[Merged] Immortality & Bayesian Statistics

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Jabba said:
Squeegee,

- I've been busy – and, responding to your post isn't easy. I've been working on a response for a few days (off and on) now, and have decided to let you know that I am working on it, and also to ask for patience.
- For now, I want to point out that I do understand that the probability of very specific events is always very small -- but then, I've been admitting that all along. Go back to post #82 (below), and you'll see what I mean.

Say that you find a deck of cards in the closet and decide to play some solitaire or something.

You sit down at the table and turn over the first card. It's an ace of spades. You place the ace back in the deck, shuffle the cards and once again, turn over the first card. This time, it's the ace of diamonds. Hmm. So, you try the same thing again. This time, you get the ace of spades again.

'Wait a minute…' You do it one more time, and this time, you get the ace of hearts.

If you’re paying attention, you’re growing suspicious about this deck you found in the closet. You’re starting to suspect that you don’t have the ordinary deck that you had assumed. But, why is that? Why are you suspicious?

You’re suspicious because the probability of drawing that 'hand' is so small if the deck is a normal deck.

Let’s try that again. But, this time, the first card you draw is a 3 of diamonds, the second is a
Jack of spades, the third is a 9 of clubs and the fourth is a 9 of hearts. In this case, you probably are not suspicious.

But, of course you realize that the prrobability of drawing that hand, given a normal deck, is just as small as the probability of drawing that previous hand…

So, what’s the problem here? Why are you not suspicious of this deck, when you were suspicious of the first one?

It turns out that there are two factors causing you to be suspicious of that first deck -- and one is missing in regard to the second deck. There is nothing about the second hand that sets it apart in such a way as to suggest another plausible hypothesis… If there were, you’d be suspicious of that second deck as well. It’s as simple as that…

- But then, I need to admit that you guys have, indeed, shaken my confidence in my Bayesian "proof"... I thought that I had an easy way around that problem, but now think that if I do have a way around that problem, it isn't easy...
- I'll be back.

--- Jabba
Click to expand...

Rich:

You tried to make a joke out of this, the last time I pointed it out to you...but until and unless you get a grasp on the idea that the likelihood of drawing four aces is exactly the same as the likelihood of drawing 2, 7, J, Q; or 4, 6, 8, 9; or any other arrangement of 4 cards. The only reason a hand of 4 aces seems "special" to you is that the rules of many different card games assign a significance, or an importance, to a hand of 4 aces that is independent of the probability of it being drawn.

You continue to confuse "having drawn a hand" with "predicting what hand will be drawn".
 
Humots said:
What do you mean by "counting k twice in each probability"?

What do you mean by "the k is already counted in P(R) and P(NR)"?

For that matter, what do you mean by "counting" and "counted"?

In addition, please indicate where I do these things in my argument.
Click to expand...
Jay,
- If you're still around, do you understand what I was saying about "counting k twice"? And if so, do you agree?
--- Jabba
 
Humots said:
What do you mean by "counting k twice in each probability"?

What do you mean by "the k is already counted in P(R) and P(NR)"?

For that matter, what do you mean by "counting" and "counted"?

In addition, please indicate where I do these things in my argument.
Click to expand...
Humots,

- By saying that k is already counted, I mean that it is the background info that determines what P(NR) and P(R) are. When you say,
P(R|k) = .01 * .01 / (.01 * .01 + .99 * .99) = .0001, all the numbers are the result of k, so you shouldn't be multiplying them by each other.

--- Jabba
 
Jabba said:
Humots,

- By saying that k is already counted, I mean that it is the background info that determines what P(NR) and P(R) are. When you say,
P(R|k) = .01 * .01 / (.01 * .01 + .99 * .99) = .0001, all the numbers are the result of k, so you shouldn't be multiplying them by each other.

--- Jabba
Click to expand...

No.

Say k is "all knowledge" and P(k) is the probability that k is true.

If you disagree that P(k) is the probability that all knowledge is true, then please give your definition.

k is not just background info, k includes
  1. all of the things that NR says is true and R says is false
  2. all the things that NR says is false and R says is true
such as Evolution.

k is the data supporting or not supporting the NR and R hypotheses.

In the statement P(H1|D) and P(H2|D), H1 is the NR hypothesis, H2 is the R hypothesis, and D is k.

So we ask, what is P(NR|k) and P(R|k)?

Note: I am trying to show that I can arrive at a "proof" of something Jabba would not accept by using Jabba's logic. How am I doing?
 
Humots said:
Jabba, you are saying:

1. P(NR|me) = P(me|NR)P(NR)/(P(me|NR)P(NR)+P(me|R)P(R)), and
  • P(NR) = .99
  • P(me|R) = .05
  • P(R) = .01

(2. - 5.) P(me|NR) approaches 0.

So, P(NR|me) approaches 0.

