Consider two discs in the plane of radius $r$ and $s$, with centers separated by a distance $l$. If we choose a point uniformly at random from each disc, what is the expected distance between the two points? (This is a generalization of this question.)

Is there a simple closed-form solution? Experimentally (via monte carlo), the answer is something like $2/3 \sqrt{r^2+s^2+(3l/2)^2}$. This has the right limiting properties, e.g. when $r,s \to 0$ or $l \to \infty$, we get simply $l$; and when $s=l=0$ we get $2r/3$, the average distance from the center of the disc, but is only approximately right when $r=s$, $l=0$, where we estimate $2\sqrt{2}/3 r \simeq 0.94 r$ compared to the exact result of $128r/45\pi \simeq 0.91 r$ for the average distance between two points in a single disc.

Aside: Writing the experimental result as $\sqrt{(2r/3)^2+(2s/3)^2 + l^2}$ suggests a nice geometric motivation for it, in which we represent each disc as a ring with 2/3 the radius (so each ring has the same average distance from its center as the original disc). Then we rotate each ring about its center and out of the plane to be perpendicular to the line connecting them (approximately preserving average distance between them). Now estimate the average distance between the rings by taking a point on each with a phase difference equal to the expected average, namely $\pi/2$. i.e. WLOG take one ring centered at the origin in the $yz$ plane, and the other centered at $(l,0,0)$, and approximate the average distance between them as the distance between the point $(0,2r/3,0)$ and $(l,0,2s/3)$, which yields the experimental result.

Aside 2: We can improve the approximation by considering the three error sources in the experimental formula: (1) rotating the discs out of the plane, (2) replacing the discs with rings, and (3) approximating the inter-ring distance by two points $\pi/2$ out of phase. The second error underestimates, since the average self-distance in a unit disc is $128/45\pi \simeq 0.91$ whereas the average self-distance in our 2/3 ring is $8/3\pi \simeq 0.84$. The third over-estimates, since the average self-distance in the unit ring is $4/\pi \simeq 1.27$ compared to the $\pi/2$ phase difference estimate of $\sqrt{2} \simeq 1.41$. The product of these two leaves our estimate slightly high (ignoring the first error), so we can introduce a "phase correction" and instead estimate as $\sqrt{(2/3)^2(r^2+s^2-2rs \cos \phi) + l^2}$, choosing $\phi$ to make the $r=s$, $l=0$ case correct, namely $\cos \phi = 1 - (64/15\pi)^2/2 \simeq 0.0778$ or $\phi \simeq 0.95 \pi/2$.

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    $\begingroup$ I think you should start with the situation s=0, l>0, r>0... The average distance between a fixed point and one taken randomly in a disc of radius r, whose center is at distance l from the fixed point... $\endgroup$ – Martigan May 23 '14 at 12:04
  • $\begingroup$ This question seems very different from the one in the link. $\endgroup$ – Yves Daoust Mar 23 '20 at 17:21

Let me take a shot at it, with a geometrical view of averaged components.

Under the assumption that in the plane the

Shaded area $A$ between direct tangents equals Average sweeping arc lengths of left side minor arcs $ {\,\bar s}$ times Average (swept) length $\bar {l}$,

$$ A= (R+r)\, l \cos\ \beta/2$$

Buffon's needle-like

Average arc length of the two left circle minor arcs shown

$$ {\,\bar s} = \frac{(R+r)}{2} \,(\pi+\beta)$$


$${\,\bar l}=\frac{2l \cos \beta/2}{(\pi+\beta)}$$

For special case $\beta=0: $

$${\,\bar l}=\frac{A}{\,\bar s}=\frac{2l}{\pi}.$$


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