Talk:Ley line/Ley lines and probability

An estimate of the probability of ley lines existing by chance
For those interested in the mathematics, the following is a very approximate estimate of the probability of "ley line"-like alignements, assuming a plane covered with uniformly distributed "significant" points.

Consider a set of n points in an area with approximate diameter d. Consider a valid line to be one where every point is within distance w/2 of the line (that is, lies on a straight track of width w).

Consider all the unordered sets (known as combinations) of k points from the n points, of which there are (see factorial for the notation used):


 * $$ \frac {n!} {(n-k)!k!} $$

What is the probability that any given set of points fits a ley line? Let's very roughly consider the line drawn through the "leftmost" and "rightmost" two points of the k selected points (for some arbitary left/right axis: we can choose top and bottom for the exceptional vertical case). These two points are by definition on this line. For each of the remaining k-2 points, the probability that the point is "near enough" to the line is roughly w/d.

So, the expected number of k-point ley lines is very roughly


 * $$ \frac {n!} {(n-k)!k!} \left({\frac{w}{d}}\right)^{k-2}$$

Proportionality in d
Now let's investigate the variation of value of the formula above in terms of d. For n >> k the formula above is very approximately


 * $$ \frac {n^k} {k!} \left({\frac{w}{d}}\right)^{k-2}$$

Now assume that area is equal to $$d^2$$, and say there is a density &alpha; of points such that $$n = \alpha d^2$$.

Then we have the expected number of lines equal to:


 * $$ \frac {\alpha^k d^{2k}} {k!} \left( {\frac{w}{d}} \right)^{k-2}$$

Gathering the terms in k we have an expected number of k-point lines in the area $$d^2$$ of:


 * $$d^{k+2} \frac {\alpha^k} {k!} w^{k-2}$$

Thus, contrary to intuition, the number of k-point lines expected from random chance increases much more than linearly with the size of the area considered.

Numerical results
Plugging in some typical values for the variables, and using the exact formula for combinations, rather than the approximation:
 * w = 50m, the width corresponding to a 1mm pencil line on a 50000:1 map
 * d = 100km
 * &alpha; = 5.5 &times; 10-8m-2 (50 points in a 30km square area)
 * k = 5

we get expected values of 942.83 4-point leys 51.48 5-point leys 2.34 6-point leys 0.09 7-point leys for this 100 km diameter area. Note that longer leys should in general include shorter leys as subsets.

Doubling the width of the area under consideration to 200km causes the number of expected lines for each value of k to explode:

60837.93 4-point leys 6680.00 5-point leys 610.94 6-point leys 47.87 7-point leys 3.28 8-point leys

Hence, the existence of long-distance ley line alignments in the English landscape should not surprise us. This result is even more extreme if scaled up to 400 km, the characteristic scale of the entire United Kingdom.