I wonder, what is P(NR|k), the probability that the Non-Religious Hypothesis is true given all background knowledge? Well,

P(NR|k) = P(k|NR)P(NR)/(P(k|NR)P(NR)+P(k|R)P(R))

From your own argument:
  • P(NR) = .99
  • P(R) = .01

Then for k = all background knowledge:
  • P(k|R) = .01 or less, since if the Religious hypothesis is true, most background knowledge (including evolution) is false
  • P(k|NR) = .99 or more, since P(NR) + P(R) = 1.0, again from your own argument that R and NR are a binary partition

We obtain
P(NR|k) = 0.99 * 0.99/0(.99 * 0.99 +0.01 * 0.01) = 0.9801 / (0.9801 + .0001) = 0.9999

What about P(R|k), the probability that the Religious Hypothesis is true given all background knowledge?

Using P(R|k) = P(k|R)P(R)/(P(k|R)P(R)+P(k|NR)P(NR))

we get, using the same values,

P(R|k) = .01 * .01 / (.01 * .01 + .99 * .99) = .0001

which is consistent with P(NR) + P(R) = 1.0.

So, what’s wrong with this picture?
Click to expand...


What is wrong with that picture is that you are using Jabba's prior probabilities as your prior probabilities (see the bolded text), when, in fact, Jabba's prior probabilities are your posterior probabilities.

Jay
 
jt512 said:
You're using a very strange deck of cards.
Click to expand...
dafydd said:
There are only four aces, but there are four twos, four sevens, four jacks and four queens.
Click to expand...

How so? Oh, right. A,A,A,A has the same probability as 2C, 7D, JH, QS...my bad.

Hasty pre-coffee posting....
 
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Slowvehicle said:
No, Daffyd, you were right. I did not, originally, specify suit...which completely concealed my point. I added suits to reflect the corrections.
Click to expand...

Ok, I was busy and I thought that I'd overlooked it.
 
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jt512 said:
What is wrong with that picture is that you are using Jabba's prior probabilities as your prior probabilities (see the bolded text), when, in fact, Jabba's prior probabilities are your posterior probabilities.

Jay
Click to expand...

Jay: in your view,

Jabba's (and my) prior probabilities are
  1. P(NR) = .99
  2. P(R) = .01
and my posterior probabilities are:
  1. P(NR|k) = 0.9999
  2. P(R|k) = .0001
Is this correct?

If so, do you mean that given the prior probabilities I used, my posterior probabilities are no surprise?

If you do, then I agree. All I'm trying to do is show Jabba that I can come to a conclusion he will not accept by using his own reasoning. Which is pure numerology.

Also, if P(R) = .01 and P(R|k) = .0001, doesn't that make the Religious hypothesis even more unlikely if all knowledge is given (again using Jabba's own reasoning)?
 
Jabba said:
carlitos,
- Sorry about that.
- When you get to be 70, I bet you have the same problem.
--- Jabba
Click to expand...

If the "same problem" is a complete inability to use logic, read for comprehension and such, you'd lose that bet. I wasn't just commenting on you getting a name wrong. You've gotten just about everything wrong here.
 
Humots said:
Jay: in your view,

Jabba's (and my) prior probabilities are
  1. P(NR) = .99
  2. P(R) = .01
and my posterior probabilities are:
  1. P(NR|k) = 0.9999
  2. P(R|k) = .0001
Is this correct?

If so, do you mean that given the prior probabilities I used, my posterior probabilities are no surprise?

If you do, then I agree. All I'm trying to do is show Jabba that I can come to a conclusion he will not accept by using his own reasoning. Which is pure numerology.

Also, if P(R) = .01 and P(R|k) = .0001, doesn't that make the Religious hypothesis even more unlikely if all knowledge is given (again using Jabba's own reasoning)?
Click to expand...

What doesn't make sense to me in your argument is this:

I wonder, what is P(NR|k), the probability that the Non-Religious Hypothesis is true given all background knowledge? Well,

P(NR|k) = P(k|NR)P(NR)/(P(k|NR)P(NR)+P(k|R)P(R))
Click to expand...


By definition, the background information k is all the information that informs the prior probabilities, P(R) and P(NR). Remember that Jabba had included k in his original equation, but eventually agreed to omit it explicitly to simplify the notation. Nonetheless, k is implicit in Bayes' Theorem. Re-inserting k into the Jabba equation, gives us

[imgw=400]http://jt512.dyndns.org/images/bayes.full.png[/imgw]​

Since the term in the lhs of your equation appears in the rhs of Jabba's equation I'm confused about what you are trying to do.

Jay
 
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wollery said:
Only if you've got a dodgy pack to start with! :jaw-dropp
Click to expand...

No, he puts the card drawn back each time, before shuffling and picking another. The fact that he refers to a 'hand' does add to the confusion, though.

Jabba said:
You sit down at the table and turn over the first card. It's an ace of spades. You place the ace back in the deck, shuffle the cards and once again, turn over the first card. This time, it's the ace of diamonds. Hmm. So, you try the same thing again. This time, you get the ace of spades again.
Click to expand...

I was wondering whether the comparison hand would be better given as "2C, 7D, JH, 7D", though I don't think it alters the probability at all.

ETA: And the original selection sequence is "AS AD AS AH".
 
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zooterkin said:
No, he puts the card drawn back each time, before shuffling and picking another. The fact that he refers to a 'hand' does add to the confusion, though.
Click to expand...
Ah, yes. I'd forgotten about his totally weird way of dealing the cards.

:rolleyes:
 
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