Re: More Hard Problems Using Bayes' Theorem, Please

From: Mitchell Porter (mitchtemporarily@hotmail.com)
Date: Sun Jul 06 2003 - 05:51:05 MDT

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    FWIW, my thought process:

    A priori probability is 50% for each barrel.

    For the red-majority barrel, the probability of
    drawing a red each time is 3/4, blue 1/4; so
    the odds of 3 blues, 7 reds is p1 = (1^3)*(3^7)/(4^10).
    For the blue-majority barrel, it's p2 = (3^3)*(1^7)/(4^10).

    Therefore the overall probability (conditioning
    on the 50-50 prior) is the average of those two
    probabilities. A posteriori we know that this is
    what happened, so the odds that it was the
    blue-majority barrel are

    p2/(p1+p2) = 27/(27+2187) = 1/82

    I now think about doing it step-by-step, as originally
    commanded. Intuitively, using the same averaging
    principle as above, if you fish out a red,
    that's a 3/4 probability that the bucket is
    red-majority. But this suggests that the final
    updated probability (of blue) will just be 3^3/4^10.
    I stop, confused. That looks more like a number
    you'd calculate 'in your head in less than thirty
    seconds'. But what went wrong the first time?

    Nothing probably; what's wrong is the idea of
    just multiplying all those retrodictive probabilities.
    They need to be combined in some more complex way.
    But what's the rule? Well, suppose that the current
    probability assigned to the blue-majority hypothesis
    is b, red-majority r. You draw out a blue. The odds
    that you would draw a blue token from the blue-majority
    barrel are 3/4, from the red-majority barrel 1/4.
    So assuming the (b,r) prior, the odds of what just
    happened are b*3/4 + r*1/4 = (3b+r)/4, and the odds
    that it came from the blue barrel are 3b/(3b+r).
    This is your new probability for the blue-majority
    hypothesis. Updating in this fashion I get

    B 1/2, R 1/2;
    (R) B 1/4, R 3/4;
    (B) B 1/2, R 1/2;
    (R) B 1/4, R 3/4;
    (R) B 1/10, R 9/10;
    (B) B 1/4, R 3/4;
    (B) B 1/2, R 1/2;
    (R) B 1/4, R 3/4;
    (R) B 1/10, R 9/10;
    (R) B 1/28, R 27/28;
    (R) B 1/82, R 81/82

    and I also observe that one can begin by saying
    that in the sequence blues and reds 'cancel' -
    the barrels have equal probabilities of producing
    a sequence with equal numbers of blues and reds -
    so one need only ask the odds on getting four reds.

    But what I still don't get is what's wrong with
    'orthodox statistics'. I obtained my p1/(p1+p2)
    through frequentist thinking, which is supposed to
    be anathema to Bayesians. On the SL4 list, Ben
    Goertzel implies that "non-Bayesian probabilists"
    are restricted in their choice of priors:
    http://sl4.org/archive/0305/6820.html

    ... but even if what he says is true, I don't
    see it reflecting on the subjectivist-vs-frequentist
    debate. Both sides need priors to get anywhere,
    and there's nothing in frequentism to *compel* the
    use of Gaussian priors. So, I still don't get the
    philosophical lesson.

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