Probability about Jesus (Christ) existence on earth

[Peter Kirby]

I still do not know how any of the conditional probability equations can apply.

When trying to compute the probability of the intersection of multiple events, the multiplication rule always, always, always applies. Specifically, the multiplication rule that can be referred to as "the conditional probability equations" always applies.

In the case of "independent" events, the "conditional probability equation" multiplication rule still applies.

For example, consider the chance that I get heads three times in a row. That is,

P(H three times in a row) = P(H1) * P(H2 | H1) * P(H3 | H1 & H2)

What is the probability of getting heads on the first toss? Assuming a fair coin, 0.5.

What is the probability of getting heads on the second toss, given that I got heads on the first toss? Assuming a fair coin, 0.5.

What is the probability of getting heads on the third toss, given that I got heads on the first and second? Assuming a fair coin, 0.5.

What is the probability of getting heads three times in a row? Assuming a fair coin, 0.125.

When dealing with "independent" events, one definition is e.g. that P(A|B) = P(A), which allows the substitution of P(A) for P(A|B) in the multiplication rule. This is how you derive the "special" version of the equation for "independent" events. You still can use the general form, which is always true.



https://people.richland.edu/james/lectu ... 5-rul.html

The following four statements are equivalent

A and B are independent events
P(A and B) = P(A) * P(B)
P(A|B) = P(A)
P(B|A) = P(B)

If someone thinks that (e.g.) six existing interpolations would influence an estimate of the 7th's probability, then they don't think they are "independent events." If someone thinks that what they're doing disagrees with the "conditional probability equations," they are doing it wrong.

https://people.richland.edu/james/lectu ... 5-rul.html

General Multiplication Rule

Always works.

P(A and B) = P(A) * P(B|A)

No exceptions.

[Bernard Muller]

to Peter,

Let's work with all probabilities being 80% implying (past) Jesus' existence (as an earthy human). The remaining 20% are about NOT implying Jesus' existence (which can be caused by either interpolation or interpretation or unreliability).
Just looking at three of those, and according to the equation P = 1 - (1-0.8)^3, we have 99.2% as the overall result.

Now let's say that the remaining 20% assigned is about the probability of only interpolation (that's generous because I take away the possibility of (wrong) interpretation and unreliability).
So again using the same equation P = 1 -[(1-0.2)^3] we have 48.5 % as the overall probability for interpolation.

That means that for 3 events, we have one chance out of two to have an interpolation, therefore it would take an average of 6 events to have likely an interpolation in one of them. That's generous because P = 1 - [(1-0.2)^6] gives us 74% only.

So the domino effect you are defending would only be applicable to 1 event for every group of 6 events, with very little lowering of the overall probability (see later).

What if the probabilities were 50%, with same calculation?
For 3 events, we would have an overall result of 87.5% (with only two events: 75%), but only 2 of these events would be necessary in order to have likely an interpolation in one of them. The lowering of the overall probability would be then more significant than for my first case.

Let's now look at your domino effect for one likely interpolation for every group of 6 events (all of them with 80% probabilities).
first event 80%, second event: 80% - (20% * 80%) = 64%, third event: 64% - (36% * 80%) = 35%, fourth event: 35% - (65% * 80%) = 0%, fifth & sixth events = 0%.
You may not agree how I calculate the lowering factor, but I think it is rather severe.
Anyway, the overall result for P = 1 - [(1-p1)*(1-p2)*...*(1-p6)] is 66.7%

So for that group of 6 events we have: P = 1 - [(1-0.8)^5 * (1-0.667)] = 99.99%

BTW, on your example, at the bottom of your graphic, you wrote:
"0.2 * 0.5 * 0.75 * 0.8 = 0.06" but according to the equation P = 1 - [(1-p1)*(1-p2)*...*(1-pN)] the overall probability is 94%.

Cordially, Bernard

[Peter Kirby]

BTW, on your example, at the bottom of your graphic, you wrote:
"0.2 * 0.5 * 0.75 * 0.8 = 0.06" but according to the equation P = 1 - [(1-p1)*(1-p2)*...*(1-pN)] the overall probability is 94%.

Right. You need to read what it is saying. The 6% chance is for four interpolations. The formula with the logical complement (with the 94% chance) is, in that example anyway, for at least one non-interpolation (out of the four). The two statements are logically equivalent.

