D = de-chi Z + X = P + Psi
a = proportion of crosses in MT
b = proportion of crosses in Delta P
p(.) = probability of dot
p(x.) = probability of crosses
p(.) in Delta(Psi)ij = b^2 + (1 - b^2)
p(.) in Delta(Psi)'ij = (1 - a) + [b^2+ (1 - b^2)]
It is assumed, for the foregoing relations, that b is the same
for eadh of the Psi wheels. Furthermore, a and b are Invariably selected
so that their product is aproximately equal to one-half. Therefore -
p(.) in Delta(Psi)'ij = b
By definition -
P(.) in Delta Pij = 1/2 + lambda
Therefore
p(.) in Delta Dij = (1/2 + lambda)b + (1/2 - lambda ) (1 - b)
= lambda(2b - 1)
6 Will send additional formulae in the near future. Even though
they represent nothing new it may be convenient to have them systernaticaily
cornpiled.