[keenerd]

[edit; whoops you were talking about the format and I was talking about the material. Oh well, hopefully this is not too out of place.]

It could be made much more approachable by removing the math entirely. If you learned programming before algebra you might have solved it like this:

    from random import random
    from collections import defaultdict
    
    tally = defaultdict(int)
    for i in range(100000):
        thief = ['raccoon', 'fox'][random() < 0.3]
        bear = ['false', 'true'][random() < 0.8]
        hair = ['raccoon', 'fox'][random() < 0.3333333]
        if thief == 'raccoon' and bear == 'true':
            continue
        if thief == 'fox' and bear == 'false':
            continue
        if thief != hair:
            continue
        tally[thief] += 1
    
    print(sorted(tally.items()))

While crude it does give the correct answer, to within 1% or so.

[mcherm]

Your approach made assumptions about independence that may or may not be justified.

[keenerd]

That everything is completely independent? This is the correct, non-naive assumption to always make unless you have a very good reason otherwise.

There are 3 binary variables. Some combinations of the variables are impossible. These impossible combinations are discarded. For example, the bear said he saw a fox. Therefor it is not possible for the bear to be truthful and the thief to be a raccoon. When that combination comes up, it is discarded. There is no dependency.

[JumpCrisscross]

> These impossible combinations are discarded. For example, the bear said he saw a fox. Therefor [sic] it is not possible for the bear to be truthful and the thief to be a raccoon.

Per the story, there is a 20% chance a truthful bear may have mistaken a raccoon for a fox. Not automatically discarding possibilities like "a truthful bear saw a fox, though the thief is a raccoon" is a hallmark of probabilistic thinking. Thinking through these examples formally has advantages over starting with code.

[keenerd]

Um, no. I did not say anything like that.

But go ahead and rigorously demonstrate any non-zero existence for "A truthful bear saw a fox, though the thief is a raccoon" with the current parameters of the model. It is not a possibility, it is nonsense.

Please prove the brute-force simulation fails to converge on the correct answer. It is figuratively running a million parallel universes and recording what happened. In no legitimate universe does the combination "bear was not mistaken, bear saw fox, thief was raccoon" ever occur.

[totallymike]

Per the discussion in the scenario, "bear was not mistaken," and "bear was truthful" are different things. One implies a bear with poor eyesight; the other implies a bear intentionally misrepresenting the facts.

Given the story, the bear mistook a raccoon for a fox 20% of the time, which is very different from being untruthful.

[qznc]

Solving a bayesian parable with frequentist methods is heresy. ;)

[keenerd]

I'm reasonably certain no camp would claim the statistical equivalent of BogoSort as a method. 98% of the loop cycles (and 98% of the random numbers generated) make no contribution to the result and are thrown away.

[c0g]

coff MCMC coff