As to the rest of it, maybe you've finally gotten the principle (though, unfortunately, it doesn't appear so). That was the point. The point was not to argue over which numbers to use. Do you finally get it? If so, good. If not, well, I'm sure some people got it.

[Peter Kirby]

Very disheartening, really. I thought the pictures would help.

[Peter Kirby]

P = 1 - (1-0.8)^3.

No.

P = 1 -[(1-0.2)^3]

No.

P = 1 - [(1-0.2)^6]

No.

P = 1 - [(1-0.8)^5 * (1-0.667)] = 99.99%

No.

All of these are wrong. You're still following the "bad" method in the first graphic. In the last case, you've modified the "bad" method in your own manner -- to invent your own "bad" method, or to make the most meaningless possible adjustment.

[Peter Kirby]

Let's work with all probabilities being 80% implying (past) Jesus' existence (as an earthy human). The remaining 20% are about NOT implying Jesus' existence (which can be caused by either interpolation or interpretation or unreliability).

Honestly, you're just not ready even to try doing something this complex, mathematically. You must walk before you try to run. There's no evidence you even understand the probability concepts involved here, despite copious links, videos, etc. That needs to be addressed first.

I totally agree that the situation would become very complex very fast, if you try to do what you want to do. I commented on this earlier in the thread, which comments this post built on. You're simply not ready to deal with that kind of complexity using mathematical probability. I think you're more than capable of dealing with it intuitively, as long as your intuition isn't being marred by an incomplete understanding of probability.

(And, of course, the numbers are made up... there's that issue.)

[Peter Kirby]

Let's now look at your domino effect for one likely interpolation for every group of 6 events (all of them with 80% probabilities).
first event 80%, second event: 80% - (20% * 80%) = 64%, third event: 64% - (36% * 80%) = 35%, fourth event: 35% - (65% * 80%) = 0%, fifth & sixth events = 0%.
You may not agree how I calculate the lowering factor, but I think it is rather severe.
Anyway, the overall result for P = 1 - [(1-p1)*(1-p2)*...*(1-p6)] is 66.7%

So for that group of 6 events we have: P = 1 - [(1-0.8)^5 * (1-0.667)] = 99.99%

This is just bizarre. You're inventing your own methods and theory again (not as cool as it sounds). Apparently, so far, you've haven't understood.

[Peter Kirby]

That means that for 3 events, we have one chance out of two to have an interpolation, therefore it would take an average of 6 events to have likely an interpolation in one of them. That's generous because P = 1 - [(1-0.2)^6] gives us 74% only.

So the domino effect you are defending would only be applicable to 1 event for every group of 6 events, with very little lowering of the overall probability (see later).

Also quite bizarre (well, part of the same bizarre train of thought), and has nothing to do with anything I said or with mathematical probability, except insofar as it is a strange distortion of it.

So the domino effect you are defending would only be applicable to 1 event for every group of 6 events

Please, no. Stop making stuff up. Don't even take my word for it. Please read a book, watch a few videos, gain an understanding.

[Peter Kirby]

So the domino effect you are defending would only be applicable to 1 event for every group of 6 events

Just so you know, the so-called "effect" is not a random event. I also doubt that whatever you understand by the words (your words) "domino effect" accurately represent what is simply called a conditional probability. A conditional probability is "applicable" to all events (in the equation), after the first event considered, which has nothing to condition on. Go read the equation. There's no ambiguity about it.

It's okay not to understand something. It's a little perverse to spend more time inventing weird ways to be wrong than trying to understand.

[Peter Kirby]

This is just very sad and frustrating. I don't understand this willingness to make up stuff when you don't even know the basics of what you're talking about. You can go talk to someone else and ask them for help with some kind of understanding. Clearly you don't trust me well enough to be honest about this topic, which I know very well since I am studying mathematics with an emphasis in probability and stats at the university (or you think I don't know what I'm talking about, take your pick). Apparently I must be BS-ing you in order to score points, in your mind -- or maybe you think I'm confused and you have it figured out, somehow. (Or maybe all the mathematicians in the world are confused, and you've got it figured out.) In any case, there doesn't seem to be any trust in the fact that I am representing the subject faithfully and accurately.

So, I'm done with this thread.