[Solved]: Understanding Monte Carlo Probabilities

Problem Detail: 

I am trying to get a good grasp on Monte Carlo (MC) algorithms, but I feel I am missing something fundamental.

What I don't understand is how MC improves its confidence of giving the correct solution by running more executions.

So please correct me if I am wrong and assume that $Pr[false] \leq 0.5$. Assume we need to find the number of executions so that $Pr[false] \leq 0.125$, thus improving confidence.

Is it true that the number of needed executions is $3$, since $0.5^3=0.125$?

Thanks in advance...

Asked By : Curious

Answered By : TCSGrad

This is the idea of probability amplification, in which one improves the probability of correctness of the algorithm by running it multiple times. You are talking about Monte-Carlo algorithms which have one-sided error. Lets take an example (from Wikipedia) having such a similar feature :

The Solovay–Strassen primality test is used to determine whether a given number is a prime number. It always answers true for prime number inputs; for composite inputs, it answers false with probability at least 1/2 and true with probability at most 1/2. Thus, false answers from the algorithm are certain to be correct, whereas the true answers remain uncertain; this is said to be a (1/2)-correct false-biased algorithm.

The probability that the above algorithm returns an incorrect answer is thus at most 1/2. Now, the key idea is to understand that successive runs of the algorithm produce answers that are independent of each other - hence, on running it twice, the probability that the algorithm returns an incorrect answer both times is at most $(\frac{1}{2})^2$. Thus, after $k$ iterations, the probability of an incorrect answer reduces to at most $(\frac{1}{2})^k$. Thus, if we adopt the scheme that the algorithm returns false if the any of the first $k$ runs give an output of false, else return true, then the probability that our scheme returns an incorrect answer is at most $1 - (\frac{1}{2})^k$, which can be driven down to $0$ for large values of $k$.

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Question Source : http://cs.stackexchange.com/questions/25793