## Curse of Probability

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

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

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

$$\epsilon \to 0$$
$$g \approx 0.03$$

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