A hunter has $2$ hunting dogs.
One day, on the trail of some animal, the hunter comes to a place where the road diverges into two paths. He knows that each dog, independent of the other, will choose the correct path with probability $p$. The hunter decides to let each dog choose a path, and if they agree, take that one, and if they disagree, to randomly pick a path. Is his strategy better than just letting one of the two dogs decide on a path?
I think the correct equation is: $$P(\text{Correct Path | Dogs Agree on Path}) \times P(\text{Dogs Agree on Path}) \\ + P(\text{Correct Path | Dogs Disagree on Path}) \times P(\text{Dogs Disagree on Path}) \\ = P(\text{Correct Path And Dogs Agree on Path}) + P(\text{Correct Path And Dogs Agree on Path}) \\ = P(\text{Correct Path}). $$ The $P(\text{Correct Path | Dogs Agree on Path}) = p\times p$; I think the $P(\text{Correct Path | Dogs Disagree on Path}) = p(1-p)$ or $1$; I think that $P(\text{Correct Path | Dogs Disagree on Path})$ could equal $1$ because either path is correct so wouldn't one dog choose the correct path? Or do both dogs need to choose the correct path? If both dogs need to choose the correct path, then wouldn't the $P(\text{Correct Path | Dogs Disagree on Path}) = 0$?
I also don't know how to solve for the $P(\text{Dogs Agree on Path})$ and the $P(\text{Dogs Disagree on Path})$.
Let us cut to the heart of the question
"He knows that each dog, independent of the other, will choose the correct path with probability $p$. The hunter decides to let each dog choose a path, and if they agree, take that one, and if they disagree, to randomly pick a path. Is his strategy better than just letting one of the two dogs decide on a path? [Emphasis mine]
Firstly, the dogs need to identify the correct path with $p >0.5$ for his strategy to be of any use !
If the dogs disagree, you can't do better than randomly choose a path so it is as good as not having them. [ Note, by the way, that they will point with equal probabilities $p(1-p)$ to each path]
It is when they agree that there is some gain.
Let $C$ = correct path, $A$ = dogs agree,
then P(C|A) $= \frac{P(C\,\cap\, A)}{P(C\;\cap\;A)+ P(C^c\;\cap\; A)} \quad= \frac{p^2}{p^2+(1-p)^2}$
And as you can see, for $p>0.5$, it is better to use two dogs
$$\begin{array}{|c|c|}\hline \\ p & \frac{p^2}{p^2+(1-p)^2}\\ \hline 0.5 & 0.5 \\ \hline 0.6 & 0.69 \\ \hline 0.7 & 0.84 \\ \hline\end{array} $$