# Bertrand’s paradox on a monitor

Martin<EMAIL_ADDRESS>
###### Abstract

We investigate Bertrand’s probabilistic paradox through the lens of discrete
geometry and old-fashioned but reliable discrete probability. We approximate
the plane unit circle with $\frac{1}{n}\times\frac{1}{n}$ boxes and count the
pairs of boxes separated by distance more than $\sqrt{3}$. For $n\to\infty$
the proportion of such pairs goes to

$\frac{1+\sqrt{3}}{8}-\frac{\pi(2-\sqrt{3})}{96}=0.33273\dots\;.$

##  1 Introduction

We discuss and implement a discrete approach to the well known Bertrand’s
probabilistic paradox. J. Bertrand proposed it in his textbook [4, pp. 5–6] in
1889. Another, less known, Bertrand’s paradox is in economics [5, 8, 11]. The
probabilistic paradox requires to compute the probability

$\mathrm{Pr_{Be}}$

that a random chord $T$ in a plane circle $C$ is longer than a side of the
inscribed equilateral triangle. If $C$ has radius $1$, then $T$ has to be
longer than $\sqrt{3}$. J. Bertrand proposed three numerically different
solutions, given below as Solutions 1–3. It is paradoxical that the solution
is far from unique.

Solution 1. For $T$ determined by its endpoints $\\{a,b\\}=C\cap T$ one gets
$\mathrm{Pr_{Be}}=\frac{1}{3}$. Solution 2. Parametrization of $T$ by the
intersection point of $T$ with the diameter $P$ of $C$ perpendicular to $T$
leads to $\mathrm{Pr_{Be}}=\frac{1}{2}$. Solution 3. Parametrization of $T$ by
its midpoint yields $\mathrm{Pr_{Be}}=\frac{1}{4}$.

But one does not have to stop here. Solution 4. W. W. Funkenbusch [7] fixes a
point $a\in C$, selects at random another point $b$ on or inside $C$, and
takes the chord $T$ going through $a$ and $b$. By computing the corresponding
area inside $C$, he gets that

$\mathrm{Pr_{Be}}=\frac{1}{3}+\frac{\sqrt{3}}{2\pi}=0.60899\dots\;.$

This constant appears in a different context also in [12]. Solution 5.
Inspired by the previous solution, we propose to fix a point $a\in C$, select
uniformly and at random a number $d\in(0,2]$, flip a fair coin
$c\in\\{h,t\\}$, and get $T$ as the upper or lower chord, depending on the
coin flip, with the endpoint $a$ and length $d$. Thus

$\mathrm{Pr_{Be}}=2-\sqrt{3}=0.26794\dots\;.$

Solution 6. O. Sheynin [22] mentions that R. De Montessus [16] proposed for
$\mathrm{Pr_{Be}}$ uncountably many values.

Let us discuss the first five solutions more abstractly. We denote the open
interior disc of $C$ by $D$, its closure by $\overline{D}=D\cup C$, the center
of $D$ by $s$ and we set

$S=C\times C\setminus\\{(a,\,a)\;|\;a\in C\\}\;.$

We denote by

$\mathcal{CH}=\\{\\{a,\,b\\}\;|\;a,\,b\in C,\,a\neq b\\}$

the set of all chords of $C$, and by $\mathcal{CH}_{l}\subset\mathcal{CH}$ the
set of chords longer than $\sqrt{3}$. A unified approach to the above
solutions selects a probability space $(\Omega,\Sigma,\mu)$, an (almost)
surjective map

$F\colon\Omega\to\mathcal{CH}$

such that $F^{-1}[\mathcal{CH}_{l}]\in\Sigma$, and then sets

$\mathrm{Pr_{Be}}=\mu(F^{-1}[\mathcal{CH}_{l}])\;.$

The following spaces are employed; by the “product probability measure” we
mean product of normalized Lebesgue measures.

Solution 1 uses $S$ with the product probability measure. Solution 2 uses
$[0,\pi)\times(0,2)$ with the product probability measure. Solution 3 uses
$D\setminus\\{s\\}$ with the normalized two-dimensional Lebesgue measure.
Solution 4 uses

$(C\times\overline{D})\setminus\\{(a,a)\;|\;a\in C\\}$

with the product probability measure. In Solution 5 we use the space

$C\times(0,2]\times\\{h,t\\}$

with the product of normalized Lebesgue measures and the counting measure on
$\\{h,t\\}$.

In Solutions 2 and 3 the map $F$ is injective. In Solution 1 the map $F$ is
two-to-one, two pairs $(a,b)$ always determine one chord. In Solution 5 the
map $F$ is two-to-one, except for $d=2$ when four triples $(a,d,c)$ determine
a single chord. In Solution 4 the map $F$ is many-to-one. No exposition of
Bertrand’s paradox we know of notices that Solution 1 differs from Solutions 2
and 3 in the non-injectivity of $F$. True, this point is not very important
because one may replace the onto and two-to-one map $F\colon S\to\mathcal{CH}$
with a bijection $G\colon S^{\prime}\to\mathcal{CH}$, where $S^{\prime}$ is a
half of $S$; this replacement preserves the value
$\mathrm{Pr_{Be}}=\frac{1}{3}$ because the measure is normalized to give
$S^{\prime}$ probability $1$.

In Solutions 1, 2, 4 and 5 the map $F$ is onto but in Solution 3 the image
$F[\Omega]$ misses the chords $T$ containing the center $s$. One may be
tempted to dismiss them on the ground that they have probability $0$, but this
is not rigorous. These chords are outside the image of $F$ and, in the chosen
model, simply have no probability. It is formally more satisfying to extend
$F$ in the obvious way to a bijection

$F\colon\Omega:=(D\setminus\\{s\\})\cup u\to\mathcal{CH}\;,$

where $u=[0,\pi)\subset\mathbb{R}^{2}$ is taken as a plane segment disjoint to
$D$, and to use in $\Omega$ again the normalized two-dimensional Lebesgue
measure. Solution 4 shows that the tacit assumption of (almost) injectivity of
$F$ is in reality not very strongly justified. One produces chords $T$ in $C$
by random experiments, and it may well happen that many of them yield the same
outcome.

We give one more Solution 7. Since both sets $\mathcal{CH}_{l}$ and
$\mathcal{CH}\setminus\mathcal{CH}_{l}$ have the same cardinality
$\mathfrak{c}$ of continuum, there is a bijection
$F\colon[0,1]\to\mathcal{CH}$ such that
$F^{-1}[\mathcal{CH}_{l}]=[0,\frac{2021}{2022}]$. So, using the normalized
Lebesgue measure on $[0,1]$, we get

$\mathrm{Pr_{Be}}=\frac{2021}{2022}=0.99950\dots\;.$

There is an extensive debate on Bertrand’s paradox: [1], [2, Chapter 1.3],
[3], [6], [7], [9], [12], [13], [15], [16], [17, pp. 62–64] (here this author
learned about the paradox for the first time), [18], [19], [20], [21], [22],
[23], [29], and [30]. Most of this list was compiled by searching in the
Mathematical Reviews for articles having “Bertrand” and “paradox” in the
title. Many more references could be added to it.

We want to contribute to this debate by proposing a realistic mathematical
model of the paradox when it is treated as a physical experiment. The idea of
experimental resolution of the paradox is an old one, but in our opinion
physical schemes proposed so far do not employ faithful mathematical models.
For example, [12] suggests to perform the random selection of $T$ by letting
cosmic rays pass through a disc chamber detector, or by imagining them doing
so. But the rays are in [12] treated as lines with zero width. The problem
with such approaches is that nothing like a ray or a circle with zero width
exists in reality. Physical chords and physical circles have nonzero, positive
widths and are made of discrete parts with positive sizes. Any
“simplification” that replaces a positive number with zero is a jump in a
(totally) different world. In each case it should be carefully justified and
one should check that contact with the original setup was not lost and that
after the simplification one is not solving a (completely) different problem.

> We think that the plethora of solutions for Bertrand’s paradox demonstrate
> that by working with chords and circles with zero width we simplify too
> much.

We return to what it means, or should mean — in our opinion — to resolve
Bertrand’s paradox in the last section.

