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Posted By: rambiz In Response To: Varianace Reduction for Dummies (rambiz)
Date: Saturday, 13 August 2016, at 11:33 a.m.
Thank you all for sharing your thoughts.
There were two major mistakes in my original post.
Firstly, as Stein Kulseth and Casper Van Tak pointed out, I had to go over all 36 dice permutations and then evaluate the resulting position. Since it is a last roll position, after going through all 36 rolls, the evaluation engine should see, one would expect, the exact probability of winning for the player on roll.
Secondly, assuming that the evaluation for some reason doesn't get the exact probabilities after the roll, I had to make sure that average luck equals to zero. In my example I had p(1-p')+(1-p)*(-p')=p-p', which is obviously not equal to zero, unless p=p' . What went wrong though? I had assigned a luck of (1-p') to all winning rolls and a luck of -p' to all losing rolls, but could I exactly determine, which rolls are winning and which are losing, the evaluations engine had been perfect in the first place. This is not how luck is calculated!
They way luck is calculated guarantees an average luck of zero.
For each of the 36 rolls luck is defined as the difference of estimated winning chance (equity if you like) after that very roll and the estimated winning chance before the roll. Say the evaluation engine goes through all rolls and spits out p1,p2,p3,...,p36 as winning chance estimates after each roll.
The estimated winning chance before the roll is then (p1+p2+p3+...+p36)/36 .
The luck assigned to the 1st roll is p1-(p1+p2+p3+...+p36)/36, the luck assigned to the 2nd roll is p2-(p1+p2+p3+...+p36)/36, and so forth.
Now the average luck is:
1/36*(p1-(p1+p2+p3+...+p36)/36)+1/36*(p2-(p1+p2+p3+...+p36)/36)+...+1/36*(p36-(p1+p2+p3+...+p36)/36)=0.
It is noted, that the above calculations make no assumptions about the accuracy of the estimates p1,p2,...,p36.
Conclusion: Variance Reduction doesn't work with a wrong luck estimator.
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