Probability of ordered elements. Suppose there are $16$ students randomly divided into $4$ different rooms, each room having $4$ seats. The students were given a question to solve. We know from our prior experience that all students can solve the question but with varying time.
A student is selected for the next task if he is the $1st$ solver of the $1st$ room or the $2nd$ solver of the $2nd$ room or the $3rd$ solver of the $3rd$ room or the $last$ solver of the $last$ room. 
I want to find out the probability that the student having the $i$th $(i=1,2,\ldots,16)$ ordered solving time was the $1st$ solver of the $1st$ room or the $2nd$ solver of the $2nd$ room or the $3rd$ solver of the $3rd$ room or the $last$ solver of the $last$ room. That is, I want to find out the probability of the $ith$ $(i=1,2,\ldots,16)$ ordered solver to be selected for the next task.
To construct the probability that the $ith$  solver among all $16$ students  is the $jth$ solver of $jth$ room $(j=1,2,3,4)$, the following three things need to happen:
$(1)$ there are $(j-1)$ fastest solvers among $(i-1)$ overall fastest solver. This can be done in $\binom{i-1} {j-1}$ ways
$(2)$ the $ith$ value is a $1$
$(3)$ there are $(4-j)$  solvers among the $(16-i)$ solvers. This can be done in $\binom{16-i}{4-j}$ ways.
The set of $4$ students can be constructed by $\binom{16}{4}$ ways.
Consequently, the probability that the $ith$  solver among all $16$ students  is the $jth$ solver of $jth$ room $(j=1,2,3,4)$ is
$$\frac{\binom{i-1}{j-1}\binom{16-i}{4-j}}{\binom{16}{4}}.$$
It is not possible that the $ith$ best solver overall was in two rooms--but he had to be in some room.  Therefore the events "the $ith$ best solver overall was in room $j$", for $j=1,2,3,4$, are an exhaustive partition of the possibilities.  Consequently their chances add up.
Hence the probability that the student having the $i$th $(i=1,2,\ldots,16)$ ordered solving time was the $1st$ solver of the $1st$ room or the $2nd$ solver of the $2nd$ room or the $3rd$ solver of the $3rd$ room or the $last$ solver of the $last$ room, that is,  the probability that the $ith$ $(i=1,2,\ldots,16)$ ordered solver to be selected for the next task is:
$$\sum_{j=1}^{4}\frac{\binom{i-1}{j-1}\binom{16-i}{4-j}}{\binom{16}{4}}. \quad\ldots (A)$$
But equation $(A)$ produces probability $\frac{1}{4}$ for any $i$. It seems that there is no difference between the following 2 criteria:
criteria 1: The probability that a student is selected for the next task is $1/4$, that is, it doesn't depend on the rank information.
criteria 2: The student having the $i$th $(i=1,2,\ldots,16)$ ordered solving time get the next task if and only if he was the $1st$ solver of the $1st$ room or the $2nd$ solver of the $2nd$ room or the $3rd$ solver of the $3rd$ room or the $last$ solver of the $last$ room. 
But If I set the problem as: A student is selected for the next task if he is the $1st$ solver of the $1st$ room or the $2nd$ solver of the $2nd$ room or the $1st$ solver of the $3rd$ room or the $2nd$ solver of the $last$ room. Then what is the probability that the $ith$ $(i=1,2,\ldots,16)$ ordered solver to be selected for the next task. Now the probabilities differ with $i$. It seems now the rank information is working. 
Why the rank information is not working when I am selecting the student for the next task if  he is the $1st$ solver of the $1st$ room or the $2nd$ solver of the $2nd$ room or the $3rd$ solver of the $3rd$ room or the $last$ solver of the $last$ room?
 A: The key element is that your set of positions in a room for a solver to be picked is exactly the set of all possible positions $\{1st,2nd,3rd,4th\}$.
The probabilities of one particular solver to be 1st,2nd,3rd or 4th in its room sum to 1. Then the number of the rooms don't play a role (it is independant from the solver's rank in the room). Said otherwise, you could shuffle the numbers written on the doors after the student have been assigned without changing the probability of a student to be picked.
In your second problem, "A student is selected for the next task if he is the 1st solver of the 1st room or the 2nd solver of the 2nd room or the 1st solver of the 3rd room or the 2nd solver of the last room.", the set of positions is not the set of all possible positions $\{1st,2d,3d,4th\}$ anymore and the probability to pe picked varies among students (for the best student, it is 0.5, for the two worst ones, it is 0).
A: First choose the room where Mr. $i $ is  going to be. The probability is $\frac {1}{4}$ for each room. Then fill that room with  other people and  consider the random variable $ X=$"number of persons in that room who are faster than $ i$".
If you decide successively for each person whether or not they will be in the room,  you see that $ X $ follows a kind of truncated binomial distribution where you stop whenever you reach $3 $ successes. The distribution depends on $ i$, but this doesn't really matter: what matters is that it doesn't depend on the room where you put Mr. $i$. Hence the probability that Mr. $i$ will proceed to step $2 $ is $$\frac {1}{4}\left (p(X=0)+p (X=1)+p (X=2)+p (X=3)\right)$$
where the bracket evaluates to $1 $ because it covers all possible values of $ X $.
In your second problem the answer has $ X=0 $ and $ X=1 $ again instead of $2 $ and $3 $, so that now it does matter that the distribution of $ X $ depends on $i$.
A: Let $a_1,a_2,...,a_{16}$ be respectively the $1^{\text{st}},2^{\text{nd}},...,16^{\text{th}}$ solver among the $16$ students.
Consider $a_i$. Let's put him/her (for simplicity, let's use "them") into a room. A room's number does not affect the probability of $a_i$ to be put into that room, so the probability of putting $a_i$ in each room is $\frac14$.
Now look at this table:

