Peter Furness, “Are leys due to chance?”

The Ley Hunter, first series, Vol. 2, No. 1 (1965)
Reprinted in: Paul Screeton, Quicksilver Heritage (1974), pp. 56–58

This article from the early days of the ley-hunting revival was the first to give a reasonably accurate theoretical estimate of the number of alignments to be expected by chance. Its author Peter Furness, now (2014) a professional statistician, was still at school at the time.

The article appeared in the short-lived first series of The Ley Hunter, edited by Philip Heselton. (The magazine was successfully revived by Paul Screeton in 1969.) Not having access to the original, I have used the text as reprinted in Quicksilver Heritage.

[Are leys due to chance?]

[by Peter Furness]

Since the early days of research, the ley hunter has never been absolutely sure that leys are not merely due to chance alignment. Some methods have been devised for showing that random points do not line up to the same degree as ley points – but these have always been experimental. Such methods are clumsy and inefficient for the simple reason that reliable results can only be obtained by repeating the experiment a large number of times.

In this article I shall only outline a theoretical solution to the problem – fuller details may be obtained by writing to me.

Consider a one inch Ordnance Survey map of area A square miles, upon which is drawn a thick line representing a ley of length y miles and breadth x miles. If a random point is dropped on the map, the probability that it will fall in the line will be the fraction (ƒ) of the area of the map that the line occupies.

i.e. Probability =ƒ=xy/A.

Similarly, the probability that m such points fall in the line is ƒ^m.

Now, if we scatter n points on the map, the probability that exactly m of these will fall in the line is given by:

P(n,m)=ƒ^m(1−ƒ)ⁿ^−m	(	n	)
		m

where (1−f)ⁿ^−m represents the probability that the remaining (n−m) points do not fall in the line, and ( n m ) is the number of ways of selecting m points out of n.

It can be shown that

(	n	)	=	n!
	m			m!(n−m)!

where n! denotes the product of all the numbers from n down to one; i.e. 3!=3×2×1=6.

If the line is first fixed on the map by two of the points (the method used in ley hunting), the probability of getting an exact alignment of m points becomes:

P(n,m)=	(	n−2	)	ƒ^m⁻²(1−ƒ)ⁿ^−m
		m−2

(Equation 1)

If there is a total of W alignments in the n points (including two-point lines) then the number N(n) of m-point lines to be expected by chance is given by:

N(m)=WP(n,m)

(Equation 2)

It can be shown that:

(

)

−

∑

ⁿ

[(

)

−1

]

P(n,m)W

₃

Rearranging:

(	n	)
	2

1+	∑	ⁿ	[(	m	)	−1	]	P(n,m)
		₃		2

(Equation 3)

If we assume that ley points are random, we can find n for a particular map of area A. If the length of the ley is y miles, what can we reasonably take as its width x? The ultimate minimum for this is that breadth of the pencil or ink line representing the ley, but inaccuracies in drawing, in symbolism on the map itself, make it reasonable to take x to be 1/100 of a mile or 17.6 yards.

The following are results of calculations performed using a typical one-inch Ordnance Survey map.

Counting gave the number of good ley points as 200. The area of land A of the map was 625 sq. mls. I took the length of a typical ley on this map to be 30 mls.

From equations 1 and 3, W=16500.

From equation 2	N(3)	=1570, i.e. 1570 3-points leys by pure chance.
similarly	N(4)	=72, i.e. 72 4-point leys by pure chance.
similarly	N(5)	=2.
similarly	N(6)	=0.05.
similarly	N(7)	=0.001.

These figures indicate that for the particular map used, any ley with less than six points could be expected by chance. The value of N(7) suggests that if leys are chance alignments, one would only find one seven-point ley in a collection of 1000 different maps!

Thus we can show with certainty that leys of more than a fixed number of points are unlikely to be due to chance.