ALIGNMENTS OP ANCIENT SITES IN ESSEX 75
Where A is the angle made with the tangent, the length of in-
tercept is d sin A. Integrating between 0 and π/2, we obtain
2d/π for the mean value of L, which reduces to L = 0.637 d.
The calculation of the mean length of intercept within
the boundaries of a square area is much more complicated,
but the resulting integral can be solved and gives a value
L=0.785 d, where in this case d is the length of a side. The
case of a rectangular area, such as the full sheet of an O.S.
Map, is more difficult still, and there is no unique solution,
but the mean of the long and short sides treated as the sides
of a square gives a value sufficiently accurate for the purposes
of this enquiry. In fact, the cases of rectangle, square and
circle, taking into account the differing areas, can be re-
lated quite simply.

Substituting the mean value of L= 2d/π in the case of a
circular area, we have: Probability of 3-spot align-
ment
In the case of a square area of side d, or a rectangle of
mean side-length d: Probability of 3-spot alignment =
Since the numerical coefficients in the two cases are so
nearly equal to 0.8 or 4/5, it will be seen that areas of all
three shapes can be dealt with in a single formula without
introducing an error of more than 2 per cent in any case.
We then have the simple probability of a 3-spot align-
ment :
It will be convenient to let the factor
and to designate the various simple probabilities by M3, M4,
M5 and so on, the subscript denoting the order of alignment,
i.e., the number of spots aligned. The number n is usually
large, of the order of 50, and no very great error will be