Conditional Probabilities Paradox

Question

I know which step is wrong in the following argument, but would like to have contributors' explanations of why it is wrong.

We assume below that weather forecasts always predict whether or not it is going to rain, so not forecast to rain means the same as forecast not to rain. We shall also assume that forecasts are not always right.

It is not generally true that the probability of rain when forecast is equal to that of its having been forecast to rain when it does rain. Indeed let us assume that $$P(\text{R}|\text{F}_{\text R}) \neq P(\text{F}_{\text R}|\text{R}).$$ But, having been forecast to rain, it will either rain or not rain ($\bar{\text{R}}$), so $$P(\text{R}|\text{F}_{\text R})+P(\overline {\text{R}}|\text{F}_{\text R})=1\ \ \ \ \ \ \mathbf{eq. 1}$$ Likewise, if it rains, it will either have been forecast to rain or (we are assuming) not forecast to rain ($\overline{\text{F}_{\text R}}$), so $$P(\text{F}_{\text R}|\text{R})+P(\overline{\text{F}_{\text R}}|\text{R})=1 \ \ \ \ \ \ \mathbf{eq. 2}$$ But we know that "If rain then not forecast to rain" is logically equivalent to "If forecast to rain then no rain". So the corresponding conditional probabilities must be equal, that is $$P(\overline{\text{F}_{\text R}}|\text{R})=P(\overline {\text{R}}|\text{F}_{\text R})\ \ \ \ \ \ \ \ \ \ \ \ \mathbf{eq. 3}$$ It follows immediately from $\mathbf {eqs 1,\ 2\ and\ 3}$ that $$P(\text{R}|\text{F}_{\text R}) = P(\text{F}_{\text R}|\text{R}).$$ which is contrary to our hypothesis.

Answer 1

The statement "If $A$ then $B$" discusses the relation between $A$ and $B$ in the $A\subseteq B$ sense. Not conditional events.

Let's throw a dice, $A$ is the result even, $B$ is the result $2$. Following the steps of the original post, we can say $$ \Pr(A\vert B)+\Pr(\bar{A}\vert B)=1,$$ and $$ \Pr(B\vert A)+\Pr(\bar{B}\vert A)=1.$$ Moreover, we know that "if B then A" is equivalent to "if not A then not B", which means $B\subseteq A$ iff $\bar{A}\subseteq \bar{B}$. But $\Pr(A\vert B)=1$ while $\Pr(B\vert A)=\tfrac{1}{3}$ which means that your next step is invalid.

Answer 2

The statement "if $A$ then $\overline B$" implies "if $B$ then $\overline A$" Is equivalent to the probability statement ... $$\bigg(P(\overline B|A) = 1\bigg ) \implies \bigg(P(\overline A|B) = 1 \bigg) \tag{1}$$

You can use the identities $$P(A|B)=1-P(\overline A|B) \\ \text{and} \\P(B)P(A|B)=P(A)P(B|A)$$ To deduce the general relationship between $P(A|\overline B)$ and $P(\overline A| B)$ $$ \begin{aligned} P(B)P(A|B) &=P(A)P(B|A) \\P(B) \left (1- P(\overline A|B) \right ) &=P(A) \left (1- P(\overline B|A) \right ) \end{aligned}$$

In this form, it is easy to see that $(1)$ must be satisfied, provided that neither $P(A)$ nor $P(B)$ are zero.

It can also be solved to give ... $$ P(\overline A|B)=1-\frac{P(A)}{P(B)} \left (1- P(\overline B|A) \right ) $$ which is an identity provided that $P(B)$ is not zero.