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If we have log of the two ratios P(z=1|L)/P(z=0|L) how can be compute P(z=1|L) from this equation? Note that $$P(z=1|L)+P(z=0|L)=1$$ hence $$P(z=1|L)=\frac{ \frac{P(z=1|L)}{P(z=0|L)}}{1+ \frac{P(z=1|L)}{P(z=0|L)}}$$ that is, $$P(z=1|L)=\frac{\mathrm e^\ell}{1+\mathrm e^\ell},$$ where $$\ell=\log\left(\frac{P(z=1|L)}{P(z=0|L)}\right).$$ Likewise,  ...

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The equivalence you mention always holds, no need for particular circumstances. By the definition of a conditional probability \begin{align} P(D_1|H) & := \frac{P(D_1,H)}{P(H)}\\ P(D_1|H)P(H) & = P(D_1,H) \\ & = P(H,D_1) \\ & = P(H|D_1) P(D_1) \end{align} Similarily ...

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I agree with the other respondent. The focus on machine learning is more geared towards ready-to-implement algorithms. However there are some Simple results like the Bias-Variance Tradeoff and other sort of discussions of "limiting" performance. From what I've seen, your question might be most practically addressed by the area of model selection (i.e. how to ...

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If the probability of heads is $p_i$ for coin $i$ that has prior probability of $1/3$, then the posterior probability for that being the correct coin assuming you get $n$ heads out of $7$ is proportional to prior times data likelihood, which is $(1/3)*p_i^n*(1-p_i)^{7-n}$. Once you get these 3 posterior probabilities for the 3 coins you can divide them by ...

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