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CidTori
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I finally found an answer in this article, using maximum entropy!

In short, it's : $$Pr(S|C_i T_j) ~ (\frac{ (Pr(S|C_i)^{-1}-1)(Pr(S|T_j)^{-1}-1 }{ Pr(S)^{-1}-1 }+1)^{-1}$$$$Pr(S|C_i T_j) \approx (\frac{ (Pr(S|C_i)^{-1}-1)(Pr(S|T_j)^{-1}-1) }{ Pr(S)^{-1}-1 }+1)^{-1}$$

Or, in terms of odds: $$Od(S|C_i T_j) \approx \frac{ Od(S|C_i) Od(S|T_j) }{ Od(S) }$$ with: $$Od(S|E) = (Pr(S|E)^{-1}-1)^{-1}$$

(coincidentallyCoincidentally, it's quite close to my "Here's"Here is what it should look like" link to Wolfram Alpha)

I finally found an answer in this article!

In short, it's : $$Pr(S|C_i T_j) ~ (\frac{ (Pr(S|C_i)^{-1}-1)(Pr(S|T_j)^{-1}-1 }{ Pr(S)^{-1}-1 }+1)^{-1}$$

(coincidentally, it's quite close to my "Here's what it should look like" link to Wolfram Alpha)

I finally found an answer in this article, using maximum entropy!

In short, it's : $$Pr(S|C_i T_j) \approx (\frac{ (Pr(S|C_i)^{-1}-1)(Pr(S|T_j)^{-1}-1) }{ Pr(S)^{-1}-1 }+1)^{-1}$$

Or, in terms of odds: $$Od(S|C_i T_j) \approx \frac{ Od(S|C_i) Od(S|T_j) }{ Od(S) }$$ with: $$Od(S|E) = (Pr(S|E)^{-1}-1)^{-1}$$

(Coincidentally, it's quite close to my "Here is what it should look like" link to Wolfram Alpha)

Source Link
CidTori
  • 181
  • 4

I finally found an answer in this article!

In short, it's : $$Pr(S|C_i T_j) ~ (\frac{ (Pr(S|C_i)^{-1}-1)(Pr(S|T_j)^{-1}-1 }{ Pr(S)^{-1}-1 }+1)^{-1}$$

(coincidentally, it's quite close to my "Here's what it should look like" link to Wolfram Alpha)