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The Outsider Test for Faith via Bayes

09 Jan

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I posted the following comment over at Richard Carrier’s blog. I thought it was relevant to my blog, so I’m reposting it here.

I arrived at the Outsider Test for Faith (OFT) looking at BT from a different angle. What it is basically trying to point out is that there’s no necessary relationship between being born into a Christian household (i.e. your parents, friends, etc. are all Christians) which is a pretty strong determining factor in most people being Christians, and Christianity actually being true. The evidence that we have is the geographical distribution of the various religions, in this case I thought that P(E) should be the probability, or percentage, of Christians who were raised as Christians; whose family, friends, etc. are also Christians. This would mean that P(E) is somewhere around 30%. 33% of the world is Christian and there are obviously some Christians who converted to Christianity from other religions.

P(E | H), then, would be the probability of the evidence given that Christianity is true, and P(E | ~H) would be the probability of the evidence at hand given some other reason (e.g. Islam is true, atheism is true, etc.). Of course, P(E) = P(E | H)*P(H) + P(E | ~H)*P(~H), the denominator for BT. I even assumed for the sake of argument that P(H), the probability that Christianity is true, is some ridiculously high number like 97% just to prevent Christians from saying that I’m biasing the argument against Christianity. So if P(H) is 97% then P(~H) is 3%. Now we have P(E) = P(E | H)*P(H) + P(E | ~H)*P(~H), or 30% = P(E | H) * 97% + P(E | ~H) * 3%. If we assume that P(E | H) is a high number, again, to bias towards Christianity being true, we end up skewing things ridiculously. The equation would then become a simple enough algebraic one to solve in the manner of 3 = 97 + .3x, but with P(E | H) being 1, it forces P(E | ~H) to be a negative number, which makes no sense.

Playing around with different values for both P(E | H) and P(E | ~H), it would seem to me that the only fair values would be to have P(E | H) = P(E | ~H), which is statistical independence. It literally means that whatever religion you were raised in has absolutely no relationship with that religion being true; it also means that it has no bearing on whether your religion is false. For both, you would need some other evidence to update your prior against. Of course, if a Christian appealed to something like “it just feels true” or “I had an experience I can’t explain, therefore Christianity is true” other religions use the same exact sort of argument, so we would end up in the same situation that P(E | H) = P(E | ~H) with E being the religious experience and H being the probability that the religion is true. So again, you would need some other evidence to update against which is basically the fundamental premise behind the OTF.

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Posted by on January 9, 2013 in Bayes

 

One response to “The Outsider Test for Faith via Bayes

 
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