The worst case has the potential to reach any positive, finite number. I will demonstrate this with a counter-example using the 11-question strategy.
Let's label the guards:
T always tells the truth.
F always lies.
Y always says yes.
N always says no.
R gives a random answer.
Ask each one "Are you both T and F?"
After which, suppose I get the following answers (zero is no; one is yes)
Now I have two groups.
Group 1 contains T and N
Group 2 contains F, Y and R
Ask one guard from group 1, "Are you T?" If the answer is yes, they are T and the other one is N. Else, visa-versa. Cool.
That leaves group 2. Ask each one "Are you F?".
The answers will be
Either R said yes, in which case the only one who said no was F. Or R said no, in which case the only one who said yes was Y. Let's take that as an example and say we can exclude Y.
We must now figure out which of the remaining two is F and which is R. Let's ask: "are you F?" to both.
in the best case scenario, F says no and R says yes.
But we're testing the worst case, so let's say R also says no. Oops. That leaves us at State 4. This could potentially go on for any finite amount of times until R finally gives a different answer to F so that you can tell the difference.
Since the worst possible case is equally bad for all answers, all answers tie for that position according to scoring.
Number of questions in the worst case
My solution is then to not ask any questions at all and guess that the 1st, 2nd, 3rd, 4th and 5th are T, F, Y, N and R respectively.
This is the best case scenario which also happens to be the next tie-breaker.
Number of questions in the best case
Is this likely? Of course not! But the next tie-breaker (Probability that the best case will be achieved) only applies if someone actually ties with me (zero questions).
I hope that it is clear that this is a genuine post and not a "troll answer". I'm not so interested in having the winning answer so much as providing an interesting answer should we take the rules as literally as we can.
I know it seems like an odd answer but read through the proof carefully and if you still have an objection, feel free to comment.
This answer does not work because
As bobble pointed out in the comments, you can short circuit the loop by using either T or F (whoever isn't in the same group as R at the time) to find out the identity of everyone who is left.
I would, however, like to leave this answer in-tact in case anyone else tries to follow the same line of reasoning.