Suppose that the participants are all indistinguishable. Each judge has a 3/4 chance of "being fooled". Let p be the event that the computer is correctly determined, and let q be the event that the computer is not selected.
We calculate (1/4p+3/4q)^3 = 1/64(p^3 +3p^2(3q) + 3p(3q)^2+(3q)^3) = 1/64 (p^3 + 9p^2q + 27pq^2 + 27q^3).
The probability, then, that the computer is chosen by random chance is 10/64 or approximately 15%.
Which means in only 14 trials, there's a greater than 90% chance of passing the test. (Again, assuming the contestant is indistinguishable from the judges.)
3 judges and 4 participants
What's the probability that any 2 or all 3 judges will pick a particular participant ("the computer") out of 4 at random?