I always believed that probabilities could be multiplied, until I encountered a statement in Machine Learning by Peter Flach about odds:
"Bayes’ rule tells us that we should simply multiply them: 1:6 times 4:1 is 4:6, corresponding to a spam probability of $0.4$."
But converting to probabilities, $1:6 = 1/7$ and $4:1 = 4/5$. Multiplying the probabilities we get $4/35$ ~$= 0.11$
Does Bayes' Theorem work differently for odds and probabilities? What seems to be the problem?
On
The book calculate $$\frac{P(a|c)}{P(b|c)}=\frac{P(a)P(c|a)}{P(b)P(c|b)}$$ What you calculate is $P(a)\cdot \frac{P(c|a)}{P(c|a)+P(c|b)} $, which has no meaning.
To get the value of 0.4:
$$P(a|c)=\frac{P(a)\cdot P(c|a)}{P(c)}=\frac{P(a)\cdot P(c|a)}{P(b)\cdot P(c|b)+P(a)\cdot P(c|a)}=\frac{\frac{1}{7}\cdot\frac{4}{5}}{\frac{6}{7}\cdot\frac{1}{5}+\frac{1}{7}\cdot\frac{4}{5}}=\frac{4}{10}$$
You should read about conditional probability and Bayes rule.
Suppose that the prior probability of a message being spam is $\dfrac{1}{7}$, and the conditional probability of a message triggering the spam filter is $k$ if it is spam but $\dfrac{k}{4}$ if it is not spam for some $k$.
Then, given the spam filter is triggered, the posterior probability the message is spam is $\dfrac{\frac{1}{7}\times k}{ \frac{1}{7}\times k+ \frac{6}{7}\times \frac{k}{4} } = \dfrac{4}{10}$.
This is equivalent to the same statement that if the prior odds for spam are $1:6$ and the likelihood ratio for the spam filter being triggered is $4:1$ then the posterior odds for spam given the spam filter is triggered is $4:6$.
Effectively you have said the earlier calculation is equivalent to $\dfrac{\dfrac{\frac{1}{7}\times k}{ \frac{1}{7}\times k+ \frac{6}{7}\times \frac{k}{4} }}{\dfrac{\frac{6}{7}\times \frac{k}{4}}{ \frac{1}{7}\times k+ \frac{6}{7}\times \frac{k}{4} }} = \dfrac {\dfrac{4}{10}}{\dfrac{6}{10}}$ or in a simplified form $\dfrac{1}{6} \times \dfrac{4}{1}=\dfrac{4}{6}$.