> 2A) Do a weighted average of 432 2-ply nodes selected by striated
> sampling with the result of a full 1-ply evaluation, using
> empirically derived weights.
With 2A, both the 2-ply and the 1-ply evaluations, considered separately,
are attempts to estimate the overall equity of the position, so it's
reasonable to combine them willy-nilly.
2B) Do a weighted average of the 432 2-ply nodes resulting from 12 of
the initial rolls and the 1-ply evaluations of the 24 other
rolls (each with weight 36)
With 2B only one weighting seems reasonable to me.
As an example, take the situation where I will win, unless I roll 2-1 and
opponent rolls 6-6:
In this case, if all our evaluations are exact, we will have
1-ply = 2-ply = 1294/1296
while 2A) gives either 1292/1296 or 1295/1296 depending on whether the
winning sequence is included in the 432 evaluated on 2-ply.
Good example. 2A results in overweighting the selected 2-ply
continuations, and this is bad.
My intuition is still that 2A would perform better than 2B. I feel
that it is beneficial to have the most accurate evaluations spread
across the range of variations.
I hope the issue can be decided empirically. It shouldn't be too
hard. 2B is actually easier to implement than 2A, because the
rolls are partitioned at the root level. (Well, for 2-ply, anyway;
maybe not for the generalization to N-ply.)