So we want to compute the probability in Bertrand’s paradox in the old-
fashioned but safe way as the limit ratio of two finite numbers, namely that
of favorable cases and that of all cases. We denote this discrete version of
$\mathrm{Pr_{Be}}$ by

$\mathrm{Pr_{Be,\,\Box}}\;.$

We work in the Euclidean plane $\mathbb{R}^{2}$ and for $n\to\infty$ consider
a sequence of square grids of axes-parallel $\frac{1}{n}\times\frac{1}{n}$
squares, so called boxes, with a vertex of the grid in the origin. We consider
physical unit circles $C(n)$ consisting of boxes whose interiors intersect the
geometric, origin-centered unit circle $C$. Let $\mathrm{Ch}(n)\subset
C(n)\times C(n)$ be the pairs $(B,B^{\prime})$ of boxes in $C(n)$ with
distance $>\sqrt{3}$, i.e., such that some points $b$ and $b^{\prime}$ in the
interior of $B$ and $B^{\prime}$, respectively, have distance
$|bb^{\prime}|>\sqrt{3}$. We denote the cardinality of a finite set $X$ by
$|X|$.

In Theorem 3.1 we compute that

$\displaystyle\mathrm{Pr_{Be,\,\Box}}$ $\displaystyle=$
$\displaystyle\lim_{n\to\infty}\frac{|\mathrm{Ch}(n)|}{|C(n)\times
C(n)|-|C(n)|}=\lim_{n\to\infty}\frac{|\mathrm{Ch}(n)|}{|C(n)|^{2}}$
$\displaystyle=$
$\displaystyle\frac{1+\sqrt{3}}{8}-\frac{\pi(2-\sqrt{3})}{96}=0.33273\dots\;.$

The chords in $\mathcal{CH}$ are unordered pairs while in $C(n)\times C(n)$
they are ordered pairs, but the factor $2$ cancels in the ratio. In Corollary
2.7 we determine the limit proportion

$\mathrm{Pr_{1,\,3,\,\Box}}:=\lim_{n\to\infty}\frac{|C_{0,\,\pi/3}(n)|}{|C(n)|}=\frac{1+\sqrt{3}}{16}=0.17075\dots$

of the boxes lying in the first quadrant of $C(n)$ in the angular positions
$(0,\pi/3)$. In Corollary 2.6 we note that there are many more (ordered or
unordered) pairs of boxes in $C(n)$ than there are boxes intersecting the disc
$D$. Thus for square grids Bertrand’s Solution 3 cannot be implemented, there
are not enough midpoints to represent bijectively all chords. Corollary 2.5
gives the formula

$|C(n)|=8n-4\sum_{\begin{subarray}{c}d\,|\,n^{2}\\\
d=2e+1\end{subarray}}(-1)^{e}$

for the size of the physical unit circle. For example,
$|C(3)|=24-4(1-1+1)=20$. But — despite its appearance — the formula is not
efficient; we do not know how to factorize integers efficiently, and therefore
the formula does not tell us how to compute for given $n$ in time polynomial
in $\log n$ the value $|C(n)|$. Propositions 2.1–2.4 are auxiliary. In Section
4 we give some concluding remarks and address suggestions of one referee.

Circles seen today are often sets of pixels on the monitor of a computer or a
TV set, selected from a fine rectangular array. Or maybe from an array with
other shape, but this is not the point. Such circles inspired us to write this
article, not the ideal geometric circle with zero width. Approximating $C$ by
$C(n)$ seems inadequate for circles drawn in media like plain paper without
the raster of graph paper, or on blackboard, slate or in sand. Is there a
realistic material such that when one draws $C$ on it, the distribution of
mass in the approximation is roughly uniform with respect to the angular
measure? In Bertrand’s Solution 1 this is taken for granted. We hope to
address this and similar questions at another occasion.

##  2 Counting vertical and horizontal boxes

By $\mathbb{N}=\\{1,2,\dots\\}$ we denote the set of natural numbers,
$\mathbb{Z}$ are the integers and
$\mathbb{N}_{0}=\mathbb{N}\cup\\{0\\}=\\{0,1,\dots\\}$. For $n\in\mathbb{N}$
we set $[n]=\\{1,2,\dots,n\\}$, $[0]=\emptyset$, for $a,b\in\mathbb{Z}$ with
$a\leq b$ we set $[a,b]_{\mathbb{Z}}=\\{a,a+1,\dots,b\\}$;
$[a,b]_{\mathbb{Z}}=\emptyset$ for $a>b$. For $n\in\mathbb{N}$ and
$i,j\in\mathbb{Z}$ we consider the $\frac{1}{n}\times\frac{1}{n}$ boxes

$B=B(n)=B_{i,\,j}(n)=[i/n,\,(i+1)/n]\times[j/n,\,(j+1)/n]\subset\mathbb{R}^{2}\;.$

$B^{o}$ denotes the interior of the box $B$. Let

$C=\\{(x,\,y)\in\mathbb{R}^{2}\;|\;x^{2}+y^{2}=1\\}$

be the geometric, origin-centered unit circle in the plane $\mathbb{R}^{2}$
and let $D$, $\overline{D}$ and $s=(0,0)\in D$ be as above in Section 1. We
define the approximations $C(n)$ of $C$ by boxes as

$C(n)=\\{B_{i,\,j}(n)\;|\;i,\,j\,\in\mathbb{Z}\wedge B_{i,\,j}(n)^{o}\cap
C\neq\emptyset\\},\ n\in\mathbb{N}\;.$

More generally, for any $n\in\mathbb{N}$ and any $X\subset\mathbb{R}^{2}$ we
define

$X(n)=\\{B(n)\;|\;B(n)^{o}\cap X\neq\emptyset\\}\;.$

We say that a box is exceptional if one of its four corners lies on $C$.

###### Proposition 2.1

.  For any fixed $\varepsilon>0$ and all $n\in\mathbb{N}$, there are at most
$O_{\varepsilon}(n^{\varepsilon})$ exceptional boxes.

Proof. $B_{i,j}(n)$ is exceptional if and only if one of the four sums
$i^{2}+j^{2}$, $(i+1)^{2}+j^{2}$, $i^{2}+(j+1)^{2}$ and $(i+1)^{2}+(j+1)^{2}$
equals $n^{2}$. Hence the number of exceptional boxes is $O(r_{2}(n^{2}))$,
where for $m\in\mathbb{N}$ the quantity $r_{2}(m)$ is the number of solutions
$(i,j)\in\mathbb{Z}^{2}$ of the equation $i^{2}+j^{2}=m$. There is the formula

$r_{2}(m)=4\sum_{d\,|\,m\,\wedge\,d=2e+1}(-1)^{e}\;,$

see [10]. Thus the number of exceptional boxes is $O(\tau(n^{2}))$, where
$\tau(m)$ is the number of divisors of $m$. It is well known ([10]) that
$\tau(m)=O_{\varepsilon}(m^{\varepsilon})$. $\Box$

A chord $T(n)$ in $C(n)$ is any ordered pair of distinct boxes,

$T(n)=(B,B^{\prime})\in C(n)\times C(n)\;\text{ and }\;B\neq B^{\prime}\;.$

The full chord $T_{B,B^{\prime}}(n)$ in $C(n)$ generated by the chord
$(B,B^{\prime})$ is the set

$T_{B,\,B^{\prime}}(n)=\\{A\;|\;\exists\,ab:\ a\in B^{o}\wedge
b\in(B^{\prime})^{o}\wedge ab\cap A^{o}\neq\emptyset\\}\;.$

of all boxes $A$ with interior intersected by a straight segment
$ab\subset\mathbb{R}^{2}$ that joins two points inside $B$ and $B^{\prime}$.
But we will not use full chords. The main reason is that one may have
$T_{A,A^{\prime}}(n)=T_{B,B^{\prime}}(n)$ for two chords $(A,A^{\prime})$ and
$(B,B^{\prime})$ even if $\\{A,A^{\prime}\\}\neq\\{B,B^{\prime}\\}$. This
happens for example when

$A=B_{-n,\,0}(n),\ A^{\prime}=B_{n,\,-1}(n),\ B=B_{-n,\,-1}(n)\;\text{ and
}\;B^{\prime}=B_{n,\,0}(n)$

— then $T_{A,A^{\prime}}(n)=T_{B,B^{\prime}}(n)$ is formed by the two layers
of boxes intersecting the segment $(-1,1)\times\\{0\\}$. We believe that this
happens infrequently, so that

$\frac{\text{the number of full chords in $C(n)$}}{\text{the number of chords
in $C(n)$}}\to\frac{1}{2}$

as $n\to\infty$.

