if the circle were fixed, and the band thrown on it at random.
Now let A (fig. 2) be a position of the random point; the
favourable cases are when HK, *the bisector of the band*, meets a circle,
centre A, radius ½*c*; and the whole number are when HK meets
a circle, centre O, radius *r* + ½*c*; hence the probability is

*p* = 2π ⋅ ½*c*2π(*r* + ½*c*) = *c**r* + ½*c*.

Fig. 2. |

This is constant for all positions of
A; hence, equating these two values of
*p*, the mean value required is
M(S) = *c*(2*r* + *c*)^{−1}π*r* ^{2}.

The mean value of the portion of
the *circumference* which falls on the band
is the same fraction *c*/(2*r* + *c*) of the
whole circumference.

If *any convex area* whose surface is
Ω and circumference L be thrown on
the band, instead of a circle, the mean
area covered is

M(S) = π*c*(L + π*c*)^{−1}Ω.

For as before, fixing the random point at A, the chance of a
random point in Ω falling on the band is *p* = 2π ⋅ ½*c*/L′, where L′
is the perimeter of a *parallel* curve to L, at a normal distance ½*c*
from it. Now

L′ = L + 2π ⋅ ½*c*.

∴ M(S)Ω = π*c*L + π*c*.

Fig. 3. |

84. Buffon's problem may be easily deduced in a similar manner.
Thus, if 2*r* = length of line, *a* = distance between
the parallels, and we conceive a circle (fig. 3) of
diameter *a* with its centre at the middle O of the
line,^{[1]} rigidly attached to the latter, and thrown
with it on the parallels, this circle must meet one
of the parallels; if it be thrown an infinite number
of times we shall thus have an infinite number of
chords crossing it at random. Their number is
measured by 2π ⋅ ½*a*, and the number which meet
2*r* is measured by 4*r*. Hence the chance that the line 2*r* meets one
of the parallels is p = 4*r*/π*a*.

Fig. 4. |

85. To investigate the probability that the inclination of the line joining any two points in a given convex area Ω shall lie within given limits. We give here a method of reducing this question to calculation, for the sake of an integral to which it leads, and which is not easy to deduce otherwise.

First let one of the points A (fig. 4)
be fixed; draw through it a chord PQ = C,
at an inclination θ to some fixed line; put
AP = *r*, AQ = *r′*; then the number of
cases for which the direction of the line
joining A and B lies between θ and θ + *d*θ
is measured by ½(*r*^{2} + *r′*^{2})dθ.

Now let A range over the space between PQ and a parallel chord
distant *dp* from it, the number of cases for which A lies in this
space and the direction of AB from θ to θ + *d*θ is (first considering
A to lie in the element *drdp*)

Let *p* be the perpendicular on C from a given origin O, and
let ω be the inclination of *p* (we may put *d*ω for *d*θ), C will be a
given function of *p*, ω; and, integrating first for ω constant, the
whole number of cases for which ω falls between given limits ω′, ω″ is

;

the integral ∫C^{3}*dp* being taken for all positions of C between two
tangents to the boundary parallel to PQ. The question is thus
reduced to the evaluation of this double integral, which, of course,
is generally difficult enough; we may, however, deduce from it a
remarkable result; for, if the integral ⅓∬C^{3}*dpd*ω be extended to
all possible positions of C, it gives the whole number of pairs of
positions of the points A, B which lie inside the area; but this
number is Ω^{2}; hence

∬C^{3}*dpd*ω = 3Ω^{2};

the integration extending to all possible positions of the chord C,—its
length being a given function of its co-ordinates *p*, ω.^{[2]}

Cor. Hence if L, Ω be the perimeter and area of any closed
convex contour, the mean value of the
cube of a chord drawn across it at random
is 3Ω^{2}/L.

Fig. 5. |

86. Let there be any two convex boundaries
(fig. 5) so related that a tangent at
any point V to the inner cuts off a constant
segment S from the outer (*e.g.* two concentric
similar ellipses); let the annular area between
them be called A; from a point X taken at
random on this annulus draw tangents XA,
XB to the inner. The mean value of the
arc AB, M(AB) = LS/A, L being the whole
length of the inner curve ABV.

