Nah. I think you're wrong.
I'm not. You may have forgotten all those statisticians you consulted who told you you were wrong, but your critics haven't.
In our formula, a and b are both accepted as true.
No, you're conflating two different concepts which you've muddled by what you admit was uncertain notation. We're not using Bayes' theorem in our illustration. This is just pure joint probability. What you're trying to write as P(c|a,b) is more properly written in my illustration as P(c) = P(a,b).
In the materialist hypothesis, P(c) = P(a). In your hypothesis, P(c) = P(a,b). P(a,b) can never exceed P(a). You argue it does.
But since you mentioned it, in the Bayesian formulation one computes the likelihood ratio P(E|H)/P(E|~H) based on assuming the conditions have been satisfied. But as you were told many times, you can't guess at both the priors
and the likelihood ratio. You have no actual numbers for any of it. If your priors are known data, then you can informally specify the likelihood ratio. But if your priors are just subjective guesses, then you must actually compute the likelihood ratio -- that is, you must have an objectively correct computation for P(E|H) and P(E|~H) which you don't have. If your P(E|~H) is really then to be written as P(E|a,b), where a and b are independent events that occur in ~H, then there is no way to get around having to compute P(a) and P(b).
This is a hypothetical...
Not in the illustration jond and I have been trying to get you to look at. We're using this formulation to try to get you to see how conditional probability works at a basic level. You're still stuck in Bayes, which we already know you don't understand.
You, or someone else, needs to show me where I'm wrong.
Where you changed horses and went from the illustration jond and I presented to Bayes, which is irrelevant for the moment.
