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?
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).