In another post I already pointed out possible strength models for a Draughts program.
All empirical, and not based upon underlying mathematical principles.
However one can always start with a model, and check what can be derived from this.
Than practice will most likely reveal that the assumptions were wrong, so back to the drawing board.
So one can start with the next model: R = RMax * ( 1 - f(depth)*g(knowledge) )
RMax and R is rating and max-rating (could be in ELO)
f(depth) is a function which describes the impact of search (depth in ply), and f(0) = 1 and f(infinite) = 0
g(knowledge) is a function which describes the impact of knowledge (difficult to quantify), and g(0) = 1 and g(infinite) = 0
If the Rating function can be split this way remains to be seen, most likely a more complex relation holds.
In any case if depth or knowledge grow to infinite the max rating is reached.
If one studies engines with equal evaluation function, one could also write: R = Rmax * ( 1 - g0 * f(depth))
The rating difference between an engine which searches d ply and another engine which searches d-x ply can be expressed as:
deltaR = RMax * ( 1 - g0 * f(d) ) - RMax * ( 1 - g0 * f(d-x) )
deltaR = RMax * g0 * ( f(d-x) - f(d)
As we don't have a model (yet) which yields a function f(d), we could start with just choosing one.
The boundary conditions are: f(0) = 1 f(infinite) = 0.
A candidate could be (but there are zillions): f(d) = exp(d0 * d), with d0 negative.
deltaR = RMax * g0 * ( exp(d0 * (d-x)) - exp(d0 * d) )
deltaR = RMax * g0 * ( exp(-d0 * x) * exp (d0 * d) - exp (d0 * d) )
deltaR = RMax * g0 * ( exp(-d0 * x) - 1 ) * exp(d0*d)
In a previous experiment with x=4, the next exponential fit was found (preliminary results, as we need more data points in the tail):
deltaR = 2384.7 * exp (-0.198 * d)
I will do some tests with x=2, and see what happens (most likely back to the drawing board