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OFF TOPIC: Probability question

Posted By: Mislav Kovacic
Date: Tuesday, 30 July 2013, at 11:35 p.m.

In Response To: OFF TOPIC: Probability question (leobueno)

So essentially, you need at least 2 out of 4 remaining judges to rule in your favor. Let's assign "P" to votes in your favor (positive), and "N" to votes against you (negative).
The probability of getting at least 2 P is 100% minus the probability of getting 0 P or 1 P.

To get 0 P, 4 independent events must all occur at the same time. Each vote is N so we multiply the individual probabilities. The probability that you get 4 N is therefore 50% x 50% x 50% x 50% = 6.25%.

To get any one permutation of one P and three N, again four individual events must occur => 1 P and 3 N. As the probability of N is the same as P = 50%, the probablity of "any one permutation" is 50% x 50% x 50% x 50% = 6.25%.
HOWEVER there are 4 different permutations where you can get 1 P and 3 N. These are PNNN, NPNN, NNPN and NNNP. All 4 of these permutations have exactly the same probability of occurring => 6.25%.
Therefore the probability of getting one P and three N is 4 x 6.25% = 25%.

Therefore as we are interested in the probability of EITHER 0 P OR 1 P we need to ADD these 2 figures together. 6.25% + 25% = 31.25%
Therefore the probability of getting AT LEAST 2 P = 100% - 31.25% = 68.75%, and that's the final answer (unless I messed up something).

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