Probability of a repeatedly right answer Suppose you forgot where the airport is on an island, north or south. You will ask people where to go. Two third of the people on the island are tourists, they give the right answer with a chance of $\frac{3}{4}$. The local inhabitants $\frac{1}{3}$ of the people always give the wrong answer. When you repeatedly ask where the airport is to the same person, the answers are independent. 
Now I am investigating how the probability that an answer is right depends on the frequency of the question. 
For example, the first time I will ask a random passer-by where the airport is, the probability that his answer (i.e. "south") is correct is:
\begin{align}
P(R)= \frac{2}{3} \cdot \frac{3}{4} + \frac{1}{3} \cdot 0 = \frac{1}{2}.
\end{align}
How does this change when I again ask the same question to this same person? What is the probability that the answer (i.e. "south") is again correct?
Well, since the events are independent. I will say that again the answer (i.e. "south") is correct will be:
\begin{align}
P(R)= \frac{2}{3} \cdot \frac{3}{4} + \frac{1}{3} \cdot 0 = \frac{1}{2}.
\end{align}
Is this supposition true?
Furthermore, what can I say about the fourth time that I again ask this same person the same question, what is the probability that this answer (i.e. "south") is the correct answer? 

If you ask the random-passerby n times there's a:


*

*$\frac 23 (\frac34)^n$ probability of him being a tourist and answering right every time.

*$\frac 23 (\frac14)^n$  probability of him being a tourist and answering wrong every time.

*$\frac 23 (1-(\frac34)^n-(\frac14)^n)$ probability of giving different answers, revealing he's a tourist.

*$\frac 13$ probability of him being a local, therefore answering wrong every time.
Let's say you asked a random guy n times. And he always answered the same. The third option is out so now you can calculate the chance of him giving the right answer every time (and obviously therefore being a tourist) as:
$$\frac{\frac 23 (\frac34)^n}{\frac 23((\frac34)^n + (\frac14)^n)+ \frac13}$$
Interestingly, you were right: it is $\frac12$ with n = 2, but then it goes lower.
Edit: Answering your original question, if he answers 4 times in a row the same, then there's a probability of $\frac{27}{70}$ of it being the correct answer.
