I am working on Monte Carlo algorithm : Given that you have an experiment MC which has a p-correct of 75%, which means it gives you the right answer 75% of the time. You run MC3 which repeats MC 3 times and keep the value that appears the more often.
So at the first time it is a binary value : Right (75%) Wrong(75%) then MC3 is P(R=2) = C(3,2)(0.75)^2(1-0.75) = 0.425
Second type of experiment : R1 (25%) R2(25%) R3(25%) W(25%), the probability to get two times the same answer (which is a right one so R1, R2, R3) is 3 * P(Ri=2) = 0.42 ==> Seems logic since we have more output the chance to get the same output twice is smaller even if we multiply it by 3 after.
Third type of experiment : R1 (37.5%), R2(37.5%), W(25%), the probability to get two times the same answer (which is a right one so R1, R2) is 2 * P(Ri=2) = 0.52 ==> And here I can't get the logic. Why in experiment 2 we have less chance to have twice same good result and in experiment 3 we have more.
I am able to proove it is normal by looking at the behavior of the function C(3,2)*(0.75/i)^2 * (1-0.75/i) but I can't figure it in my head, I don't see the logic. Is someone able to explain it with words ?
Your $MC_3$ is correct with a higher probability than what you calculated. $MC_3$ is correct whenever at least two of the $MC$ trials return correct (as opposed to exactly two). In your notation, we want $R \geq 2$, not $R=2$. The probability that $MC_3$ is correct is $$P(MC_3) = \binom{3}{2}P(MC)^2(1-P(MC))+ P(MC)^3$$
$$P(MC_3) = 3\cdot \left(\frac{3}{4}\right)^2\cdot \frac{1}{4} + \left(\frac{3}{4}\right)^3$$
$$P(MC_3) = \frac{27}{64} + \frac{27}{64} = \frac{27}{32} = 84.375 \%$$
It's a bit unclear what you mean in the "second type" and "third type" experiments ... I suggest you use MathJax and define as many of your variables/functions as possible.