A key tool in our computations is Abel’s summation formula ([24, Thm. 0.3]).
It has a nice and short proof, which we include here. Our proof differs from
that in [24] and we actually give a more general theorem. Recall that every
bounded function $f\colon[a,b]\to\mathbb{R}$ with only finitely many
discontinuities is Riemann-integrable; we only use Riemann integration ([24]
uses Stieltjes integrals). For a sequence
$(a_{n})=(a_{1},a_{2},\dots)\subset\mathbb{R}$ and a real number $x$ we set
($n\in\mathbb{N}$)

$A(x)=\sum_{n\leq x}a_{n}\;,$

where the empty sum is as usual defined to be $0$. For a function
$f\colon[a,b]\to\mathbb{R}$ we use the notation $[f]_{a}^{b}=f(b)-f(a)$.

######  Proposition 2.2 (Abel’s summation)

.  Let $l,m\in\mathbb{N}_{0}$ with $l<m$, $(a_{n})\subset\mathbb{R}$ be a
sequence and $f\colon[l,m]\to\mathbb{R}$ be a continuously differentiable
function. Then

$\sum_{l<n\leq
m}a_{n}f(n)=[A(x)f(x)]_{l}^{m}-\int_{l}^{m}A(t)f^{\prime}(t)\;\mathrm{dt}\;.$

Proof. Both expressions equated in the identity are additive. This means that
if $k,l,m\in\mathbb{N}_{0}$ with $k<l<m$, then the value of the expression on
the interval $[k,m]$ equals to the sum of its values on the intervals $[k,l]$
and $[l,m]$. Thus it suffices to prove the identity for every
$l\in\mathbb{N}_{0}$ and $m=l+1$. Then the left-hand side is $a_{m}f(m)$ and
the right-hand side is

$\displaystyle[A(x)f(x)]_{l}^{m}-A(l)\int_{l}^{m}f^{\prime}$ $\displaystyle=$
$\displaystyle[A(x)f(x)]_{l}^{m}-A(l)[f(x)]_{l}^{m}=[A(x)]_{l}^{m}f(m)$
$\displaystyle=$ $\displaystyle a_{m}f(m)$

too. $\Box$

See [25, p. 13/14] for another application of this technique proving sum-
integral identities.

We orient $C$ counter-clockwisely and call a box $B\in C(n)$ vertical (resp.
horizontal) if $C$ enters $B$ in the vertical (resp. horizontal) side of $B$.
Each non-exceptional box in $C(n)$ is either vertical or horizontal, and if a
box in $C(n)$ is both then it is exceptional. For any two angles
$\alpha\leq\beta$ in $\mathbb{R}$ with $\beta-\alpha\leq 2\pi$ we define

$C_{\alpha,\,\beta}\subset C$

to be the open arc of points $p\in C$ such that the length of the arc
$P_{0}p\subset C$, where $P_{0}:=(1,0)$, that goes counter-clockwisely from
$P_{0}$ to $p$ lies, modulo $2\pi$, in the interval $(\alpha,\beta)$. We
consider the approximation $C_{\alpha,\beta}(n)$ of the arc $C_{\alpha,\beta}$
by boxes and denote by $C_{\alpha,\beta,\mathrm{v}}(n)$ (resp.
$C_{\alpha,\beta,\mathrm{h}}(n)$) the subset of vertical (resp. horizontal)
boxes.

###### Proposition 2.3

. For any fixed $\alpha,\beta\in\mathbb{R}$ with $0\leq\alpha\leq\beta\leq\pi$
and all $n\in\mathbb{N}$, we have the estimate

$|C_{\alpha,\,\beta,\,\mathrm{v}}(n)|=[-\cos t]_{\alpha}^{\beta}\cdot
n+O(1)\;.$

For $\pi\leq\alpha\leq\beta\leq 2\pi$ this estimate holds with the function
$\cos t$. For any fixed $\alpha,\beta\in\mathbb{R}$ with
$-\pi/2\leq\alpha\leq\beta\leq\pi/2$ and all $n\in\mathbb{N}$, we have the
estimate

$|C_{\alpha,\,\beta,\,\mathrm{h}}(n)|=[\sin t]_{\alpha}^{\beta}\cdot
n+O(1)\;.$

For $\pi/2\leq\alpha\leq\beta\leq 3\pi/2$ this estimate holds with the
function $-\sin t$.

Proof. Let an $n\in\mathbb{N}$ and $\alpha,\beta\in[0,\pi]$ with
$\alpha\leq\beta$ be given. Let $I\subset[-n+1,n]_{\mathbb{Z}}$ be the
interval of the integers $i$ such that the line $x=i/n$ intersects the arc
$C_{\alpha,\beta}$. Since the boxes $B$ in $C_{\alpha,\beta,\mathrm{v}}(n)$
1-1 correspond, except possibly for the rightmost one, to $I$ (the right side
of the box $B$, where $C_{\alpha,\beta}$ enters $B$, lies on the line $x=i/n$,
$i\in I$), we have the bounds

$|I|\leq|C_{\alpha,\,\beta,\,\mathrm{v}}(n)|\leq|I|+1\;.$

For $i\in[-n,n]_{\mathbb{Z}}$, let $\alpha_{i}\in[0,\pi]$ be the angle at the
vertex $s=(0,0)$ in the triangle $p_{i}\,s\,(1,0)$, where $p_{i}\in C$ is the
intersection of the upper closed semicircle of $C$ with the line $x=i/n$.
Clearly, $i/n=\cos\alpha_{i}$. Thus

$|I|=\sum_{i\in I}1=n\sum_{i\in
I}(\cos\alpha_{i+1}-\cos\alpha_{i})=n(\cos\alpha_{k+1}-\cos\alpha_{j})\;,$

where $j=\min I$ and $k=\max I$. Since $\cos\alpha\in[k/n,(k+1)/n]$, the
numbers $\cos\alpha$ and $\cos\alpha_{k+1}$ differ by at most $1/n$. Similarly
for $\cos\beta$ and $\cos\alpha_{j}$. Thus by the above displayed bounds,

$|C_{\alpha,\,\beta,\,\mathrm{v}}(n)|=n(\cos\alpha-\cos\beta)+O(1)\;,$

which is the first stated estimate. For the other range of $\alpha$ and
$\beta$ and for the second stated estimate for horizontal boxes we argue
similarly. $\Box$

###### Proposition 2.4

.  For any fixed $\varepsilon>0$, $\alpha,\beta\in\mathbb{R}$ with
$0\leq\alpha\leq\beta\leq\pi/2$ and all $n\in\mathbb{N}$, we have the estimate

$|C_{\alpha,\,\beta}(n)|=[\sin t-\cos t]_{\alpha}^{\beta}\cdot
n+E_{n}-\Delta_{n}\;,$

where

$E_{n}=O(1)\;\text{ and }\;0\leq\Delta_{n}\leq\min\big{(}[\sin t-\cos
t]_{\alpha}^{\beta}\cdot n+E_{n},\,O_{\varepsilon}(n^{\varepsilon})\big{)}\;.$

In the other three quadrants of $C$ this estimate holds with the appropriate
signs of $\sin t$ and $\cos t$, $+$ if the function increases and $-$ if it
decreases.

