From Buffon's Needle to Buffon's Noodle

https://mbmccoy.dev/posts/buffons-noodle/

by _alternator_ • 3 months ago

37 13 3 months ago

enricozb• 2 months ago

Pretty neat! However, if you wanted to know the _probability_ of a noodle crossing any line in the long noodle case (L/W > 1), the expression is more complex (and I believe would require an integral) :).

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.

_alternator_5• 2 months ago

This is the crux of the observation. For needles of length less than W, the probability that it crosses a floorboard is equal to the average number of floorboards it crosses. (Exercise for the reader ;))

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.

gowld2• 2 months ago

That's the menace of expected-value problems. They are easy-to-solve variants of more interesting hard-to-solve problems.

_alternator_5• 2 months ago

Is the probability actually more interesting though? I find the symmetry of this type of result extremely compelling, beautiful even. Buffon himself restricted his attention to the case where the needle was short enough that "probability" and "expectation" had the same answer. Put simply, math is best when complicated-seeming things suddenly become simple.

card_zero2• 2 months ago

I'm wary of asking questions (my curiosity is bounded), but what changes if you limit the range of allowed angles to multiples of, say, 10°? How about 90°, does pi go away then?

_alternator_5• 2 months ago

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.

card_zero2• 2 months ago

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?

topaz0• 2 months ago

That was a delicious read. I love this kind of mathematical writing.

penteract• 2 months ago

Very cool. I was expecting it to make circles bigger rather than making needles smaller. Take a near-circle consisting of N lines. As N tends to infinity, the near-circle would have a diameter close to N*L/π, so would touch N*L/πW + O(1) lines twice each.

gowld2• 2 months ago

Why is the simulation always exactly 2 for a closed polygon, but has error for the almost-full circle?

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!

_alternator_5• 2 months ago

The simulation is always 2 for the closed circle of width W. Actually if you run it enough, the simulation will occasionally end up not perfectly equal to 2 because it's a 50 sided polygon, not a true circle. ;)

But for a perfect circle of diameter W, it will alway hit exactly two vertical lines.

From Buffon's Needle to Buffon's Noodle

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