[Math] Help simplifying Bayes’ theorem for multiple conditions

bayes-theoremprobability

I'm trying to use Bayes' theorem to calculate the likelihood, in my example, of religion H being true considering evidence for and against it. I think I understand Bayes' theorem well enough to do this in two steps, calculating the odds that H is true considering evidence against it, and then taking that result as a new prior probability to factor in evidence for it.

I'm sure there's a way to combine those two steps into one, but the explanations I found were not that clear and used a notation that I'm not very familiar with.

So for example prior probability H is true due to the number of possible true religions P(H) is 0.001, the likelihood we'd find particular arguments against it if it were true P(Aa|H) is 0.01, the likelihood that we'd find particular arguments against it if it were not true P(Aa|H') is 0.6, the likelihood we'd find particular arguments for it if it were true P(Af|H) is 0.8, and the likelihood that we'd find particular arguments for it if it were not true P(Af|H') is 0.2. And in my example the arguments against it and arguments for it are both observed.

So then I do the math in two stages, first considering arguments against it:

 P(H|Aa) = P(Aa|H) * P(H) / [P(Aa|H) * P(H) + P(Aa|H’) * P(H’)]
P(H|Aa) = 0.01 * 0.001 / [0.01 * 0.001 + 0.6 * 0.999]
P(H|Aa) = 0.000017

So now I take that and calculate considering arguments for it:

 P(H|Af) = P(Af|H) * P(H) / [P(Af|H) * P(H) + P(Af|H’) * P(H’)]
P(H|Af) = 0.8 * 0.000017 / [0.8 * 0.000017 + 0.2 * 0.999983]
P(H|Af) = 0.000068

This gives me a 0.0068% probability that the religion is true given the above assumptions. (I hope I didn't mess up that math.)

But what would be a single equation to do both steps at once?

Edit: I think I may have found the simplification:

P(H|Aa, Af) = P(Aa|H) * P(Af|H) * P(H) / [P(Aa|H) * P(Af|H) * P(H) + P(Aa|H’) * P(Af|H’) * P(H’)]

P(H|Aa, Af) = 0.01 * 0.8 * 0.001 / [0.01 * 0.8 * 0.001 + 0.6 * 0.2 * 0.999]

P(H|Aa, Af) = 0.000067

Which is the same result (slight difference probably due to my rounding in the intermediate step of the separate calculations). Is there any reason I shouldn't be doing this? I am looking at evidence for and against as "if H is true, how likely would I expect to see Aa? Not likely. But Aa is present, so that's new information."

Best Answer

Maybe this general formula for multiple conditions helps?

$$P(A|B, C)= \frac{P(B|A, C) * P(A|C)}{P(B|C)} $$

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