Proof. This follows from the previous proposition and Proposition 2.1: $[\sin
t-\cos t]_{\alpha}^{\beta}\cdot n+E_{n}$ is the sum of estimates of the
numbers of vertical and horizontal boxes in $C_{\alpha,\,\beta}(n)$ and
$\Delta_{n}$ is the number of the boxes which are both vertical and horizontal
(and hence are exceptional). $\Box$

###### Corollary 2.5

.  For every $n\in\mathbb{N}$ the cardinality of the approximate unit circle
$C(n)$ equals

$|C(n)|=8n-4\sum_{\begin{subarray}{c}d\,|\,n^{2}\\\
d=2e+1\end{subarray}}(-1)^{e}\;.$

Proof. We count the boxes in $C(n)$ more carefully:

$\displaystyle|C(n)|$ $\displaystyle=$ $\displaystyle|C_{\mathrm{v}}(n)\cup
C_{\mathrm{h}}(n)|=|C_{\mathrm{v}}(n)|+|C_{\mathrm{h}}(n)|-|C_{\mathrm{v}}(n)\cap
C_{\mathrm{h}}(n)|$ $\displaystyle=$
$\displaystyle|C_{\mathrm{v}}(n)|+|C_{\mathrm{h}}(n)|-|C_{\mathrm{e}}(n)|\;,$

where $C_{\mathrm{v}}(n)$ (resp. $C_{\mathrm{h}}(n)$) is the set of vertical
(resp. horizontal) boxes in $C(n)$ and $C_{\mathrm{e}}(n)$ is the set of those
boxes in $C(n)$ in which $C$ enters (counter-clockwisely) in a vertex. Now
$|C_{\mathrm{v}}(n)|=4n$ because exactly two boxes in $C_{\mathrm{v}}(n)$
correspond to each of the $2n-1$ vertical lines $x=i/n$ with
$i=-n+1,-n+2,\dots,n-1$ (the point of entrance of $C$ in the box lies on the
line), only one box in $C_{\mathrm{v}}(n)$ corresponds to each of the two
vertical lines $x=-1$ and $x=1$, and this exhausts all boxes in
$C_{\mathrm{v}}(n)$. By symmetry, also $|C_{\mathrm{h}}(n)|=4n$. Clearly,
$C_{\mathrm{v}}(n)\cap C_{\mathrm{h}}(n)=C_{\mathrm{e}}(n)$ and
$C_{\mathrm{e}}(n)$ bijectively corresponds to the solutions
$(i,j)\in\mathbb{Z}^{2}$ of the equation $i^{2}+j^{2}=n^{2}$. Thus
$|C_{\mathrm{e}}(n)|=r_{2}(n^{2})$. Using the formula in [10] for the number
$r_{2}(m)$ of these solutions, we obtain the stated formula. $\Box$

As we already mentioned, at the present state of knowledge it is not known how
to use this formula to compute $|C(n)|$ in time polynomial in $\log n$.

Let $B^{*}$ be a fixed box, say $B^{*}=B_{0,0}(n)$, which will serve as an
origin. An even more “physical” approximations of $C$ by boxes would be

$C[n]=\\{B(n)\;|\;\exists\,ab:\ a\in(B^{*})^{o}\wedge b\in
B(n)^{o}\wedge|ab|=1\\},\ n\in\mathbb{N}\;.$

We hope to investigate them in [14]. Here we confine to the approximations
$C(n)$ which are easier to handle.

###### Corollary 2.6

.  For any fixed $\varepsilon>0$ and all $n\in\mathbb{N}$,

$|C(n)\times C(n)|=64n^{2}+O_{\varepsilon}(n^{1+\varepsilon})\;.$

For $D=\\{(x,y)\in\mathbb{R}^{2}\;|\;x^{2}+y^{2}<1\\}$ one has that

$|D(n)|=\pi n^{2}+\Delta_{n}\;,$

where $0\leq\Delta_{n}\leq|C(n)|<8n$. Thus there are many more chords (ordered
or unordered) in $C(n)$ than boxes in $D(n)$.

Proof. The first formula follows from Corollary 2.5 and from the bound
$\tau(m)=O_{\varepsilon}(m^{\varepsilon})$ on the number of divisors function
([10]), already mentioned in the proof of Proposition 2.1. To prove the second
formula, we consider besides $D(n)$ also the set $U(n)$ of boxes $B(n)$ such
that $B(n)^{o}\subset D$, thus $U(n)\subset D(n)$. Since $\bigcup
U(n)\subset\overline{D}\subset\bigcup D(n)$, the area of the disc
$\overline{D}$ equals $\pi$ and $D(n)\setminus U(n)=C(n)$, we have the
relations

$\frac{|U(n)|}{n^{2}}\leq\pi\leq\frac{|D(n)|}{n^{2}}\;\text{ and
}\;|D(n)|-|U(n)|=|C(n)|\;.$

By this and Corollary 2.5 we get the second formula. $\Box$

Bertrand’s Solution 3 therefore cannot be implemented for $C(n)$ and $D(n)$,
there are not enough midpoints for the chords.

In the next corollary the indices $1$ and $3$ refer to the arc between the
hours $1$ and $3$ on a standard round clock face.

###### Corollary 2.7

.

$\mathrm{Pr}_{1,\,3,\,\Box}:=\lim_{n\to\infty}\frac{|C_{0,\,\pi/3}(n)|}{|C(n)|}=\frac{1+\sqrt{3}}{16}=0.17075\dots\;.$

Proof. We set $\varepsilon:=1/2$ and get by Proposition 2.4 and by the formula
in Corollary 2.5 that

$\mathrm{Pr}_{1,\,3,\,\Box}=\lim_{n\to\infty}\frac{[\sin t-\cos
t]_{0}^{\pi/3}\cdot n+O(\sqrt{n})}{8n+O(\sqrt{n})}=\frac{1+\sqrt{3}}{16}\;.$

$\Box$

This is inspired by [2, Chapter 1.3] where exposition of Bertrand’s paradox
starts with the remark that the second hand on clocks appears between hours
$1$ and $3$ with probability $\frac{1}{6}=0.16666\dots\;$. In the world of
$C(n)$ clocks spend slightly more time between these two hours.

##  3 Counting long physical chords in $C(n)$

In this section we compute $\mathrm{Pr_{Be,\Box}}$ by counting the long chords

$\mathrm{Ch}(n)=\\{(B,\,B^{\prime})\in C(n)^{2}\;|\;\exists\,ab:\ a\in
B^{o}\wedge b\in(B^{\prime})^{o}\wedge|ab|>\sqrt{3}\\}$

and then computing the limit

$\mathrm{Pr_{Be,\,\Box}}=\lim_{n\to\infty}\frac{|\mathrm{Ch}(n)|}{|C(n)|^{2}}\;.$

We identify $C$ with the interval $[0,2\pi)$ and for $a\in C$ define the arc
antipodal to $a$ as

$\\{a\\}^{\mathrm{ap}}=(a+2\pi/3,\,a+4\pi/3)\ (\mathrm{mod}\ 2\pi)\
(\subset[0,\,2\pi))\;.$

This definition is motivated by the equivalence that a chord $ab$ with $a,b\in
C$ has length $|ab|>\sqrt{3}$ if and only if $b\in\\{a\\}^{\mathrm{ap}}$.
Bertrand’s Solution 1 that $\mathrm{Pr_{Be}}=\frac{1}{3}$ follows at once. For
any $X\subset C$ we define

$X^{\cap}=\bigcap_{a\in X}\\{a\\}^{\mathrm{ap}}\;.$

Thus

$a\in X\subset C\wedge b\in X^{\cap}\Rightarrow|ab|>\sqrt{3}\;.$

For $a\in C$ and $d\in(0,\sqrt{3}]$ we define more generally

$\\{a\\}^{\mathrm{ap},\,d}=\\{b\in C\;|\;|ab|>d\\}$

to be the arc of the other endpoints $b$ of chords $ab$ with length $|ab|>d$.
For any $X\subset C$ we also define

$X^{\cup}=\bigcup_{a\in X}\\{a\\}^{\mathrm{ap}}\;\text{ and
}\;X^{\cup,\,d}=\bigcup_{a\in X}\\{a\\}^{\mathrm{ap},\,d}.$

Clearly, $X^{\cap}\subset X^{\cup}\subset X^{\cup,d}$ and

$a\in X\subset C\wedge b\in C\wedge|ab|>d\Rightarrow b\in X^{\cup,\,d}\;.$

###### Theorem 3.1

.  The limit proportion of long chords among all chords in the approximate
unit circle $C(n)$ is

$\mathrm{Pr_{Be,\,\Box}}=\lim_{n\to\infty}\frac{|\mathrm{Ch}(n)|}{|C(n)|^{2}}=\frac{1+\sqrt{3}}{8}-\frac{\pi(2-\sqrt{3})}{96}=0.33273\dots\;.$

Proof. We split the computational proof in seven steps.

