True Or False Probability Let's say you are taking a true or false test and the distribution of true and false answers questions is not necessarily 50/50, say probability of right answer being false is $p$. Is it true if for a fair coin I let a flip of heads means answer true and flip of tails answer false, I always have 50/50 chance of getting answer right.
My reasoning would be probability of getting answer right would be equal to
$$\frac{1}{2}p+\frac{1}{2}(1-p)=\frac{1}{2}$$
 A: You are right. You guess correctly if the test says false and your coin says false or if the test says true and your coin says true.
A: Assume you answer true when you get a tail and false when you get a head.
Let $F$ be the event that the correct answer is false, and $H$ be the event of flipping a head.  
So the probability of a fair coin matching the correct answer is: $
\quad\mathsf P\Big((H\cap F)\vee(H^\complement\cap F^\complement)\Big)
\\[1ex] = \mathsf P(H)\mathsf P(F)+\mathsf P(H^\complement)\mathsf P(F^\complement)
\\[1ex] = \tfrac 1 2 p + \tfrac 1 2 (1-p)
\\[1ex] = \tfrac 1 2 + \frac 1 2\xcancel{(p-p)} \require{cancel}
\\[1ex] = \dfrac 1 2 
$
It doesn't work when your coin is biased because, then the bias of the answer isn't cancelled.
Suppose the probability of a tail is $q$.  Then: $\quad \mathsf P(H)\mathsf P(F)+\mathsf P(H^\complement)\mathsf P(F^\complement)
\\[1ex] = q p + (1-q) (1-p)
\\[1ex] = 1 - q + (2 q - 1)p
$
A: Yes, it's right, and here's why it's counter-intuitive (at least, why it seems that way to me).
Since I'd rather not overuse True/False, let's say the questions are a series of $A$ or $B$ choices. The counter-intuitive part is that it doesn't matter how the right answers are distributed among the $A$ or $B$ choices, only the number of choices and that you're equally likely to choose to $A$ as $B$.
Let's take an extreme example: The right answer is always $A$. But you don't know that, so you happily make random guesses. Then, statistically, you expect to get half of the answers right, since you expect to have randomly chosen $A$ half the time.
So, as long as there are two answer choices and you're randomly guessing, you're expected to get half right, however the correct answers are distributed between the $A$ and $B$ choices.  
