In a "match-the-following" test with $N$ questions and $N$ possible answers, what is the expectation value of my score if I guess randomly? I apologize if the heading wasn't descriptive enough. I wasn't able to formulate the question in a sentence. Anyway, here's a detailed explanation.
Let's say there's a test with $N$ questions and $N$ possible answers are given. Each answer matches uniquely to a particular question but since I have no knowledge about the topic, I'm going to match the answers randomly. If I get $\frac{1}{N}$ points for each correct answer and $0$ for each incorrect answer, what's the expected value of my score?
 A: Since you are harboring some doubts, I am giving a more detailed answer using what are known as indicator variables
Let $X_i$ be an indicator random variable that $=1$ if your answer to the $i_{th}$ question matches, and $0$ otherwise.
Since only one of all $n$ answers will match,  $\Bbb{P}(X_i) = \frac1{n} $
Now the expectation of an indicator variable is just the probability of the event it indicates,
so $\Bbb{E}(X_i) = \Bbb{P}(X_i) = \frac1{n}$
and by linearity of expectation, which applies even if the variables are not independent,
$\Bbb{E}(X) = \Bbb{E}(X_1) + \Bbb{E}(X_2) + .... \Bbb{E}(X_n)$
$= n\cdot\dfrac 1{n} = 1$

Edit
I misinterpreted the question to mean you wanted the expected number of answers that were correct. The expected score will, of course, be $\dfrac1{n}$
A: The expected value of a quantity is just the (weighted) average over the sample space of that quantity. Here the sample space consists of all $N!$ bijections from questions to answers, and the assumption that you know absolutely nothing about the topic implies that all possibilities get equal weight (which is called a uniform distribution), necessarily $1/N!$ per bijection. But one does not really have to consider these permutations separately, because averages behave linearly: if you know for each question how many points you will get on average for that question, then average score you get for all questions is just the sum of those averages.
It is not difficult to see that for a given question, all $N$ possible answers  have the same number of bijections that make the correspondence, namely $N!/N=(N-1)!$ (these are just the $(N-1)!$ distinct bijections from the remaining $N-1$ questions to the remaining $N-1$ answers). So whatever you guess for that question, it will have a $1/N$ probability of being the right answer, and a $(N-1)/N$ probability of being wrong. Since you are rewarded $1/N$ points for a correct answer and $0$ points for a wrong answer, your average score for a single question will be $(1/N)*(1/N)+((N-1)/N)*0=1/N^2$. As said your average overall score will be the sum of that for all $N$ questions, which is an average score of $1/N$ (for exactly one correct answer on average).
So for the expected value of your score it makes absolutely no difference how you answer the questions, as long as you make a guess for every question. This does not mean that there are not other statistics that differ according to how you answer. If you want to ensure a minimal score, you can respond to all questions with the same answer, which ensures you exactly one correct answer (this is like a clock that does not move, and is guaranteed to give the right time once every 12 or 24 hours, depending on its type of display). On the other hand if you guess an actual bijection, there is a slight change that you have a perfect score of $1$, but there is also a substantial probability (near 37% unless $N$ is very small) that you get a score of $0$. So you do have an opportunity to manifest a propensity for or aversion of risk.
