Cont: Proof of Immortality VIII

You have a history of not reading the articles that you cite and then being spanked with them later.

It's even deeper than that. Most of Jabba's errors are conceptual. He doesn't seem to know how to model something properly. He doesn't understand how the parts of a problem correspond to the elements of the notation in which the model is expressed, and conversely how the constraints in the method relate to requirements for the problem. But a curious thing happens when you identify this problem to him: he gives you the chapter-and-verse explanation of how the correct way is to do something. But he never goes on to fix the broken connections. It's as if being able to quote a correct statement somehow addresses whether some other proposition conforms to it. It's as if his argument today is, "Here is an introductory article on Bayesian methods. Therefore my proof is correct." The first statement is true, but it doesn't cure the problems with his proof.

Also today we see a familiar tactic in fringe argumentation. Jsfisher is being told he has to correct Jabba's proof using the proffered article. The question is not "How is my argument wrong?" -- which has already been copiously answered. It's "How does this article show that my argument is wrong?" which is a wholly different question. Limiting how one's critics may respond is a hallmark of dishonest debate. Limiting it to sources the claimant foists is even worse. This fits the pattern of demanding a specified rebuttal from existing authority, which Jabba has tried before. Jabba wants to know where in the literature it specifically forbids him from doing what he's doing. If it's wrong, as we claim, then he suggests there should be some specific page number in some authority that says in so many words, "You can't do this with that." It begs the question that the appropriate literature is, or contains, a comprehensive laundry list of dos and don'ts that could conceivably arise in any practical application of the science. It proposes to negate any refutation based on expertise and ad hoc derivation, which is what jsfisher and others have provided. Obviously that's not how knowledge works.

All of this is part of a larger scheme to limit debate, which in Jabba's case consists of this plus also limiting his responses to one person at a time, foisting ground rules, etc. He seems to want to subject his theory to enough discussion to purport that it has been thoroughly examined, but not enough to actually require it to be valid.
 
- I just took about 30 minutes halfway answering your request, and accidentally erased it. :eek:
- I'll try again.
- First, I assume that the symbol hilited above is a typo, and should have been "+."

- P(H|E) = P(E|H)*P(H)/(P(E|H)*P(H) + P(E|~H)P(~H))
- P(H|E) = 10-100 *.99/(10-100 *.99 + .62 * .01)
- P(H|E) = 10-100 /(10-100 + .0062)
- P(H|E) = 10-100 /.0062
- P(H|E) = 10-100- P(H|E) = 0
- And,
- P(~H|E) = .0062/(.0062 + 10-100*.99)
- P(~H|E) = 1.

- I accept that 10-100 isn't 0, but I do think that it's good enough for government work (I'm retired from government work, but you know what they say about old dogs).

Surely you have noticed that absolutely no one accepts your 'close enough for government work' figure?

Good sir, you really need to address that one. Your entire argument relies on it and no one is buying it
 
- First, I assume that the symbol hilited above is a typo, and should have been "+."

It's painfully obvious to the casual observer that this assumption is wrong; the expression is typed correctly, and means that both P(E|H) and P(E|~H) are equal to 1. If you tried typing values into the equation jt512 asked you to, you'd quickly see that this must be true. But instead...


- P(H|E) = P(E|H)*P(H)/(P(E|H)*P(H) + P(E|~H)P(~H))
- P(H|E) = 10-100 *.99/(10-100 *.99 + .62 * .01)
- P(H|E) = 10-100 /(10-100 + .0062)
- P(H|E) = 10-100 /.0062
- P(H|E) = 10-100- P(H|E) = 0
- And,
- P(~H|E) = .0062/(.0062 + 10-100*.99)
- P(~H|E) = 1.

... you've chosen, yet again, to make up some numbers and type them into a different equation. The only question at this stage is, who do you think you're fooling here?

- I accept that 10-100 isn't 0, but I do think that it's good enough for government work (I'm retired from government work, but you know what they say about old dogs).

10-100, government work notwithstanding, is neither of the two things you so desperately want it to be:
(a) A number so small that no other possible number can be smaller, and
(b) Based on any realistic calculation.
Your made-up example based on made-up numbers is therefore rejected. Again.

Dave
 
P(H|E) = 10-100...

We can stop right there. How did you calculate the highlighted number?

I accept that 10-100 isn't 0...

Your acceptance is irrelevant. 10-100 is not zero as a matter of arithmetic fact. You seem to insinuate that the number should be zero in the model, but you've provided no valid argument for that. In contrast you seem to be once again arguing for the nonsensical concept of "virtual zero," except giving us an arbitrary finite number that's supposed to represent the concept.