1\. Deducing and proving equation (3) below. Let $n\in\mathbb{N}$. For
$j\in[4]$ we define

$\mathrm{Ch}_{j\mathrm{q}}(n)=\\{(B,\,B^{\prime})\in\mathrm{Ch}(n)\;|\;B\in
C_{(j-1)\pi/2,\,j\pi/2}(n)\\}$

to be the set of long chords $(B,B^{\prime})$ with $B$ in the $j$-th quadrant
of $C(n)$. The sets $\mathrm{Ch}_{j\mathrm{q}}(n)$, $j\in[4]$, partition
$\mathrm{Ch}(n)$ and, by symmetry, have equal cardinalities. Thus

$|\mathrm{Ch}(n)|=4\cdot|\mathrm{Ch}_{1\mathrm{q}}(n)|$ (1)

and it suffices to count $\mathrm{Ch}_{1\mathrm{q}}(n)$. We employ a function
$k=k(n)\in\mathbb{N}$, $n\in\mathbb{N}$ and $k\leq n$, such that $k\to\infty$
with $n$ but more slowly than $n$; we specify it at the end. We assume that
$n>6$ and $k<n/6$ and refer to it as the basic assumption on $n$ and $k$. We
divide $C$ in $12k$ equally long subarcs

$C_{i}=C_{(i-1)\gamma,\,i\gamma},\;\text{ where }\;i\in[12k]\;\text{ and
}\;\gamma=\gamma(k)=\pi/6k\;.$

We describe the sets $C_{i}^{\cap}(n)$, $i\in[3k]$, and their partitions in
quadrants — these are needed because the estimates of Proposition 2.4 apply
only within a single quadrant.

For $i\in[k]$ we set

$I_{1}=I_{1}(i,\,k)=[4k+i+1,\,6k]_{\mathbb{Z}}\;\text{ and
}\;J_{1}=J_{1}(i,\,k)=[6k+1,\,8k+i-1]_{\mathbb{Z}}\;.$

For $i\in[k+1,2k]_{\mathbb{Z}}$ we similarly set

$I_{2}=[4k+i+1,\,6k]_{\mathbb{Z}},\ J_{2}=[6k+1,\,9k]_{\mathbb{Z}}\;\text{ and
}\;K_{2}=[9k+1,\,8k+i-1]_{\mathbb{Z}}\;,$

and for $i\in[2k+1,3k]_{\mathbb{Z}}$ we set

$I_{3}=[4k+i+1,\,9k]_{\mathbb{Z}}\;\text{ and
}\;J_{3}=[9k+1,\,8k+i-1]_{\mathbb{Z}}\;.$

We set $K_{1}=K_{3}=\emptyset$. For $j\in[3]$ and
$i\in[(j-1)k+1,jk]_{\mathbb{Z}}$ we set

$L(i,k)=I_{j}\cup J_{j}\cup K_{j}=[4k+i+1,\,8k+i-1]_{\mathbb{Z}}$

and

$L^{\prime}(i,k)=[4k+i,\,8k+i]_{\mathbb{Z}}\;.$

For any set $I\subset[12k]$ we denote

$C_{I}=\bigcup_{i\in I}C_{i}\subset C\;.$

Thus $C_{I}(n)=(\bigcup_{i\in I}C_{i})(n)=\bigcup_{i\in I}C_{i}(n)\subset
C(n)$. It follows from the above definitions that for every $i\in[3k]$,

$C_{i}^{\cap}(n)=C_{L(i,k)}(n)\;\text{ and
}\;C_{i}^{\cup}(n)=C_{L^{\prime}(i,k)}(n)\;.$ (2)

(To see this, note that if $I\subset C$ is an open interval and $\overline{I}$
is its closure, then $I(n)=\overline{I}(n)$.)

We claim that for all $n,k\in\mathbb{N}$ satisfying the basic assumption,

$\displaystyle|\mathrm{Ch}_{1\mathrm{q}}(n)|$ $\displaystyle=$
$\displaystyle\sum_{i=1}^{3k}|C_{i}(n)|\cdot|C_{L(i,\,k)}(n)|+O(kn)+O(n^{2}/k)\;.$
(3)

To prove it, we first consider chords $(B,B^{\prime})$ counted in the sum of
products on the right-hand side of equation (3): there is an $i\in[3k]$ such
that $B\in C_{i}(n)$ and $B^{\prime}\in C_{L(i,k)}(n)$. Since
$C_{L(i,k)}(n)=C_{i}^{\cap}(n)$ (by equations (2)), it follows that
$(B,B^{\prime})\in\mathrm{Ch}_{1\mathrm{q}}(n)$ and these chords are counted
also in the left-hand side. But there is an overcount, a chord
$(B,B^{\prime})$ is counted in the right-hand side more than once iff $B\in
C_{i}(n)\cap C_{i^{\prime}}(n)$ for some $i<i^{\prime}$ in $[3k]$. This may
happen only if $i^{\prime}=i+1$ because $\gamma>\pi/n$ (the maximum length of
a circular arc contained in a $1\times 1$ square is $\pi$). Since

$|C_{i}(n)\cap C_{i+1}(n)|=|\\{i\gamma\\}(n)|\leq 1\;\text{ and
}\;|C_{L(i,\,k)}(n)|=O(k\cdot n/k)=O(n)$

by Proposition 2.4, this overcount is bounded by the term $O(kn)$.

We also have to consider chords
$(B,B^{\prime})\in\mathrm{Ch}_{1\mathrm{q}}(n)$ counted in the left-hand side
of equation (3) but not in the right-hand side. This means that, setting
$d=\sqrt{3}-2\sqrt{2}/n$ (each box has diameter $\sqrt{2}/n$), there is an
$i\in[3k]$ such that $B\in C_{i}(n)$ but

$\displaystyle B^{\prime}\in C^{\cup,\,d}_{i}(n)\setminus C^{\cap}_{i}(n)$
$\displaystyle=$ $\displaystyle(C^{\cup,\,d}_{i}(n)\setminus
C^{\cup}_{i}(n))\cup(C^{\cup}_{i}(n)\setminus C^{\cap}_{i}(n))$
$\displaystyle=:$ $\displaystyle X_{i,\,n}\cup Y_{i,\,n}\;.$

For every $i\in[3k]$ and every $n>6$,

$|X_{i,\,n}|=O(1)\;\text{ and }\;|Y_{i,\,n}|=O(n/k)\;.$

The first bound follows from Proposition 2.4 because $|X_{i,n}|$ is at most
the number of boxes whose interiors intersect $C_{i}^{\cup,d}\setminus
C_{i}^{\cup}$, which consists of two arcs of $C$ with lengths
$O(\sqrt{3}-d)=O(1/n)$. The second bound also follows from Proposition 2.4
because $|Y_{i,\,n}|$ is at most the number of boxes whose interiors intersect
two arcs of $C$ with lengths $\gamma=O(1/k)$ (see equalities 2). The first
bound is subsumed in the second because $k\leq n$. Again by Proposition 2.4,
$|C_{i}(n)|=O(n/k)$. Thus the surplus of the left-hand side in (3) over the
sum of products in the right-hand side is compensated by the term

$O(k\cdot(n/k)^{2})=O(n^{2}/k)\;.$

Equality (3) is therefore proven.

