Bayes' theorem p(A|B) = p(B|A) p(A) / p(B) In general p(A|B) (usually read 'probability of A given B’) = the probability of finding observation A, given that some piece of evidence B is present p(A) = the probability of the outcome occurring, without knowledge of the new evidence p(B) = the probability of the evidence arising, without regard for the outcome p(B|A) = the probability of the evidence turning up, given that the outcome obtains

Video Also mentions base rate fallacy *try not to laugh with hypothesitis

An example 4 possible situations: There is a race between 2 horses: Fleetfoot and Dogmeat  Fleetfoot and Dogmeat have raced against each other on 12 previous occasions: Dogmeat won 5 and Fleetfoot won 7 Therefore, the estimated probability of Dogmeat winning the next race is 5/12 = 0.417 = 41.7% Fleetfoot, on the other hand, appears to be a better bet at 7/12 = 58.3% BUT 3/5 of Dogmeat's previous wins, it had rained heavily However, it had rained only once on the days that he lost On the day of the race in question, it is raining Which horse should I bet on? 4 possible situations: Raining Not Raining Dogmeat wins 3 2 Dogmeat loses 1 6 Source: http://www.kevinboone.net/bayes.html

So, Bayes’ theorem in our example… p(A|B) = p(B|A) p(A) / p(B) ‘A’ = the outcome in which Dogmeat wins ‘B’ = the piece of evidence that it is raining p(A|B) = the probability of Dogmeat winning given that it is raining = what we want to find out p(B|A) = the probability that it is raining, given that Dogmeat wins p(B|A) = 3/5 = 0.6 (since it was raining 3/5 times that Dogmeat won) p(A) = 5/2 = 0.417 (because Dogmeat won on 5/12 occasions) p(B) = 4/12 = 0.333 (since we know it rained on 4 days in total) Therefore, p(A|B) = p(B|A)*p(A) / pB) = 0.6 * 0.417 / 0.333 = 0.75

Patten et al. (2017)

Friston et al. (2014)

The brain as a phantastic organ Friston et al. (2014)