# Finding the expected score of a set of questions

Suppose there are 4 multiple choice questions. Each question has 6 possible answers to choose from and only one of the answers is correct. I answer the questions by rolling a fair die for each question and choose the die's facing value as the answer to the questions. Each correct answer is $5$ marks and incorrect answer is $-1$ mark. I want to find the expected score I will get.

So, I could let $X\sim Binomial(4,\frac{1}{6})$.

Then, $E(X)=4(\frac{1}{6})=\frac{2}{3}$. This is the average number of questions that I will be correct.

Since each correct answer is $5$ points and wrong answer is $-1$ point, $\frac{2}{3} \cdot 5 + (6-\frac{2}{3}) \cdot (-1) = -3.333$. So I say that my expected score is $-3.333$.

But on a second thought, if I let each question be $Y_i$. So there are four questions, so $Y_1, ..., Y_4$. And the expected score for each question is: $E(Y_i)= \frac{1}{6} \cdot 5 + \frac{5}{6} \cdot (-1) = 0$

Then for four questions, $4 \cdot E(Y_i) = 4 \cdot 0 = 0$. In this case, then my expected score is $0$.

Now, I am confuse. How should I think about the problem to get the expected score?

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So your expected score is $5\times \frac 2 3 - 3 \frac 1 3 = \frac {10}3-\frac {10}3 = 0$
Why isn't $\frac{2}{3}$ the expected number of correct questions that I will get but the average score of all the 4 questions? Each of the question has $\frac{1}{6}$ chance of getting correct. There are $4$ questions. So, $n=4$ and $p=\frac{1}{6}$. Let $X$ be the number of correct questions, then $X \sim Bin(n, p)=Bin(4, \frac{1}{6})$. So, $E(X)=np=\frac{2}{3}$ is the expected number of correct questions, isn't it? – xenon Apr 22 '12 at 19:57
Yes, and therefore the expected number incorrect is $3\frac13$ – Mark Bennet Apr 22 '12 at 20:03
Why is the expected number of incorrect question $3 \cdot \frac{1}{3}$ but not $4-\frac{2}{3}=3.333$? – xenon Apr 22 '12 at 20:08