2\. Symmetries $S_{1}(n)=S_{3}(n)$ and
$S_{2}^{\prime}(n)=S_{2}^{\prime\prime\prime}(n)$ in (11). For $j\in[3]$ we
define

$\displaystyle S_{j}(n)$ $\displaystyle=$
$\displaystyle\sum_{i=(j-1)k+1}^{jk}|C_{i}(n)|\cdot|C_{L(i,k)}(n)|$ (5)
$\displaystyle=$
$\displaystyle\sum_{i=(j-1)k+1}^{jk}|C_{i}(n)|\cdot\big{(}|C_{I_{j}(i,\,k)}(n)|+|C_{J_{j}(i,\,k)}(n)|+|C_{K_{j}(i,\,k)}(n)|\big{)}+$
$\displaystyle+\;O(n)\;,$

so

$|\mathrm{Ch}_{1\mathrm{q}}(n)|=\sum_{j=1}^{3}S_{j}(n)+O(kn)+O(n^{2}/k)\;.$
(6)

The error term $O(n)=O(k\cdot n/k)$ on the right-hand side of equation (5) —
recall that $|C_{i}(n)|=O(n/k)$ — compensates like earlier its overcount of
boxes in $(\dots+\dots+\dots)$ which occurs even though $I_{j}$, $J_{j}$ and
$K_{j}$ are disjoint. We define

$\displaystyle S_{2}^{\prime}(n)$ $\displaystyle=$
$\displaystyle\sum_{i=k+1}^{2k}|C_{i}(n)|\cdot|C_{I_{2}(i,\,k)}(n)|\;,$ (7)
$\displaystyle S_{2}^{\prime\prime}(n)$ $\displaystyle=$
$\displaystyle\sum_{i=k+1}^{2k}|C_{i}(n)|\cdot|C_{J_{2}(i,\,k)}(n)|\;\text{
and }$ (8) $\displaystyle S_{2}^{\prime\prime\prime}(n)$ $\displaystyle=$
$\displaystyle\sum_{i=k+1}^{2k}|C_{i}(n)|\cdot|C_{K_{2}(i,\,k)}(n)|\;.$ (9)

Hence

$S_{2}(n)=S_{2}^{\prime}(n)+S_{2}^{\prime\prime}(n)+S_{2}^{\prime\prime\prime}(n)+O(n)\;.$
(10)

We claim that for every $n\in\mathbb{N}$,

$S_{1}(n)=S_{3}(n)\;\text{ and
}S_{2}^{\prime}(n)=S_{2}^{\prime\prime\prime}(n)\;.$ (11)

This follows from the definition of these sums in equations (5), (7) and (9)
and from the involution $(x,y)\mapsto(y,x)$ of the plane. For every $i\in[k]$
it induces a bijection between

$C_{i}(n)\times C_{L(i,\,k)}(n)\;\text{ and }C_{3k-i+1}(n)\times
C_{L(3k-i+1,\,k)}(n)\;,$

and for every $i\in[k+1,2k]_{\mathbb{Z}}$ a bijection between

$C_{i}(n)\times C_{I_{2}(i,\,k)}(n)\;\text{ and }C_{3k-i+1}(n)\times
C_{K_{2}(3k-i+1,\,k)}(n)\;.$

3\. Expressing $S_{1}(n)$, $S_{2}^{\prime}(n)$ and $S_{2}^{\prime\prime}(n)$
by sine and cosine in (12)–(17). We substitute for each cardinality $|\cdots|$
on the right-hand sides of equations (5) with $j=1$, (7) and (8) the estimate
of Proposition 2.4. To this end we introduce some notation. Recall that for
$i\in[3k]$, $i\gamma=i\pi/6k$. We consider the functions

$f_{1}(t)=\sin t-\cos t,\ f_{2}(t)=-\sin t-\cos t,\ f_{3}(t)=-\sin t+\cos t$

and

$f_{4}(t)=\sin t+\cos\;;$

we use $f_{j}(t)$ in the $j$-th quadrant. We label quadrants counter-
clockwisely, starting with the first one where $x,y>0$. For any function
$f\colon[0,2\pi)=C\to\mathbb{R}$ and any interval
$I=[a,b]_{\mathbb{Z}}\subset[12k]$, $a,b\in[12k]$, we write

$[f]_{I}=\big{[}f\big{]}_{(a-1)\gamma}^{b\gamma}\;.$

Note that if $I=[a,b]_{\mathbb{Z}}$ is any of the above intervals
$I_{j},J_{j}$ and $K_{2}$, $j\in[3]$, then always $a-1\leq b$.

Thus using Proposition 2.4, for any fixed $\varepsilon>0$ and all
$k,n\in\mathbb{N}$ satisfying the basic assumption we get that

$\displaystyle S_{1}(n)$ $\displaystyle=$ $\displaystyle S_{1}\cdot
n^{2}+O_{\varepsilon}(kn^{1+\varepsilon})\;,$ (12) $\displaystyle
S_{2}^{\prime}(n)$ $\displaystyle=$ $\displaystyle S_{2}^{\prime}\cdot
n^{2}+O_{\varepsilon}(kn^{1+\varepsilon})\;\text{ and }$ (13) $\displaystyle
S_{2}^{\prime\prime}(n)$ $\displaystyle=$ $\displaystyle
S_{2}^{\prime\prime}\cdot n^{2}+O_{\varepsilon}(kn^{1+\varepsilon})\;,\;\text{
where}$ (14) $\displaystyle S_{1}$ $\displaystyle=$
$\displaystyle\sum_{i=1}^{k}[f_{1}(t)]_{(i-1)\gamma}^{i\gamma}\cdot\big{(}[f_{2}(t)]_{I_{1}}+[f_{3}(t)]_{J_{1}}\big{)}\;,$
(15) $\displaystyle S_{2}^{\prime}$ $\displaystyle=$
$\displaystyle\sum_{i=k+1}^{2k}[f_{1}(t)]_{(i-1)\gamma}^{i\gamma}\cdot[f_{2}(t)]_{I_{2}}\;\text{
and}$ (16) $\displaystyle S_{2}^{\prime\prime}$ $\displaystyle=$
$\displaystyle\sum_{i=k+1}^{2k}[f_{1}(t)]_{(i-1)\gamma}^{i\gamma}\cdot[f_{3}(t)]_{J_{2}}\;.$
(17)

4\. Computing the sum $S_{1}$ in (25). We evaluate sums $S_{1}$,
$S_{2}^{\prime}$ and $S_{2}^{\prime\prime}$ in equations (15), (16) and (17),
respectively. We begin with sum (15). For any real $x\geq 0$ we define
($i\in\mathbb{N}$)

$A(x)=\sum_{i\leq x}[f_{1}(t)]_{(i-1)\gamma}^{i\gamma}=f_{1}(\lfloor
x\rfloor\gamma)+1\;,$ (18)

with the empty sum defined as $0$. One has that

$[f_{2}(t)]_{I_{1}(i,\,k)}+[f_{3}(t)]_{J_{1}(i,\,k)}=2+g(t)_{t=i}$ (19)

where

${\textstyle g(t)=\sin(\frac{2\pi}{3}+\gamma t)+\cos(\frac{2\pi}{3}+\gamma
t)-\sin(\frac{4\pi}{3}+\gamma t-\gamma)+\cos(\frac{4\pi}{3}+\gamma
t-\gamma)\;.}$

Using in equation (15) expressions (18) and (19), we get by Proposition 2.2,
invoked on the second line, that

$\displaystyle S_{1}$ $\displaystyle=$
$\displaystyle\sum_{i=1}^{k}[f_{1}(t)]_{(i-1)\gamma}^{i\gamma}\cdot(2+g(i))=2[f_{1}(t)]_{0}^{\pi/6}+\sum_{i=1}^{k}[f_{1}(t)]_{(i-1)\gamma}^{i\gamma}\cdot
g(i)$ (20) $\displaystyle=$ $\displaystyle
3-\sqrt{3}+[A(t)g(t)]_{0}^{k}-\int_{0}^{k}A(t)\cdot
g^{\prime}(t)\;\mathrm{dt}$ $\displaystyle=$ $\displaystyle
3-\sqrt{3}+[f_{1}(\lfloor t\rfloor\gamma)\cdot
g(t)]_{0}^{k}-\int_{0}^{k}f_{1}(\lfloor t\rfloor\gamma)\cdot
g^{\prime}(t)\;\mathrm{dt}$ $\displaystyle=$ $\displaystyle
3-\sqrt{3}+\frac{1}{2}-\int_{0}^{k}f_{1}(\lfloor t\rfloor\gamma)\cdot
g^{\prime}(t)\;\mathrm{dt}$ $\displaystyle=$ $\displaystyle
3-\sqrt{3}+\frac{1}{2}-\int_{2\pi/3}^{5\pi/6}f_{1}(\lfloor(u-2\pi/3)/\gamma\rfloor\gamma)\cdot(\cos
u-\sin u)\;\mathrm{du}+$
$\displaystyle+\;\int_{4\pi/3}^{3\pi/2}f_{1}(\lfloor(u-4\pi/3)/\gamma\rfloor\gamma)\cdot(\cos
u+\sin u)\;\mathrm{du}\;.$

