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

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

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.

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.

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: $\quad \mathsf P(H)\mathsf P(F)+\mathsf P(H^\complement)\mathsf P(F^\complement) \\[1ex] = q p + (1-q) (1-p) \\[1ex] = 1 - q + (2 q - 1)p$