There are at least three different lines of inquiry here:- Hypothesis testing. If the [null] hypothesis is that p(heads) is 1, you can't prove this, only disprove it. So: "doesn't sway". Not very interesting, but there it is.- Simple Bayesian. The probability of his claim given that it comes up heads, p(C|H), is the prior of his claim, p(C), times p(H|C), divided by p(H). Well, p(H|C) is 1 (that is the claim), and p(H), if I fudge things a little bit, is about 1/2, so p(C|H) should be about double p(C)---assuming p(C) is very low to start with.[0][2]- Complex Bayesian. There's a hidden probability in the simple case, because p(C) is encompassing both my belief in coins generally and also my belief about Tom's truthtelling. So really I have p(C) "p that the claim is true" but also p(S) "p that Tom stated the claim to me". Thus also p(S|C) "p that if the claim were true, Tom would state this claim to me" and p(C|S) "p of the claim being true given that Tom stated it to me"; but also the highly relevant p(S|not C) "p of that if the claim were NOT true, Tom would state this claim to me ANYWAY" and a few other variants. When you start doing Bayesian analysis with more than two variables you nearly always need to account for both p(A|B) and p(A|not B) for at least some of the cases, even where you could sometimes fudge this in the simpler problems.SO this brings us to a formulation of the original question as: what is the relationship between p(C|S,H) and p(C|S)? The former as p(H|C,S)p(C,S)/(p(C,S,H) + p(not C,S,H)) and then p(H|C,S)p(C,S)/(p(H|C,S)p(C,S) + p(H|not C,S)p(not C,S)) and if I take p(H|C,S) as 1 (given) and p(H|not C,S) as 1/2 (approximate), I'm left with p(C,S)/(p(C,S) + 0.5p(not C,S)) For the prior quantity p(C|S), a similar set of rewrites gives me p(C,S)/(p(C,S) + p(not C,S)) Now I'm in the home stretch, but I'm not done.Here we have to break down p(C,S) and p(not C,S). For p(C,S) we can use p(C)p(S|C), which is "very small" times "near 1", assuming Tom would be really likely to state that claim if it were true (wouldn't you want to show off your magic coin?). The other one's more interesting. We rewrite p(not C,S) as p(not C)p(S|not C), which is "near 1" times "is Tom just messing with me?".Because a crucial part of this analysis, which is missing in the hypothesis-test version or in the simpler Bayesian model, but "obvious" to anyone who approaches it from a more intuitive standpoint, is that it matters a lot whether you think Tom might be lying in the first place, and whether he's the sort that would state a claim like this just to get a reaction or whatever. In the case where you basically trust Tom ("he wouldn't say that unless he at least thought it to be true") then the terms of p(C,S) + p(not C,S) might be of comparable magnitude, and multiplying the second of them by 1/2 will have a noticeable effect. But if you think Tom likely to state a claim like this, even if false, just for effect (or any other reason), then p(C,S) + p(not C,S) is hugely dominated by that second term, which would be many orders of magnitude larger than the first, and so multiplying that second term by 1/2 is still going to leave it orders of magnitude larger, and the overall probabilityâ€”even with the extra evidenceâ€”remains negligible.[0] This clearly breaks if p(C) is higher than 1/2, because twice that is more than 1. If we assume that the prior p(H) is a distribution over coins, centred on the fair ones and with a long tail going out to near-certainty at both ends, the claim "this coin is an always-heads coin"[1] is removing a chunk of that distribution in the H direction, meaning that p(H|not C) is actually slightly, very slightly, greater than 1/2. This is the "fudge" I refer to above that lets me put the p(H) as 1/2. Clearly if my prior p(C) is higher than "very small" this would be inconsistent with the prior p(H) I've described.[1] I'm further assuming that "always" means "reallllllly close to always", because otherwise the claim is trivially false and the problem isn't very interesting.[2] Note that this is not actually a "naive Bayesian" approach---that's a technical term that means something more complicated.

 A little dense, sorry; I reposted on my blog with slightly better formatting (and at least breaking out some of the math onto separate lines):
 Nice walkthrough of bayesian probability, then bayesian epistemology. All that's missing is a link to http://yudkowsky.net/rational/technical for those who'd like furhter readings on the subject.
 This answer makes the most sense to me, though I dont understand it.Some quantity x which is your trust in Tom, plus some quantity y which is the probability of the coin multiplied by some factor of increasing confidence
 For a different approach: from a scientific/logical perspective the result has no bearing on the truth of his claim. The only possibility which we can remove from consideration is the claim that "The coin will always land tails."

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