Bayes’ twin puzzle

The Rev. Price is a believer in the Rev. Bayes and his rule. But here I imagine him getting his reasoning tangled. Can you help him out?

One day, the Rev. Price’s trusted house-keeper, Alice, tells him that he has seen the Rev. Bayes in a compromising situation with Betty the whore. The Rev. Would have thought, ‘a priori’ that this was very improbable, but tends to believe Alice. Thus for H=’Bayes was with Betty’ and E=’Alice reports H’, the Rev. Price estimates that:

The ratio of the priors is P(H)/P(¬H) = 1/100,

And the Bayes factor (the ratio of the likelihoods) is P(E|H)/P(E|¬H) = 100,

Hence, applying Bayes rule the ratio of the posterior probabilities is;

P(H|E)/P(¬H|E) = (P(E|H)/P(E|¬H)).(P(H)/P(¬H)) = 1.

That is, the Rev. Price regards the situation as inconclusive between H (Bayes was with Betty) and ¬H (Bayes was not with Betty).

The next time the two Revs. meet, the Rev. Bayes says E’=’My twin brother, the black sheep of the family, is in town.’.  The Rev. Price, as a committed Bayesian, considers the likelihood of E’ for the two cases, H and ¬H.  If H were true then Bayes would wish to deny it, and E’ would be a like tale. On the other hand, Bayes’ not going with Betty is presumably an everyday circumstance, and there is no reason to think that that this would cause Bayes’ brother to visit. Hence the Rev. Price estimates the new Bayes factor as:

P(E’|H)/P(E’|¬H) = 100.

Applying Bayes rule he gets:

P(H|E,E’)/P(¬H|E,E’) = 100.

That is, he now regards H as being 100 time more probable than ¬H, and must either abandon Bayesianism or his belief in Bayes.

Can you fix his logic and save his belief in Bayes?

Dave Marsay 

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