some assumptions. As the sibling comment points out, if there’s a single allowed angle theta then the expected number of intersections is cos(theta) * L/W (-pi/2 < theta < pi/2). You can get from this fact to the standard Buffon’s needle result by integrating wrt theta to find the average probability over thetas with a uniform distribution on (-pi/2 < theta < pi/2): \int 1/pi * cos(theta) * L/W d theta.
Now suppose you have two angles, theta_1 and theta_2. The expected number of intersections for each of them is as above, and if the needle falls at one or the other with equal probability then the overall expectation is 1/2 * cos(theta_1) * L/W + 1/2 * cos(theta_2) * L/W. Passing to the case with n distinct angles with equal probabilities we have \sum_i 1/n cos(theta_i) * L/W.
Now if we make the further assumption that the angles are evenly distributed over (-pi/2 < theta < pi/2), i.e. they are the angles of the sides of a regular n-gon, then we can interpret that sum as a Riemann sum. If we write it as
1/pi \sum_i pi/n cos(theta_i) * L/W
Then pi/n is the delta_i term in the riemann sum, and the limit is
lim_{n -> inf} 1/pi \sum_i pi/n cos(theta_i) * L/W = 1/pi \int cos(theta) * L/W d theta.
We can pull the L/W out, leaving \int_-pi/2^pi/2 cos(theta) d theta = sin(pi/2) - sin(-pi/2) = 2, giving the final result of 2/pi * L/W.
Essentially, as we increase the number of allowable angles we are approximating an integral of the cosine function (times constants) from -pi to pi, which is where the pi creeps in. The angles don’t need to be strictly evenly spaced for this to work—if they are independent randomly selected from the uniform distribution then it will also work, as you’re then performing a monte carlo integration.
If you don't allow rotations, but somehow still take a polygonal limit to circles, I suspect you'll end up with the same answer. But the limit is necessarily restricted relative to highly symmetric polygons going this route.
In general, rotational symmetry gives a ton of power to simplify the math, and leads to highly general results like arbitrary "noodles" having the same average crossing count as needles of the same length.