If this is what you're doing, then sorry, that's not how math works. If you claim the likelihood of your current existence given materialism is this finite number, then you must explain how you computed it. Suggesting that you chose it to be arbitrarily close to zero doesn't repair your proof, either algebraically or conceptually.

....but I do think that it's good enough for government work (I'm retired from government work, but you know what they say about old dogs).

Don't try to be folksy and distract from this serious conceptual problem in your proof. Your critics don't accept that 10-100 is "good enough" for any purpose, because you stubbornly refuse to explain where it came from. They are not obliged to agree with your foisted numbers, or with the larger problem that you think a valid proof would allow you to guess at numbers you don't know.
 
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So you can't say the body is irrelevant in one case and required in the other case. That would mean you had a different E for H than for ~H. You're changing horses. E, your current existence, necessarily includes your body regardless of the extraneous claims of H or ~H.
 
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- I accept that 10-100 isn't 0, but I do think that it's good enough for government work
Only if the government work needed is a bridge to be built and the government instead tries to build a tunnel to the moon.

(I'm retired from government work, but you know what they say about old dogs).
That they're unable to learn anything. Thank you for admitting what has been painfully obvious to everyone else.

How did you calculate 10-100? It looks as if you've simply made it up like someone who hasn't learned anything.
 
I accept that 10-100 isn't 0, but I do think that it's good enough for government work.

Only if the government work needed is a bridge to be built and the government instead tries to build a tunnel to the moon.

Dr. Gregory House: Oxygen saturation is 94%, check her heart.
Dr. Eric Foreman: Her oxygen saturation is normal.
Dr. Gregory House: It's off by one percentage point.
Dr. Eric Foreman: Within range. It's normal.
Dr. Gregory House: If her DNA was off by one percentage point, she'd be a dolphin.
 
Dave,
- If reincarnation is true, my self will inhabit many different bodies. And again, maybe brains receive, rather than produce their selves.

...

aaaaaaaaaaaand we're back to brain is a radio again. Endlessly asserted, never evidenced. Isn't the thread title "PROOF of immortality"?
 
aaaaaaaaaaaand we're back to brain is a radio again. Endlessly asserted, never evidenced. Isn't the thread title "PROOF of immortality"?

It's a never-ending shell game. To get past the abject lack of evidence for persistent souls, Jabba asserted that disembodied souls effectively have no attributes. No memory, no perception, no knowledge transfer, no nothing. They exist in name only between incarnations. That way he can say we should expect to see no evidence of them even though they may still be there. That fairly necessitates the body to be the seat of memory, perception, self-awareness, etc. All that stuff has to live somewhere, and if not in the soul then where?

Now confronted with the problem that arises from that and from his definition of "current existence" as requiring a specific body, he has to try to move all the necessary elements of self back into the soul. So now the body is just a lifeless mass until it gets fitted with a soul that brings with it all that the person needs to have memory, perceive, and all those other human activities.

As usual, Jabba frantically equivocates back and forth, trying to find language that obscures the necessary distinctions between elements of his model enough to fool someone.
 
- I just took about 30 minutes halfway answering your request, and accidentally erased it. :eek:
- I'll try again.
- First, I assume that the symbol hilited above is a typo, and should have been "+."

- P(H|E) = P(E|H)*P(H)/(P(E|H)*P(H) + P(E|~H)P(~H))
- P(H|E) = 10-100 *.99/(10-100 *.99 + .62 * .01)
- P(H|E) = 10-100 /(10-100 + .0062)
- P(H|E) = 10-100 /.0062
- P(H|E) = 10-100- P(H|E) = 0
- And,
- P(~H|E) = .0062/(.0062 + 10-100*.99)
- P(~H|E) = 1.

- I accept that 10-100 isn't 0, but I do think that it's good enough for government work (I'm retired from government work, but you know what they say about old dogs).

No, it isn't. How did you calculate 10-100?
 
Jabba,

It is true.

P(E) is the denominator in Bayes' Theorem. Since you like the word "likelihood" so much, you'll be happy to know that P(E) is called the marginal likelihood. It is the weighted average of P(E|H) and P(E|~H), where the weights are P(H) and P(~H), respectively. Thus the denominator in Bayes' Theorem is

P(E) = P(E|H)P(H) + P(E|~H)P(~H).