For now, let's consider room $1$:

Now, because there are still only $1$ person in the room, we put $3$ people out of $15$ ($a_1,...,a_{i-1},a_{i+1},...,a_{16}$) into the room. Let:


*

*the number of possibilities of there being $0$ people in room $1$ which are faster than $a_i$ be $p_0$

*the number of possibilities of there being $1$ people in room $1$ which is faster than $a_i$ be $p_1$

*the number of possibilities of there being $2$ people in room $1$ which are faster than $a_i$ be $p_2$

*the number of possibilities of there being $3$ people in room $1$ which are faster than $a_i$ be $p_3$.


Let's put those into the table:

Now, consider that if we put $a_i$ in another room, the room's number does not affect the number of possibilities of putting any specific number of people faster than $a_i$ in the room.
So, we can do those steps for the other rooms (columns):

Highlighting the cases in which $a_i$ is chosen for the next task:

So, the probability that $a_i$ is chosen for the next task is
$$\begin{align}&\frac{P(\text{highlighted possibilities})}{P(\text{all possibilities in table})}\\
=&\frac{p_0+p_1+p_2+p_3}{4p_0+4p_1+4p_2+4p_3}\\
=&\frac{p_0+p_1+p_2+p_3}{4(p_0+p_1+p_2+p_3)}\\
=&\frac14\text,\end{align}$$
independently of the value of $i$.
In the second question, the probability is affected by the value of $i$; one of the examples has been provided by @Evergalo: $a_1$ has probability $0.5=\frac24=\frac12$, $a_{15}$ and $a_{16}$ has probability $0$. That is because there are $2$ rooms (out of $4$) with the criteria "fastest in room" and none with the criteria "slowest in room".
In terms of my answer, the probability of $i$ being chosen for the next task is
$$\frac{p_0+p_1+p_0+p_1}{4(p_0+p_1+p_2+p_3)}\\
=\frac12\cdot\frac{p_0+p_1}{p_0+p_1+p_2+p_3}\text,$$
which is affected by the value of $i$.
