Here's an old discussion I had on an intrade board about the "probability of Rossi being real"
http://intrade.freeforums.org/i-miss-intrade-t29.html Re: I miss Intrade<http://intrade.freeforums.org/i-miss-intrade-t29.html#p138> [image: Post] <http://intrade.freeforums.org/post138.html#p138>by *intrader <http://intrade.freeforums.org/memberlist.php?mode=viewprofile&u=70>* ยป Mon May 27, 2013 2:12 am Third time is the charm: P(A|B) = P(B|A) * P(A) / P(B) or P(B) = P(B|A) * P(A) / P(A|B) A = E-Cat & Rossi is real B = Cold fusion (or something close to it) is discovered If E-Cat is real, it looks like cold fusion to me (or something close to it). P(B|A) = 0.5 I think we all can go with the prior probability that E-Cat & Rossi was probably not real (history of fraud / was convicted / etc) P(A) = 0.01 Now, what is the probability that if cold fusion exists that it's going to be Rossi that makes a real e-cat? Interestingly, the more we disparage Rossi (relative to his colleagues) here, the more likely cold fusion exists. Unfortunately, I think only people like Rossi are actually looking at cold fusion. So if it does exists, I think it's reasonable to say it'll be Rossi or perhaps someone like Rossi that might discover it. So, P(A|B) = 0.05 (I think it's fair to say at least 20 other people are looking at it). However, if it looks like more people of Rossi's caliber or better are looking at Cold Fusion, then that bodes well for CF. So, go ahead and punch in your own number there. Counter intuitive, kinda, but that's bayes for you. So, P(B) = (0.5 * 0.01) / 0.05 = 25% cold fusion exists.
