Probability about Jesus (Christ) existence on earth

[Bernard Muller]

P(A) = 0.9 - "the scenario where interpolations had occurred each with 10% probability" (chance interpolations)
P(B) = 0.1 - "the scenario where interpolations had occurred each with 100% probability" (orchestrated interpolations)
P(N1 | ~X1) = 0.95 - "95% chance of no evidence if not interpolated"
P(~N1 | ~X1) = 0.05 - "5% chance of evidence if not interpolated"
P(N1 | X1) = 0.0 - "0% chance of no evidence if interpolated"
P(~N1 | X1) = 1.0 - "100% chance of evidence if interpolated"

For 8 X's & 7 ~N's: 0.0%
For 8 X's & 6 ~N's: 0.0%
For 8 X's & 5 ~N's: 0.0%
For 8 X's & 4 ~N's: 0.0%
For 8 X's & 3 ~N's: 0.0%
For 8 X's & 2 ~N's: 0.0%
For 8 X's & 1 ~N's: 0.0%
For 8 X's & 0 ~N's: 0.0%

It looks if we have strong evidence all the items are interpolated, then the algorithm shows we do not need to assume to have priorly known for sure interpolations for any item. But the algorithm shows no possibilty the 8 items would be all interpolations which conflicts with the statement of the first sentence.

Keeping the same inputs, except changing two as such:
P(N1 | X1) = 1 - "100% chance of no evidence if interpolated"
P(~N1 | X1) = 0.0- "0% chance of evidence if interpolated"

For 8 X's & 7 ~N's: 0.000%
For 8 X's & 6 ~N's: 0.000%
For 8 X's & 5 ~N's: 0.000%
For 8 X's & 4 ~N's: 0.000%
For 8 X's & 3 ~N's: 0.000%
For 8 X's & 2 ~N's: 0.000%
For 8 X's & 1 ~N's: 0.000%
For 8 X's & 0 ~N's: 13.83%

This case is absurd but there is a glitch for zero known for sure interpolations.

Keeping the same inputs, except changing two as such:
P(N1 | X1) = 0.9 - "90% chance of no evidence if interpolated"
P(~N1 | X1) = 0.1- "10% chance of evidence if interpolated"

For 8 X's & 7 ~N's: 87.42%
For 8 X's & 6 ~N's: 78.45%
For 8 X's & 5 ~N's: 65.60%
For 8 X's & 4 ~N's: 50.03%
For 8 X's & 3 ~N's: 34.72%
For 8 X's & 2 ~N's: 23.91%
For 8 X's & 1 ~N's: 27.06%
For 8 X's & 0 ~N's: 90.77%

Again, a glitch for to the end for zero known for sure interpolations. Also a 10% difference for P(N1 | X1) makes huge variation in the output.

Going back to:
P(A) = 0.9 - "the scenario where interpolations had occurred each with 10% probability" (chance interpolations)
P(B) = 0.1 - "the scenario where interpolations had occurred each with 100% probability" (orchestrated interpolations)
P(N1 | ~X1) = 0.95 - "95% chance of no evidence if not interpolated"
P(~N1 | ~X1) = 0.05 - "5% chance of evidence if not interpolated"
P(N1 | X1) = 0.8 - "80% chance of no evidence if interpolated"
P(~N1 | X1) = 0.2 - "20% chance of evidence if interpolated"

But changing P(A) to 0.99 and P(B) to 0.01
For 8 X's & 7 ~N's: 95.76%
For 8 X's & 6 ~N's: 86.25%
For 8 X's & 5 ~N's: 63.57%
For 8 X's & 4 ~N's: 32.67%
For 8 X's & 3 ~N's: 11.89%
For 8 X's & 2 ~N's: 3.62%
For 8 X's & 1 ~N's: 1.03%
For 8 X's & 0 ~N's: 0.29%

With assuming only a probability of 1% for all items to be interpolations, we need at least 5 items to be priorly known for sure to be interpolations.