You have stated that P(E) = 1 and that P(H) and P(~H) are both non-zero. If you deny that that implies that P(E|H) = P(E|~H) = 1, then, in the equation below, plug in your favorite values for P(H) and P(~H) and find any values other than 1 for P(E|H) and P(E|~H) for which the equation (below) is true. We will be looking forward to your response.

1 = P(E|H)P(H) + P(E|~H)P(~H).

- I just took about 30 minutes halfway answering your request, and accidentally erased it. :eek:
- I'll try again.
- First, I assume that the symbol hilited above is a typo, and should have been "+."

- P(H|E) = P(E|H)*P(H)/(P(E|H)*P(H) + P(E|~H)P(~H))
- P(H|E) = 10-100 *.99/(10-100 *.99 + .62 * .01)
- P(H|E) = 10-100 /(10-100 + .0062)
- P(H|E) = 10-100 /.0062
- P(H|E) = 10-100- P(H|E) = 0
- And,
- P(~H|E) = .0062/(.0062 + 10-100*.99)
- P(~H|E) = 1.

- I accept that 10-100 isn't 0, but I do think that it's good enough for government work (I'm retired from government work, but you know what they say about old dogs).

No, it was not a typo, and your response does not address the contradiction in your work. To repeat, for, i don't know, the seventh time: You claim that P(E) = 1, and that P(H) and P(~H) are both non-zero. Together, those things imply that P(E|H) = P(E|~H) = 1 (that means both equal 1). But you claim that P(E|H) and P(E|~H) are both less than 1. To disprove the contradiction, you have to find two numbers, P(E|H) and P(E|~H), not both equal to 1, that satisfy the following equation. Recalling that P(E) is the denominator of Bayes' Theorem, we have (again):

P(E) = P(E|H)P(H) + P(E|~H)P(~H) .

Using your values for P(E), P(H) and P(~H), we have,

1 = P(E|H)(.99) + P(E|~H)(.01) .

If we let P(E|H) = P(E|~H) = 1, then the equation is satisified, but then the evidence of your existence is the same under both hypotheses, and so is not evidence for either hypothesis over the other, ie, your existence does not discriminate betwen the hypotheses. So, for your mathematics to be coherent and for E to have any evidential value, you have to find two numbers, P(E|H) and P(E|~H) that aren't both 1 which make the above equation true. We will be looking forward to your response.
 
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No, it was not a typo, and your response does not address the contradiction in your work. To repeat, for, i don't know, the seventh time: You claim that P(E) = 1, and that P(H) and P(~H) are both non-zero. Together, those things imply that P(E|H) = P(E|~H) = 1 (that means both equal 1). But you claim that P(E|H) and P(E|~H) are both less than 1. To disprove the contradiction, you have to find two numbers, P(E|H) and P(E|~H), not both equal to 1, that satisfy the following equation. Recalling that P(E) is the denominator of Bayes' Theorem, we have (again):

P(E) = P(E|H)P(H) + P(E|~H)P(~H) .

Using your values for P(E), P(H) and P(~H), we have,

1 = P(E|H)(.99) + P(E|~H)(.01) .

If we let P(E|H) = P(E|~H) = 1, then the equation is satisified, but then the evidence of your existence is the same under both hypotheses, and so is not evidence for either hypothesis over the other, ie, your existence does not discriminate betwen the hypotheses. So, for your mathematics to be coherent and for E to have any evidential value, you have to find two numbers, P(E|H) and P(E|~H) that aren't both 1 which make the above equation true. We will be looking forward to your response.
jt,
- I don't perceive the implication that you do. To me P(E|H) + P(E|~H) = 1...
 
It's actually a pretty huge number, when you think about it. There's an infinite quantity of numbers smaller than that.

Couple of years back I stumbled across Vsauce's breakdown of varies "kinds" of infinity and was a legit eye opener for me. Things like the Aleph numbers, countable versus uncountable infinity... it gets really weird.

It really just hammers home the utter absurdity of trying to argue differences of scale on the level Jabba is.
 
I don't perceive the implication that you do.

You don't perceive a lot of the implications that other people do. That's your biggest problem. You've convinced yourself that you're not only competent in this field but exemplary. When people correct you, you insult them by spouting Googled-up truisms and ignoring what they're trying to tell you. For any practical purpose relating to your proof, you are unteachable.

To me P(E|H) + P(E|~H) = 1...

And that's why you don't understand how E, your "current existence," is useless for distinguishing between H and ~H. You like likelihoods. Are P(E|H) and P(E|~H) likelihoods? If so, must they sum to 1? Must they, in fact, have any reciprocal relationship?
 
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