Recall that a Monte Carlo algorithm is $ p $-correct if it gives a correct answer with at least one probability $ p $. In the case of decision problems, where the answer is binary, the repetition of the MC algorithm can increase the confidence that a correct answer is obtained.
However, in the event that more than one possible answer is correct, this confidence can actually decrease. I wonder how many times k $ Should such an algorithm be executed to obtain a confidence in the response less than 50%?
Here is what I did:
Assume that $ k = $ 3 and I have a 75% correct MC algorithm that generates 5 possible answers, of which 4 are correct and 1 is incorrect. I think that in this case, the "correction" of 75% of the algorithm is shared among all the correct answers possible, so that each correct answer has a "probability" of 75% / 4.
Suppose the right answers are $ a, b, c, d $ and a bad answer is $ e $.
In this case, if I list all possible combinations of outputs of an MC algorithm that I run $ k = $ 3 Sometimes I get a list of tuples $ (a, a, a) $, $ (a, a, b) $… and so on, each being associated with a probability of occurrence. So for example $ (a, a, a) $ a probability $ (0.75 / 4) ^ 3 $ to happen and $ (a, a, e) $ a probability $ (0.75 / 4) ^ 2 * 0.25 $.
So, if I build a table associating each possible tuple with its probability of occurrence and I sum the probabilities of all the events for which the algorithm would give the correct answer by a majority vote (the equalities are divided by randomly choosing), this should give me the final result "trust" of the algorithm. In this case, I get numbers around 0.65-0.75 (because of the randomness of the divisions).
However, I have not been able to find a k $ this value is less than 50%.
Ideas if I do something wrong? Help is very much appreciated.