Author here. If I understand the question, the answer is that the average number of lines that the "noodle" intersects depends only on the length of the noodle. If you change the angles between the segments, the average stays the same.

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.

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card_zero 14 hours ago | parent [-]

I was thinking along these lines: suppose it's a needle, but it can't rotate. It always falls at the same angle. Then there's no noodle, and no apparent connection to circles. Is pi still involved? Next, suppose there are two perpendicular angles that are permitted, and the needle always falls at one of those. That means you can have square noodles, but rotations still aren't allowed, so the squares must always be aligned the same way, and the only suggestion of a circle is if you consider a square to be an approximation to a circle. Then three angles, hexagonal noodles. Does an approximation to pi therefore slowly creep in as you increase the sides on the polygon?

	▲	_alternator_ 13 hours ago \| parent [-]
		For the first question, the answer is just cos(\theta)*L/W, where theta is the angle off horizontal (assuming the floorboards are vertical). So a trig function shows up, if not pi. 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.