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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 $$\frac{1}{2}p+\frac{1}{2}(1-p)=\frac{1}{2}$$

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  • $\begingroup$ This may just be really simple, just seemed strange to me so thought would ask. $\endgroup$ – Kamster Feb 28 '15 at 5:31
  • $\begingroup$ Also because it doesn't seem to work if my coin was biased so couldn't really intuitively see why this was true $\endgroup$ – Kamster Feb 28 '15 at 5:32
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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.

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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.

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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 $

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