Notes on sample 1:

T1:  2 does not have BK
T2:  if 3 does not have A, BK in gang
T3:  2 does have EF
T4:  2 does not have I
T5:  3 does not have I : I in gang
T6: -
T7:  2's entire hand is EFOQR
T8:  1 has D, 2 has Q, so 3 has A, so BK in gang (from turn 2)

There is no guarantee with the judge's data that there is such a short logical conclusion.  The general solution is brute force, since the parameters are small enough.