Suppose there is a line with fixed length $L$, and $x$ is the position of the points on this line ranging from $0$ to $L$, i.e. $0 \leq x \leq L$, then what is the probability distribution of fitting two circles with diameter $d$ ($d$ is always less than half of the length, $d < L / 2$). The first circle can be randomly positioned any point (including both ends), the centre of the first circle, $x_1$, can have values $0$, $L$ and any value between them. but the second circle, with its centre at $x_2$, cannot overlap - only it can be adjacent, so always $\vert x_1-x_2 \vert \geq d$.

How can I find a generic statement for the probability of distribution of these two circles?

$x$ is the variable for the $1$-dimensional coordination of circles centres.

For example, assuming $L=1$, and $d=0.2$, if the $x$ for the first circle happens to be at $x=0.1$, the second circle can be located at any point between $x=0.2$ and $x=1$ (to avoid overlapping), so $80\%$ of $L$. if $x_1$ ($x$ for 1st circle's centre)$=0.2$, the $x_2$ can have only $70\%$, but if $x_1=0.4$, $x_2$ has again $80\%$ chance.

I'm looking for a probability model to describe this problem in general in terms of $L$ and $d$.

  • 1
    $\begingroup$ Very obscure. What is the link between your circles and x ? what do you mean by fitting the circles ???? $\endgroup$ – Arnaud Mégret Apr 21 '17 at 10:03
  • $\begingroup$ $d=d$ ? ${}{}{}$ $\endgroup$ – Joseph O'Rourke Apr 22 '17 at 1:39
  • $\begingroup$ sorry, I meant d smaller than half L. provided more explanation to be clear. $\endgroup$ – Rui Apr 22 '17 at 11:05
  • $\begingroup$ Is $x2$ also supposed to be on $L$? $\endgroup$ – Magdiragdag Apr 22 '17 at 13:38

So you shall have that

$$ 0 \le x_{\,2} \le x_{\,1} - d\quad \vee \quad x_{\,1} + d \le x_{\,2} \le L $$ which, dividing by $L$, can be reduced to a "adimensional" form as $$ 0 \le {{x_{\,2} } \over L} \le {{x_{\,1} } \over L} - {d \over L}\quad \vee \quad {{x_{\,1} } \over L} + {d \over L} \le {{x_{\,2} } \over L} \le 1 $$

Geometrically this is represented by the attached sketch

enter image description here

Therefore you get a probability equal to the area of the two white triangles, i.e. $$ P = \left( {1 - {d \over L}} \right)^{\,2} = {{\left( {L - d} \right)^{\,2} } \over {L^{\,2} }} $$

By the way note that this problem is the contrary of the meeting problem (see e.g. this post)


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