73 THE ESSEX NATURALIST
introduced by regarding (n—1) and (n—2) as equal to (n).
Approximately, then, M3 = 1/2 R n2.
By similar reasoning it can be shown that M4 = 1/2R2 n2,
and in general, Mp = 1/2n2 • Rp-2 approximately where p is the
order of the alignment.
However, it will be observed that if a third spot should be
on one of the strips joining two other spots, the total number
of strips available for possible alignments will be reduced by
two; similarly, four spots on a strip reduce the total number
of strips by five. In general, the reduction is given by PC2—1
or [1/2p(p—1)—1]. where p is the order of alignment. A
correction must therefore be made to the probability ex-
pression given above.
Let m3, m4, ms be the true probable number of alignments
of the orders shown by the subscripts, then
m3 =R (1/2n2—2m2—5m4—9m5—14m6. . . .)
and similarly for m4 and m5, and so on.
It can be shown that the following approximate algebraic
solution of these equations is adequate for the values of R,
n and m encountered in these spot-alignment problems.
In order to allow for all the higher order alignments, we
substitute PC2 for (pC2—1) as the correction factor :
Approximately, then, m3 = R (1/2n2—3m3)
or m, (1 + 3R)=R. 1/2n2
therefore, m,
Similarly, m4
and, in general, mp
where m = probable number of alignments each having p
spots thereon.
s = spot diameter.
d = diameter of circular area or side of square or mean
side of rectangle.
n = total number of spots.
and
These formulae have been used to calculate the probabilities
of random alignments given in the body of the paper.