How to decrease the confidence of a Monte Carlo algorithm in its answer? Recall that a Monte Carlo algorithm is $p$-correct if it outputs a correct answer with probability at least $p$. In the case of decision problems, where the answer is binary, repeating the MC algorithm can increase the confidence that a correct answer is obtained. 

However, in the case where more than 1 possible answer is correct, this confidence can actually decrease. I'm wondering how many times $k$ does such an algorithm have to be run to obtain a confidence in the answer that is less than 50%. 

Here's what I've done:

Suppose that $k=3$ and I have a 75%-correct MC algorithm that outputs 5 possible answers, 4 of which are right, and 1 is wrong. I think in this case, the 75% "correctness" of the algorithm is split between all possible right answers, such that each correct answer has a "probability" of 75%/4.

Suppose the right answers are $a, b, c, d$ and a wrong answer is $e$. 

In this case, if I list all possible combinations of outputs of an MC algorithm that I run $k=3$ times, I get a list of tuples $(a, a, a)$, $(a, a, b)$... and so on, each of which is associated with a probability of occurring. So for instance $(a, a, a)$ has probability $(0.75/4)^3$ of occurring and $(a, a, e)$ has probability $(0.75/4)^2 * 0.25$.

So if I build a table associating each possible tuple to its probability of occurring, and sum the probabilities of all events where the algorithm would output the correct answer by majority voting (ties are split by choosing randomly), this should give me the final "confidence" of the algorithm. In this case, I get numbers around 0.65-0.75 (because of the randomness in tying splits).

However, I haven't been able to find a $k$ that gets this value to below 50%.

Any ideas if I'm doing something wrong? Help is much appreciated.
You can compute the probability that outcome $e$ is the unique plurality answer, when repeating the algorithm $k$ times, as follows:
$$
\sum_{\substack{a+b+c+d+e=k \\ a,b,c,d,e \geq 0 \\ e &gt; a,b,c,d}} \frac{n!}{a!b!c!d!e!} \left(\frac{1}{4}\right)^e \left(\frac{3}{16}\right)^{k-e}.
$$
This can be computed explicitly. When $k = 70$, the probability is roughly $0.502$, and this is the minimal $k$ for which the probability exceeds $1/2$.