Variant of the hardest logic puzzle ever This question was reposted on puzzling.stackexchange and hexomino's answer there might be the correct one( although I am still not completely convinced). The following is the link to the reposted question:
https://puzzling.stackexchange.com/questions/101720/variant-of-the-hardest-logic-puzzle-ever
I have landed on a new planet and there are 4 people there. One of them is a truth teller and they always speak the truth. The other is a liar and they always lie. The other 2 are random and they sometimes say yes and sometimes say no, all at random. Each of them knows everything about all the others. I wish to find out the identities of all of them by asking the minimum number of questions possible. What should be my approach ?
This is a variation of "The hardest logic puzzle ever" . The only difference is is that in the original problem, there is only one random instead of two. The following is an excellent video that details, both the original question and the answer :
https://youtu.be/LKvjIsyYng8
Coming back to my question, how many minimum questions will i need to  ask to find the identities of all the 4 and what should be the questions?
I have solved it partially and am detailing my attempt below. Notice that I am able to solve it for cases 1 and 2 but not for cases 3 and 4.

My attempt :
Let us assume that the persons are standing in a line and are facing towards me.
I ask the first person about the second person, " Would you have said yes if I had asked you if the person standing to your left is a random?"
Then, I ask the third person about the fourth person, "Would you have said yes if I had asked you if the person to your left is a random?"
Possible replies :
Case1:  Yes No ( the 1st person says yes & the 3rd says no)
Case 2:  No Yes
Case 3:  Yes Yes
Case 4:   No  No
I am able to solve it for the cases 1 and 2 i.e when one of them says yes and the other says no. I will illustrate why I am able to solve, by using case 1. However, the same logic holds for case 2.
Case 1:
Lemma 1: At least one person between the first person and the second person is a random. This is because :
a) The first person themselves is a random and chose to say
yes  randomly, or
b) The first person is a truthteller and if they are saying yes
then that means that the second person is surely a
random.
c) The first person is a liar and their answer to the above
question can be yes only if the second is a random (it is
easy to figure out why but if it is still unclear then please see the video above to understand why).
Lemma 2: The fourth person is not a random. This is because :
a) The third person  themselves is a random and chose to say  no randomly. ( And since we know that at least one
person between the first and the second person is a
random,  then this means that the fourth person cannot
be the other random) or,
b) The third person is a truthteller and if they are saying no
then that means that the fourth person is surely not a
random, or
c) The third person is a liar and their answer to the
above question can be "no"  only if the fourth person is
not a random (again, it is easy to figure out why but if
it is still unclear then please see the video above to
understand why).
Therefore, now that we have figured out that the 4th person is not a random, we can simply ask them, "Is 2+2=4?". Based on their answer, we can find if they are a truth teller or a liar and then use them to find the identities of everybody else.
We can have the same approach for case 2. But I cannot figure out how to solve cases 3 and 4.
 A: I think that there is no minimum with the following reasoning (please feel free to point out any flaws in my reasoning):
Label the four individuals as $A$, $B$, $C$, $D$ and consider the following alternative scenario, which I'll call Scenario 1

$A$ thinks that they are a truthteller, $B$ is a liar and $C$ and $D$ are random.
$B$ thinks that they are a liar, $A$ is a truthteller and $C$ and $D$ are random.
$C$ thinks that they are a truthteller, $D$ is a liar and $A$ and $B$ are random.
$D$ thinks that they are a liar, $C$ is a truthteller and $A$ and $B$ are random.

In this version of the problem, we can swap $A$ for $C$ and $B$ for $D$ and the problem remains the same. Hence, for any questions asked, there is no way to distinguish $(A,B)$ from $(C,D)$.
Now let us consider the following Scenario 2

Replace $C$ and $D$ in Scenario 1 with randoms but all of their answers will be as if we are in Scenario 1 (for any finite number of questions this may always happen by chance).

And Scenario 3

Replace $A$ and $B$ in Scenario 1 with randoms but all of their answers will be as if we are in Scenario 1 (again, can happen by chance).

Given that we cannot distinguish the two cases in Scenario 1, it is also impossible to guarantee that we can distinguish Scenario 2 from Scenario 3 with any finite number of questions.
That is to say, for any finite number of questions, the responses in Scenario 2 can match up with the responses in Scenario 3 and we cannot guarantee to distinguish them.
A: I must say, I'm starting to like these logic puzzles! Since you've asked just about cases 3 and 4, I'll only go through them in this answer.
Case 3 (Yes, Yes)
Person 1 saying YES means one of three things:

*

*They are randomly saying yes (Person 1 is a random)

*They are telling the truth, because Person 2 is a random (Person 1 is the truth-teller and Person 2 is a random)

*They are lying, because Person 2 is a random (Person 1 is the liar and Person 2 is a random)

Person 3 saying YES means one of three things:

*

*They are randomly saying yes (Person 3 is a random)

*They are telling the truth, because Person 4 is a random (Person 3 is the truth-teller and Person 4 is a random)

*They are lying, because Person 4 is a random (Person 3 is the liar and Person 4 is a random)

Analysis of Case 3
Person 1 and 2 cannot both be randoms, and Person 3 and 4 cannot both be randoms either.
Case 4 (No, No)
Person 1 saying NO means one of three things:

*

*They are randomly saying no (Person 1 is a random)

*They are telling the truth, because Person 2 is the liar (Person 1 is the truth-teller and Person 2 is the liar; Person 3 and 4 are both randoms)

*They are lying, because Person 2 is the truth-teller (Person 1 is the liar and Person 2 is the truth-teller; Person 3 and 4 are both randoms)

Person 3 saying NO means one of three things:

*

*They are randomly saying no (Person 3 is a random)

*They are telling the truth, because Person 4 is the liar (Person 3 is the truth-teller and Person 4 is the liar; Person 1 and 2 are both randoms)

*They are lying, because Person 4 is the truth-teller (Person 3 is the liar and Person 4 is the truth-teller; Person 1 and 2 are both randoms)

Analysis of Case 4
Here, we can't use any of your lemmas. The good thing, however, is that if you have identified either the truth-teller or liar beforehand, you will be able to determine everyone's identities.