We evaluate the last two integrals with precision $O(1/k)$. We use the
formulas $\sin(x+y)=\sin x\cos y+\cos x\sin y$, $\cos(x+y)=\cos x\cos y-\sin
x\sin y$, $\sin(x+O(1/k))=\sin x+O(1/k)$, $\cos(x+O(1/k))=\cos x+O(1/k)$
(recall that $\gamma=\pi/6k$) and get that

$\displaystyle
f_{1}(\lfloor(u-2\pi/3)/\gamma\rfloor\gamma)=f_{1}(u-2\pi/3-O(1/k))$
$\displaystyle{\textstyle=\sin u\cdot(-\frac{1}{2})+\cos
u\cdot(-\frac{\sqrt{3}}{2})-\cos u\cdot(-\frac{1}{2})+\sin
u\cdot(-\frac{\sqrt{3}}{2})+O(1/k)}$
$\displaystyle={\textstyle-\frac{1+\sqrt{3}}{2}\sin u+\frac{1-\sqrt{3}}{2}\cos
u+O(1/k)}$ (21)

and

$\displaystyle
f_{1}(\lfloor(u-4\pi/3)/\gamma\rfloor\gamma)=f_{1}(u-4\pi/3-O(1/k))$
$\displaystyle{\textstyle=\sin u\cdot(-\frac{1}{2})+\cos
u\cdot\frac{\sqrt{3}}{2}-\cos u\cdot(-\frac{1}{2})+\sin
u\cdot\frac{\sqrt{3}}{2}+O(1/k)}$
$\displaystyle{\textstyle=\frac{\sqrt{3}-1}{2}\sin u+\frac{1+\sqrt{3}}{2}\cos
u+O(1/k)\;.}$ (22)

After substituting expansions (21) and (22) in the respective integrands in
equation (20) we see that the first integral in equation (20) is

${\textstyle-\int\sin u\cos
u\;\mathrm{du}+\frac{1+\sqrt{3}}{2}\int\sin^{2}u\;\mathrm{du}+\frac{1-\sqrt{3}}{2}\int\cos^{2}u\;\mathrm{du}+O(1/k)}\;,$

with $\int=\int_{2\pi/3}^{5\pi/6}$, and the second one is

${\textstyle\sqrt{3}\int\sin u\cos
u\;\mathrm{du}+\frac{\sqrt{3}-1}{2}\int\sin^{2}u\;\mathrm{du}+\frac{1+\sqrt{3}}{2}\int\cos^{2}u\;\mathrm{du}+O(1/k)}\;,$

with $\int=\int_{4\pi/3}^{3\pi/2}$. The former expression evaluates to (recall
that $\sin(2x)=2\sin x\cdot\cos x$ and
$\cos(2x)=\cos^{2}x-\sin^{2}x=1-2\sin^{2}x=2\cos^{2}x-1$)

$\displaystyle{\textstyle\frac{1}{4}[\cos(2u)]_{2\pi/3}^{5\pi/6}+\frac{1+\sqrt{3}}{8}[2u-\sin(2u)]_{2\pi/3}^{5\pi/6}+\frac{1-\sqrt{3}}{8}[2u+\sin(2u)]_{2\pi/3}^{5\pi/6}+}$
$\displaystyle+\;O(1/k)$ $\displaystyle{\textstyle=\frac{1}{4}\cdot
1+\frac{1+\sqrt{3}}{8}\cdot\frac{\pi}{3}+\frac{1-\sqrt{3}}{8}\cdot\frac{\pi}{3}+O(1/k)}$
$\displaystyle{\textstyle=\frac{1+\pi/3}{4}+O(1/k)}\;,$ (23)

and the latter to

$\displaystyle{\textstyle\frac{\sqrt{3}}{4}[-\cos(2u)]_{4\pi/3}^{3\pi/2}+\frac{\sqrt{3}-1}{8}[2u-\sin(2u)]_{4\pi/3}^{3\pi/2}+\frac{1+\sqrt{3}}{8}[2u+\sin(2u)]_{4\pi/3}^{3\pi/2}+}$
$\displaystyle+\;O(1/k)$
$\displaystyle{\textstyle=\frac{\sqrt{3}}{8}+\frac{\sqrt{3}-1}{8}\cdot\left(\frac{\pi}{3}+\frac{\sqrt{3}}{2}\right)+\frac{1+\sqrt{3}}{8}\cdot\left(\frac{\pi}{3}-\frac{\sqrt{3}}{2}\right)+O(1/k)}$
$\displaystyle{\textstyle=\frac{\pi\sqrt{3}}{12}+O(1/k)}\;.$ (24)

Substituting values (23) and (24) for the first and second integral in
equation (20), respectively, we therefore get that

$\displaystyle S_{1}$ $\displaystyle=$ $\displaystyle
3-\sqrt{3}+\frac{1}{2}-\frac{1+\pi/3}{4}+\frac{\pi\sqrt{3}}{12}+O(1/k)$ (25)
$\displaystyle=$
$\displaystyle\frac{39-12\sqrt{3}+\pi(\sqrt{3}-1)}{12}+O(1/k)\;.$

5\. Computing the sum $S_{2}^{\prime}$ in (29). We evaluate the second sum
(16). Now

$[f_{2}(t)]_{I_{2}(i,\,k)}=:1+h(t)_{t=i}$ (26)

where

$h(t)=\sin(2\pi/3+\gamma t)+\cos(2\pi/3+\gamma t)\;.$

Using in equation (16) expressions (18) and (26), we get by Proposition 2.2,
invoked on the second line, that

$\displaystyle S_{2}^{\prime}$ $\displaystyle=$
$\displaystyle\sum_{i=k+1}^{2k}[f_{1}(t)]_{(i-1)\gamma}^{i\gamma}\cdot(1+h(i))=[f_{1}(t)]_{\pi/6}^{\pi/3}+\sum_{i=k+1}^{2k}[f_{1}(t)]_{(i-1)\gamma}^{i\gamma}\cdot
h(i)$ (27) $\displaystyle=$
$\displaystyle\sqrt{3}-1+[A(t)h(t)]_{k}^{2k}-\int_{k}^{2k}A(t)\cdot
h^{\prime}(t)\;\mathrm{dt}$ $\displaystyle=$
$\displaystyle\sqrt{3}-1+[f_{1}(\lfloor t\rfloor\gamma)\cdot
h(t)]_{k}^{2k}-\int_{k}^{2k}f_{1}(\lfloor t\rfloor\gamma)\cdot
h^{\prime}(t)\;\mathrm{dt}$ $\displaystyle=$
$\displaystyle\sqrt{3}-\frac{3}{2}-\int_{k}^{2k}f_{1}(\lfloor
t\rfloor\gamma)\cdot h^{\prime}(t)\;\mathrm{dt}$ $\displaystyle=$
$\displaystyle\sqrt{3}-\frac{3}{2}-\int_{5\pi/6}^{\pi}f_{1}(\lfloor(u-2\pi/3)/\gamma\rfloor\gamma)\cdot(\cos
u-\sin u)\;\mathrm{du}\;.$

We substitute expansion (21) in the integrand in equation (27) and see that
the integral in equation (27) is again

