Shroud of Turin redivivus

:joy: “Photographic quality” :laughing:

Here’s the shroud compared to a photograph and to artwork of the time.

Anyone who thinks the shroud image is of “photographic quality” (or even of more photographic quality than other pictures of the period) needs their head examined.

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But, you see, you are comparing them to the actual image on the “shroud”. Not to the digitally manipulated “negative” that is usually presented by believers, and which I bet most people think is what actually appears on the cloth.

While looking for ‘enhanced’ pictures of the shroud I found some-one claiming that the picture on the shroud is so good that not only can you see coins placed on Jesus’s eyes, but you can tell that the coins are Roman and even identify the coins used.

:nut_and_bolt: :peanuts: :chestnut: :coconut:

That’s probably considered another data point for the “Most Researched Artifact in History Ever!” claim.

When you take a picture of it the shroud is of photographic quality. When you show its negative image as you have it is not.

Left is the negative image on the Shroud. The right is the image after a photo is taken.

Let’s see. Below is the way I initially defined P(A)
Except that P(A) is not the probability that the codex shows the same features that are shown on the shroud, not at all. Rather, it is the probability that the codex displays the 6 features that O.K. is referring to in his piece.

Well. it seems that my formulation was somewhat clumsy, given that neither you nor @Gisteron got it.

So let me try this one:
P(A) is the probability that the codex displays the 6 features that O.K. is referring to in his piece by pure chance, independently of the shroud.
With this new, hopefully better definition, do you now agree that P(A) rises if the codex is based on the shroud but doesn’t if the shroud is based on the codex?

By your new definition, A is independent of whether or not the Codex is based on the Shroud. This new definition now explicitly states that P\left(A\right)=P\left(A\,\middle|\,H_1\right)=P\left(A\,\middle|\,H_2\right). This is what “independent” means: That the conditional probability for the two events so called are equal to their respective individual probabilities.

Can’t speak for Roy, but since you pinged me over this also (in a sentence saying I hadn’t gotten your previous description, when I did, and rephrased it for you, and pointed out what was flawed both with your formulation and with what you meant actually), I’ll throw in my humble opinion that, no, this “new and hopefully better definition” of yours now openly conflicts with what you say, instead of merely being an incomplete picture of the situation like before.

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