First, let's keep 1,2,3 as whatever for this next step, and use TLR to refer to Truthteller, Liar, and Ranom)
Okay, thought about it for a minute. Our first question failed to resolve the "yes/no" situation, as I'd hoped, so let's try a different tact - Now that we know part of the benefit of the initial question, we might be able to squeeze a bit more out of it by making it less trivial. If we ask if person 1 is the random guy (a verifiable piece of information we can use to verify other bits later on), we have 12 possible results:
(We'll call
1 is a liar (2 is truthful, 3 is random)
1 2 3 |
a b a | A is tue
a b b |
b a a | B is true
b a b |
1 is a liar (2 is random, 3 is truthful)
1 2 3 |
a a b | A is tue
a b b |
b a a | B is true
b b a |
1 is random (2 is liar, 3 is truthful)
1 2 3 |
a b a | A is tue
b b a |
a a b | B is true
b a b |
1 is random (2 is liar, 3 is truthful)
1 2 3 |
a a b | A is tue
b a b |
a b a | B is true
b b a |
No
Now, we still know that the odd person out is NEVER going to be random - the random guy will always share his answers with one of the other two.
So for out five distinct "results" and potential outcomes for 1,2,3
aab - RTL, RLT, TRL, LRT
abb - TRL, TLR, LTR, LRT
aba - TLR, RLT, LTR, RTL
bab - TLR, RLT, LTR, RTL
bba - RTL, RLT, TRL, LRT
And now, for each of these, we need to figure out which of the possible four configurations is correct... which is the next step. And we have an important additional bit of knowledge linking a or b to true/false no matter which situation we get - if we find the truthteller/liar/random dude, we'll instantly have another identified thanks to this first question (and then we'll be done). But we still need to identify at least one...
But I'm taking a break now.