Why are probabilities equal for the very first and the very last student in this problem? Problem: A group of students undergo an exam. Each student randomly takes one piece of papper out of a bowl with n pieces of paper. Each piece of paper contains one unique question and if the question is answered than given student passed the exam. Suppose I as a student have a choice - either I will be the very first student to take my question or the very last. I'm also informed that out of n questions I can answer only k questions, where $ k<n$. I want to maximize my chance of getting question that I know. What I should do, should I be the first to take a question or the very last?
Answer in my textbook says that in both cases chances are equal, without proving this statement. I was able to prove it for case when k=n-1, but failed to generalize for every possible k. Here is how my proof for k=n-1: k=n-1 means that I don't know one question.
If I'm the first to pick a question out of the bowl, then I have $\frac{1}{n}$ chance of getting question that I don't know.
Suppose I will be the last to pick a question. There are $P_n$ of all possible orders in which pieces of paper can be taken. In order to calculate number of combinations where I get the question that I can't answer we just need to calculate $P_{n-1}$ because one element is fixed, namely the unknown question, it must be at the very end. So my chance of getting the unknown question is 
$P_{n-1}$/$P_n$, that is equal to $ \frac{(n-1)!}{n!}=\frac{(n-1)!}{(n-1)!*n}=\frac{1}{n} $, meaning that I have equal chance of getting the unknown question in both cases. From this we can also conclude that chances of getting questions that I DO know must be equal too.
But as I already said, I don't see how to prove it for other values of k and will appreciate any help.
 A: When you pick a question out of the box, no matter whether you are the first or the last or someone in the middle, the probability of that question being any particular question is $1/n$. If you know the answer to $k$ questions, the probability of you getting a question you know the answer is, therefore, $k/n$.
This is true by symmetry. If the probability of each question is not $1/n$, that means there is a question that is more likely to be picked last (or at any step). Which one would that be? All the questions are treated equally in the previous steps.
A: Imagine that instead of picking questions out of a bowl, they are written on cards and shuffled together. Then each student in turn takes the top card of the deck. Since the order of the cards is random, this is equivalent to each student randomly selecting a piece of paper from the bowl, as both processes simply generate a random permutation of the questions.
Now, is the top card of the deck any more likely to be a question you know than the bottom card?
A: The questions in the urn have numbers from $1$ to $n$. The whole process of selecting questions amounts to creating a random bijective map  $\sigma: [n]\to[n]$: $\ $ At the $k^{\rm th}$ step the student in turn selects question $\sigma(k)$. The basic assumption is that all $n!$ possible bijective maps $\sigma:\>[n]\to[n]$ are equiprobable. Assume that you know the answer to question $j$. It is a simple exercise in combinatorics to show that exactly ${1\over n}$ of all maps $\sigma$ have $\sigma(1)=j$, and in the same way exactly ${1\over n}$ of all maps $\sigma$ have $\sigma(n)=j$. Similarly, if you know the answers to the questions $j_1$, $\ldots$, $j_k$: Since the corresponding events $\sigma(1)=j_1$, $\ldots$, $\sigma(1)=j_k$ are disjoint the corresponding probabilities just add up.