Finally with the set of inputs data which I would find the most realistic, as such:
P(A) = 0.99 - "the scenario where interpolations had occurred each with 10% probability" (chance interpolations)
P(B) = 0.01 - "the scenario where interpolations had occurred each with 100% probability" (orchestrated interpolations)
P(N1 | ~X1) = 0.95 - "95% chance of no evidence if not interpolated"
P(~N1 | ~X1) = 0.05 - "5% chance of evidence if not interpolated"
P(N1 | X1) = 0.2 - "20% chance of no evidence if interpolated"
P(~N1 | X1) = 0.8 - "80% chance of evidence if interpolated"

For 8 X's & 7 ~N's: 99.90%
For 8 X's & 6 ~N's: 97.32%
For 8 X's & 5 ~N's: 56.43%
For 8 X's & 4 ~N's: 4.42%
For 8 X's & 3 ~N's: 1.65%
For 8 X's & 2 ~N's: 0.0%
For 8 X's & 1 ~N's: 0.0%
For 8 X's & 0 ~N's: 0.0%

We need to assume priorly at least 5 Items to be known for sure interpolations in order to presume the other items are also interpolations (even if they don't have evidence showing that).

Cordially, Bernard

[MrMacSon]

P(A) = 0.9 - "the scenario where interpolations had occurred each with 10% probability" (chance interpolations)
P(B) = 0.1 - "the scenario where interpolations had occurred each with 100% probability" (orchestrated interpolations)
.

• ??
  
  How does 0.9 = 10% probability ?? or 0.1 = 100% probability ??

[Bernard Muller]

How does 0.9 = 10% probability ?? or 0.1 = 100% probability ??

That's not from me, but Peter.
I understand by that it is assumed there is a 10%/100% probability of known for sure interpolations whenever they occur.
But Peter will surely have a better answer: he is the one who wrote first: " "the scenario where interpolations had occurred each with 100% probability" (orchestrated interpolations)"

Which makes me ponder, what happens if a mythicist thinks there is 50% chance of the "the scenario where interpolations had occurred each with 100% probability" (orchestrated interpolations)"

Going back to:
P(A) = 0.9 - "the scenario where interpolations had occurred each with 10% probability" (chance interpolations)
P(B) = 0.1 - "the scenario where interpolations had occurred each with 100% probability" (orchestrated interpolations)
P(N1 | ~X1) = 0.95 - "95% chance of no evidence if not interpolated"
P(~N1 | ~X1) = 0.05 - "5% chance of evidence if not interpolated"
P(N1 | X1) = 0.8 - "80% chance of no evidence if interpolated"
P(~N1 | X1) = 0.2 - "20% chance of evidence if interpolated"

But changing P(A) to 0.5 and P(B) to 0.5
For 8 X's & 7 ~N's: 99.95%
For 8 X's & 6 ~N's: 99.84%
For 8 X's & 5 ~N's: 99.42%
For 8 X's & 4 ~N's: 97.96%
For 8 X's & 3 ~N's: 93.04%
For 8 X's & 2 ~N's: 78.79%
For 8 X's & 1 ~N's: 50.81%
For 8 X's & 0 ~N's: 22.23%

But 50% might be not enough for those who wishes (or "know") all these 8 items are all interpolations.
So now,
Let's change P(A) to 0.2 and P(B) to 0.8 (80%)
For 8 X's & 7 ~N's: 99.99%
For 8 X's & 6 ~N's: 99.96%
For 8 X's & 5 ~N's: 99.86%
For 8 X's & 4 ~N's: 99.48%
For 8 X's & 3 ~N's: 98.16%
For 8 X's & 2 ~N's: 93.69%
For 8 X's & 1 ~N's: 80.51%
For 8 X's & 0 ~N's: 53.64%

Cordially, Bernard

[Bernard Muller]

to Peter,
I took the liberty to revise your definition of your inputs. I hope they are correct.

X's: suspected interpolations (which all imply a common claim).
N (among the X's): by considering the evidence, more likely to be NOT an interpolation than one.
~N (among the X's): by considering the evidence, more likely to be an interpolation than none.

P(A) = "Probability of the scenario where interpolations had occurred through several interpolators"
P(B) = 1 - P(A) = "Probability of the scenario where interpolations had occurred through the same interpolator"

P(N1 | ~X1) = "chance of no or little evidence if not interpolated"
P(~N1 | ~X1) = 1 - P(N1 | ~X1) = "chance of significant evidence if not interpolated"
P(N1 | X1) = "chance of no or little evidence if interpolated"
P(~N1 | X1) = 1 - P(~N1 | ~X1) ="chance of significant evidence if interpolated"

P(X1 I A) = average probability of all N's in A scenario (<0.5)
P(X1 I A) = 1 – P(X1 I A) = average probability of all ~N's in A scenario (>0.5)
P(X1 I B) = average probability of all ~N's in A scenario (>0.5)

Here is my spreadsheet with equations according to OpenOffice Apache



Cordially, Bernard