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Conclusion

Posted By: rambiz
Date: Sunday, 14 August 2016, at 2:57 p.m.

In Response To: Conclusion (Bob Koca)

It seems that there is a bound on each luck estimate individually, not on the aggregate luck estimate. Otherwise, the straightforward approach below yields the same result as Casper's.

System Model and Problem Description
It is a last roll situation.There are only two outcomes: 1 for winning or 0 for losing.
The exact probability of winning is p.
A one time experiment without variance reduction (VR is defined below) has a variance of p(1-p).
Now, instead of 1 and 0, we introduce a new random variable, call it X, which has the outcomes, (1-L1) with probability p and (0-L0) with probability (1-p). L1 and L0 are the luck estimate assigned to the events winning and losing, respectively.
We want to choose L1 and Lo in a way that:
1. No Bias:The new random variable has exactly the same expectation as the old one, namely p.
2. Less Variance The variance of the new random variable is smaller than p(1-p).

Solution
To achieve the no bias condition, as seen in other posts before, we must have pL1+(1-p)L0=0. (Equation 1)
The variance of the new random variable (X) is, for example calculated by the handy formula Var(X)=E(X^2)-E(X)^2, where E(.) is the expectation operator. And we get:
Var(X)=(1-L1)^2+(1-p)Lo^2-p^2 < p(1-p) (Inequality 1)
Now from (Equation 1) we have L0=-pL1/(1-p). Substitution into (Inequality 1) and simplification yields:
L1(L1-2(1-p))<0 ---> L1<2(1-p) and -L0<2p.

It is noted that (1-p) is the exact luck corresponding to a 1 and -p is the exact luck corresponding to a zero, as asserted by Casper.

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