What's the expected score for guessing on a word bank/"matching" test? Some quizzes/tests have a "matching" or "word bank" section, which is set up as follows:


*

*There are $n$ questions that the student must answer.

*The answer to each of the $n$ questions is one of $k \geq n$ choices provided in a "word bank."

*Each answer will be used at most once. In the case where $k = n$, each answer will be used exactly once.


Without knowing anything about the subject material, what would be the expected score $S$ for intelligently guessing—that is, guessing without using any answer more than once?
Superficially, $\color{gray}{S = n/k}$ might seem like the correct answer ($n$ questions, and a $1/k$ chance of getting any given question right), but that doesn't really make sense because the questions aren't independent. Answering A to question (1) and A to question (2) would be illogical.
What would be an elegant way to solve this?
 A: Your intuition is correct, $\mathbb{E}[S]= n/k$ is the right answer!
Even though the questions aren't independent, we can apply linearity of expectation. In particular, let $X_{i} = 1$ if we guess the $i$th answer correctly, and let $X_{i} = 0$ otherwise. Then $S = \sum_{i=1}^{n} X_{i}$, so by linearity of expectation, $\mathbb{E}[S] = \sum_{i=1}^{n} \mathbb{E}[X_{i}]$. 
Well, what's $\mathbb{E}[X_{i}]$? If the student is guessing randomly (regardless of whether he's repeating answers or not), the probability $X_{i}=1$ is $1/k$, so $\mathbb{E}[X_{i}] = 1/k$, and it follows that $\mathbb{E}[S]$ is $n/k$. 
You'll note that the answer is the same even if the student guesses 'unintelligently'; i.e., possibly using the same answer more than once. This might seem a bit counterintuitive, but note that it's not actually as 'unintelligent' as it may seem. Sure, if you use the same answer more than once, you're definitely not going to get all the questions correct, but if $k=n$ and you answer $A$ to all the questions, you're guaranteed to get at least one question right!
