OK, so what should W% look like? Provided that the Heaviside function H, matches the condition H(0) = 1/2 (which most of the common examples do), we can use the following expressions which are certain to fit the three fixed points.
R = RS+RA
X = (RS-RA) / R
W% = ( H(X) + H(1) - 1 ) / ( 2 * H(1) - 1 )
From here, the approach is pretty straight forward: choose a version of the Heaviside function for H, find the appropriate coefficients, and go. As an exercise, it is worth confirming that very small coefficients do ensure that W% is approximately linear (specifically (1+x) / 2 ).
The first order approximations I've experimented with get within a game or two of the right answer; second order gets a faction of a game, third order a smaller fraction - not big enough to bother with.
In a 9.7 run/game context, we get
W%
= ( -1.01111) * ( .5 * ERFC( 1.558019 X + 0.240296 X ^ 3 ) - 0.99451 )
This is accurate to about 1/8 of a game. Between 9.2 and 10.2 R/G, these are accurate to about 3/5 of a game. Between 8.7 and 10.7 R/G, these are accurate to about 7/5 of a game.
In the range between 5 and 15 runs, the X coefficient fits well to 0.532476 Ln(R) + .34973; the X^3 term looks like 0.164906 Ln(R) - 0.12689, which produces estimates accurate to 1/9 of a game per 162.
September 18, 2004 5:01 PM
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