Fisherman riddle: Combining probabilities This is more a probabilities problem than a riddle. The riddle is:
I am in a village, where a fisherman lives. The fisherman tells me that there is a 70% possibility that it will rain tomorrow. I know that fisherman's statement's validity is 80%. What is the actual probability that it will rain?
My approach:
We have 2 factors: Let a be the fisherman's prediction and b the fisherman's validity. We are looking for a function $f(a,b)$ with:


*

*$f,a,b\in[0,1]$

*$f(a,1)=a$

*$f(a,0)=1-a$

*$f(a,b)=f(b,a)$


Am I correct? 


*

*I am sure about (1) and (2). 

*Is number (3) correct? Or maybe $f(a,0)$ could be any random number in [0,1]? 

*Could (4) be correct? I reached there by thinking that $f(1,b)=b$ and $f(0,b)=1-b$.


How can I continue from here?
Edit: Extra thoughts.
Regarding b-fisherman's validity: What does it mean if b=0? 


*

*One opinion could be that b=0 means that fisherman is always wrong. So f(a,0)=1-a. 

*One second opinion would be that fisherman's prediction can be either true or false, without any further clues. So f(a,0)=0.5.


Could in that second case the function be $$f(a,b)=ab+ 0.5(1-b)$$
Moreover, I think that always $f(0.5,b)=0.5$
 A: Let's look at it this way:
Let $X$ be a Bernoulli random variable that indicates whether the fisherman is correct:  $X = 1$ with probability $p = 0.8$ if he is correct, and $X = 0$ otherwise, with probability $1-p = 0.2$.
Let $R$ be another Bernoulli random variable indicating whether it rains tomorrow; $R = 1$ indicates rain, $R = 0$ indicates no rain.
Then $R \mid X = 1 \sim Bernoulli(0.7)$.  But we have no information how the conditional random variable $R \mid X = 0$ is distributed.  Therefore, we cannot answer the question about the unconditional distribution of $R$.
A: (Of course this is all not very serious.)
I'd argue as follows: 
The fisherman thinks: With 70% probability it will rain, and with 30% probability it will not rain. His subjective probability is only 80% correct. This means that only in 80% of his 70% "rain" it will actually rain, and in 20% of his 30% "not rain" it will  rain anyway. In this way we arrive at the following probability of rain tomorrow:
$$0.8\cdot 0.7+ 0.2\cdot 0.3=0.62= 62\%\ .$$