The following lemma will first be proved:—

Fig. 6. |

If there be any convex arc AB (fig. 6), and if N_{1} be (the measure
of) the number of random lines which meet
it once, N_{2} the number which meet it
twice,

2 arc AB = N_{1} + 2N_{2}.

For draw the chord AB; the number of
lines meeting the convex-figure so formed is N_{1} + N_{2} = arc + chord
(the perimeter); but N_{1} = number of lines meeting the chord =
2 chord;

∴ 2 arc + N_{1} = 2N_{2} + 2N_{2}, ∴ 2 arc = N_{1} + 2N_{2}.

Now fix the point X, in fig. 5, and draw XA, XB. If a random
line cross the boundary L, and *p*_{1} be the
probability that it meets the arc AB once,
*p*_{2} that it does so twice,

2AB/L = *p*_{1} + 2*p*_{2};

and if the point X range all over the
annulus, and *p*_{1}, *p*_{2} are the same probabilities
for all positions of X,

2M(AB)/L = *p*_{1} + 2*p*_{2}.

Fig. 7. |

Let now IK (fig. 7) be any position of
the random line; drawing tangents at I, K,
it is easy to see that it will cut the arc AB twice when X is in the
space marked α, and once when X is in either space marked β;
hence, for this position of the line, *p*_{1} + 2*p*_{2} = 2(α + β)/A = 2S/A,
which is constant; hence M(AB)/L = S/A.

Hence the mean value of the arc is the same fraction of the perimeter that the constant area S is of the annulus.

If L be not related as above to the outer boundary, M(AB)/L = M(S)/A, M(S) being the mean area of the segment cut off by a tangent at a random point on the perimeter L.

The above result may be expressed as an integral. If *s* be the
arc AB included by tangents from any point (*x*, *y*) on the annulus,

∬*sdxdy* = LS.

It has been shown (*Phil. Trans.*, 1868, p. 191) that, if θ be the
angle between the tangents XA, XB,

∬θ*dxdy* = π(A − 2S).

The mean value of the tangent XA or XB may be shown to be M(XA) = SP/2A, where P = perimeter of locus of centre of gravity of the segment S.

87. When we go on to species of three dimensions further speculative
difficulties occur. How is a random line through a given
point to be defined? Since it is usual to define a vector by two
angles (viz. φ the angle made with the axis X by a vector *r* in the
plane XY, and θ (or ½π − θ) the angle made by the vector ρ with *r*
in the plane containing both ρ and *r* and the axis Z) it seems natural
to treat the angles φ and θ as the equiprobable variables. In other
words, if we take at random any meridian on the celestial globe
and combine it with any right ascension the vector joining the centre
to the point thus assigned is a random line.^{[3]} It is possible that
for some purposes this conception may be appropriate. For many
purposes surely it is proper to assume a more symmetrical distribution
of the terminal points on the surface of a sphere, a distribution
such that each element of the surface shall contain an approximately
equal number of points. Such an assumption is usually
made in the kinetic theory of molecules with respect to the direction
of the line joining the centres of two colliding spheres in a “molecular
chaos.”^{[4]} It is safe to say with Czuber, “No discussion can
remove indeterminateness.” Let us hope with him that “though
this branch of probability can for the present claim only a theoretic
interest, in the future it will perhaps also lead to practical results.”^{[5]}

88. *Illustrations of probability and expectation*.—The close relation
between probability and expectation is well 'illustrated by geometrical
examples. As above stated, when a given space S is
included within a given space A, if *p* is the probability that a point

- ↑ The line might be anywhere within the circle without altering the question.
- ↑ This integral was given by Morgan Crofton in the
*Comptes**rendus*(1869), p. 1469. An analytical proof was given by Serret,*Annales scient. de l'école normale*(1869), p. 177. - ↑ Cf. Bertrand,
*op. cit.*§ 135. - ↑ See
*e.g.*Watson,*Kinetic Theory of Gases*, p. 2; Tait,*Trans.**Roy. Soc., Edin.*(1888), xxxiii. 68. - ↑
*Wahrscheinlichkeitstheorie*, p. 64.