A question related to Buffon's needle

Question

The following is an elementary probability question related to a generalization of the famous "Buffon's needle experiment" which allows one to estimate $\pi$ by counting how many times a randomly tossed needle crosses a line on a lined sheet of paper. If we replace the needle with a rigid wire in the shape of any piecewise smooth plane curve, I believe it is well-known that the expected number of line crossings depends only on the length of the wire and not on its specific shape.

I am seeking an elementary proof of this fact in the case where the wire consists of two line segments joined end-to-end. The only parameters here are the lengths of the two line segments and the angle at which they are joined; I would like to prove that the expected number of crossings depends only on the sum of the lengths. If it helps, I am happy to assume that both line segments are very small compared to the spacing between the lines on the paper. Any ideas?

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Added: Several have argued that this follows simply from the linearity of expectation, but I am not convinced. Suppose it were the case that the expectation for a single segment of length $\ell$ was given by $\ell^2$. Then if $X$ and $Y$ are the random variables representing the two needles making up the wire we would have $E(X+Y) = E(X) + E(Y) = \ell_X^2 + \ell_Y^2$, and this is not a function of $\ell_X + \ell_Y$ (though it is a function of $\ell_X$ and $\ell_Y$). Of course we secretly know that $E(X) = C \ell_X$, but my goal in asking this question is to prove this fact without actually calculating anything.

Answer 1

Let $E(\ell)$ denote the expected number of crossings achieved by a segment of length $\ell$. By linearity of expectation, we know that any union of two segments of lengths $\ell_1,\ell_2$ will achieve an expected crossing number of $E(\ell_1)+E(\ell_2)$, irrespective of the angle they meet at.

In particular, this holds for the line segment of length $\ell_1+\ell_2$. So we must have $$ E(\ell_1)+E(\ell_2)=E(\ell_1+\ell_2) \, ; $$ since this last quantity depends only on the total length of the segments, it follows that the expected crossing number of any union of two segments depends only on the total length of the segments.

Answer 2

The result simply follows from linearity of expectation. The expected total number of crossings is the sum of the expected number of crossings for each portion of the wire (even as those portions become infinitesimally small).

In fact, this leads to an elementary proof of the original Buffon result. Let the separation between the lines be $L$, and let the length of the straight wire be $X$. Curl the wire $n$ times around, into a tight circle of radius $\frac{X}{2\pi n} \ll L$. Then the number of crossings will be $2n$ if the center of the circle lands within a distance $\frac{X}{2\pi n}$ of a line, which will happen with probability $\frac{X}{\pi n L}$. The expected number of crossings is therefore $2n\cdot\frac{X}{\pi n L}=\frac{2X}{\pi L}.$ This result is correct for the original straight wire as well.

Answer 3

Let's reconsider the "throwing at random" part in terms of an arbitrary shape needle of overall length S. Focus on an arbitrarily short section at one end, Section #1. Throw the wiggly needle down and count the crossings only by this Section #1.

Meanwhile another observer has decided to "throw at random" by taking the position and orientation of Section #1 (from the throw above) and applying the necessary translation and rotation (from the shape of the wiggly needle) to find the position and orientation of, say, Section #2. Then he counts the crossings of only Section #2.

Repeat for all the short sections.

The proof of the Buffon's needle theorem for "short" needles shows the probability of a crossing depends in a linear fashion on the length of the short needle. So we can just add up the results for all the Section #1, Section #2, .... by adding the lengths.

QED ?