${\textstyle-\int\sin u\cos
u\;\mathrm{du}+\frac{1+\sqrt{3}}{2}\int\sin^{2}u\;\mathrm{du}+\frac{1-\sqrt{3}}{2}\int\cos^{2}u\;\mathrm{du}+O(1/k)}$

but now $\int=\int_{5\pi/6}^{\pi}$. Now this expression evaluates to

$\displaystyle{\textstyle\frac{1}{4}[\cos(2u)]_{5\pi/6}^{\pi}+\frac{1+\sqrt{3}}{8}[2u-\sin(2u)]_{5\pi/6}^{\pi}+\frac{1-\sqrt{3}}{8}[2u+\sin(2u)]_{5\pi/6}^{\pi}+}$
$\displaystyle+\;O(1/k)$
$\displaystyle{\textstyle=\frac{1}{4}\cdot\frac{1}{2}+\frac{1+\sqrt{3}}{8}\cdot\left(\frac{\pi}{3}-\frac{\sqrt{3}}{2}\right)+\frac{1-\sqrt{3}}{8}\cdot\left(\frac{\pi}{3}+\frac{\sqrt{3}}{2}\right)+O(1/k)}$
$\displaystyle{\textstyle=\frac{\pi}{12}-\frac{1}{4}+O(1/k)}\;.$ (28)

Substituting value (28) for the integral in equation (27), we get that

${\textstyle
S_{2}^{\prime}=\sqrt{3}-\frac{3}{2}-\left(\frac{\pi}{12}-\frac{1}{4}\right)+O(1/k)=\sqrt{3}-\frac{5}{4}-\frac{\pi}{12}+O(1/k)\;.}$
(29)

6\. Computing the sum $S_{2}^{\prime\prime}$ in (30). We proceed to the third
and last sum (17). Since the interval $J_{2}(i,k)$ is independent of $i$, the
sum $S_{2}^{\prime\prime}$ can be computed easily and precisely:

$\displaystyle S_{2}^{\prime\prime}$ $\displaystyle=$
$\displaystyle\sum_{i=k+1}^{2k}[f_{1}(t)]_{(i-1)\gamma}^{i\gamma}\cdot[f_{3}(t)]_{J_{2}(i,\,k)}$
(30) $\displaystyle=$ $\displaystyle[\sin t-\cos t]_{\pi/6}^{\pi/3}\cdot[-\sin
t+\cos t]_{\pi}^{3\pi/2}$ $\displaystyle=$ $\displaystyle 2(\sqrt{3}-1)\;.$

7\. Conclusion of the computation. Using equations (1), (6), (10), (11), (12),
(13), (14), (25), (29), (30) and setting $k=k(n)=\lfloor\sqrt{n}\rfloor$, we
deduce that for any fixed $\varepsilon>0$ and all large $n\in\mathbb{N}$,

$\displaystyle|\mathrm{Ch}(n)|$ $\displaystyle=$ $\displaystyle{\textstyle
4\cdot\big{(}2\cdot\frac{39-12\sqrt{3}+\pi(\sqrt{3}-1)}{12}+2\cdot(\sqrt{3}-\frac{5}{4}-\frac{\pi}{12})+2(\sqrt{3}-1)\big{)}\cdot
n^{2}+}$ (31) $\displaystyle+\;O_{\varepsilon}(n^{3/2+\varepsilon})$
$\displaystyle=$ $\displaystyle\frac{24+24\sqrt{3}+\pi(2\sqrt{3}-4)}{3}\cdot
n^{2}+O_{\varepsilon}(n^{3/2+\varepsilon})\;.$

Finally, by equation (31) and Corollary 2.6 we see that

$\displaystyle\mathrm{Pr_{Be,\,\Box}}=\lim_{n\to\infty}\frac{|\mathrm{Ch}(n)|}{|C(n)|^{2}}$
$\displaystyle=$ $\displaystyle\frac{24(1+\sqrt{3})+\pi(2\sqrt{3}-4)}{3\cdot
64}$ $\displaystyle=$
$\displaystyle\frac{1+\sqrt{3}}{8}-\frac{\pi(2-\sqrt{3})}{96}=0.33273\dots\;.$

$\Box$

This is close to but different from Bertrand’s Solution 1
$\mathrm{Pr_{Be}}=\frac{1}{3}$.

##  4 Suggestions of a referee and concluding remarks

A referee, dismissive of the approach here, wrote the following.

$(\dots)$
“It is is easy to see that in the limit $n\to\infty$ this gives rise to the
following distribution on the circle: with the usual parametrization
$\varphi\mapsto e^{i\varphi}$, $\varphi\in[0,2\pi)$ of the unit circle, the
probability density function is

$f(\varphi)=\frac{|\sin(\varphi)|+|\cos(\varphi)|}{8}.$

In the limit problem, one needs to choose the two endpoints of the chord
independently, each according to this distribution. So the probability in
question is

$\int_{0}^{2\pi}\int_{2\pi/3}^{4\pi/3}f(x)f(x+y)\,\mathrm{d}x\mathrm{d}y.$

It is a straightforward (although not pleasent [sic]) task to compute this
integral.”
$(\dots)$
“I did not check the author’s computations (that take up roughly 12 pages) but
I am certain that they are unnecessarily complicated and the result could be
obtained in a few pages (by proving a weak convergence for the discrete
approximations and computing the above integral for the limit problem).”
$(\dots)$

In the sentence “It is is easy to see that $\dots$” we should get a credit and
Proposition 2.4 should be mentioned because the density $f(\varphi)$ follows
from the asymptotics in Proposition 2.4, not the other way around (as is much
more common in probabilistic computations). Were the values of the arithmetic
function $r_{2}(m)$ — see the proof of Proposition 2.1 — not small enough,
$f(\varphi)$ would be different, if it existed at all. The obvious bound
$r_{2}(m)=O(m^{1/2})$ does not suffice, nor even $r_{2}(m)=O(m^{1/4})$. In
reality it is not true that “It is is easy to see that $\dots$”.

MAPLE (2022 edition) cannot compute the above double integral in a closed
form, instead it expresses it, by computing the inner integral, in the form

$I:=\int_{0}^{2\pi}I(x)\,\mathrm{d}x$

with a complicated integrand $I(x)$ defined by considering various cases. More
precisely, this result was only obtained for some forms of the integrand
$f(x)f(x+y)$. For other trivial variations, MAPLE could not compute anything
or was running inconclusively for one hour before it was terminated. It
remains to be seen if this reflects some inherent complexity of the integral
or if it results from imperfection of algorithms in MAPLE.

Fortunately, MAPLE can compute approximations of the integral,

$I=0.33273773412864161039\dots\;.$

These decimal digits agree with our value of $\mathrm{Pr_{Be,\Box}}$. We have
therefore a non-rigorous “proof” of the fact that

$I=\frac{1+\sqrt{3}}{8}-\frac{\pi(2-\sqrt{3})}{96}\;.$

To make it rigorous would mean (i) to prove “a weak convergence for the
discrete approximations” or (ii) to compute the double integral in closed
form. We are less optimistic about the simplicity of a proof in (i) than the
referee but this can be decided only by providing a detailed and rigorous
proof in (i). We thank the referee for the suggested alternative approach to
asymptotics of sums counting long chords or similar quantities.

Returning to our view of Bertrand’s paradox, we suggest to restate the problem
and, especially, what it means to solve it in the following terms. One should
not be searching, like some analyst of old trying to find the true sum of the
series $1-1+1-1+1-1+\dots$, for the true probability of long chords in the
unit geometric circle.

> Instead, to resolve Bertrand’s paradox, one should devise a framework that
> preserves as much as possible from the spirit of the original intuitive
> problem (suppose we throw erratically, “randomly” sticks on a circle drawn
> by chalk on the sidewalk, what is the chance that $\dots$) and at the same
> time makes the solution unique.

The approximation by boxes in this article is a possible solution of this
metaproblem. Of course, one cannot get rid of non-uniqueness completely, it is
only shifted to the higher level.

To close, as for some ideas for further research we only mention treating the
Buffon needle problem in our framework ([14]). The method of discretization is
not new in geometric probability — this author learned it from his colleague
P. Valtr in [26, 27, 28], but its application here to Bertrand’s paradox is
new.

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Department of Applied Mathematics
Faculty of Mathematics and Physics
Charles University
Malostranské nám. 25
118 00 Praha
Czechia