Curse of Probability

\(P(\lnot L(A, B) | G(A)) = P(\lnot L(A, B)) = 1-\epsilon\)
\(P(L(A, B) | G(A)) = P(L(A, B)) = \epsilon\)
\(P(L(A, B) \wedge \lnot G(B) | G(A)) = P(L(A, B)) * P(\lnot G(B)) = (1-g)\epsilon\)
\(P(L(A, B) \wedge G(B) | G(A)) = P(L(A, B)) * P(G(B)) = g\epsilon\)
\(P(L(A, B) \wedge L(B, A) \wedge G(B) | G(A)) = P(L(A, B)) * P(L(B, A)) * P(G(B)) = g\epsilon^2\)
\(P(L(A, B) \wedge L(B, A) \wedge \lnot G(B) | G(A)) = 0\)

\(P(L(A, B) \wedge L(B, A) | G(A) \wedge G(B)) = P(L(A, B)) * P(L(B, A)) = \epsilon^2\)
\(P(L(A, B) | G(A) \wedge G(B) \wedge L(B, A)) = P(L(A, B)) = \epsilon\)
\(P( G(A) \wedge G(B) \wedge L(A, B) \wedge L(B, A)| G(A) \wedge G(B) \wedge L(A, B) \wedge L(B, A)) = 1\)

\(P(G(A) \wedge G(B) \wedge L(A, B) \wedge L(B, A)) = g^2\epsilon^2\)

\(\epsilon \to 0\)
\(g \approx 0.03\)

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2015-02-05
However, there is no curse of ability.