Deductive Arguments And Multiplying Probabilities

[Moot]

I agree that when we start adding arguments together, we can make a cumulative case, but this is where the math gets a bit over my head. Say that we have three arguments with two premises each and every premise has a .6 probability of being true, how probable is it then that one of the arguments have two true premises and how do we calculate it (maybe you explained this in your last post)?

I just started trying to write it down, but it gets pretty hectic if you don't assume the arguments are mutually exclusive, and of course you don't want to assume that. There are lots of permutations of the argument 1 premises being true and argument 2 premises being false etc. that need to be taken care of. Gets a bit out of control even with only 3 arguments. Maybe I'll try it another time when I have more time to spare. Get some structures like this:

P (A U B U C) = P(A) + P(B) + P(C) - P(A ∩ B) - P(A ∩ C) - P(B ∩ C) + P(A ∩ B ∩ C)

Thanks. Interesting stuff. I guess the conclusion we can draw is that it's not perfectly clear how deductive arguments supports a conclusion.

Well only because it is complicated :p. It is just a tedious  mechanical task of doing the math, not a deep mystery.

Oh, sure (I guess I phrased that a bit weird). I meant that it can get quite complicated. Especially in cases where the falsity of a premise contradicts the conclusion.

[Moot]

Sorry to bring this backwards a little, but I have a question regarding an earlier part of the discussion.  When you had the premise:

1. If my neighbour's dog is in the park, it's wearing a dog-sweater.

And you said that 2/3 times when your neighbors dog is in the part it is wearing a sweater so the probability of the premise is .6, is that really good enough ?  Isn't it a defeater for the premise to point to a single time when the dog has been in the park without its sweater ?

I don't see why it would be a defeater. It's enough to say that the premise is probably true. Or that it's more plausible than it's negation.

[Johan Biemans (jbiemans)]

I don't see why it would be a defeater. It's enough to say that the premise is probably true. Or that it's more plausible than it's negation.

So showing a live example where the premise is false is not a defeater for the premise, provided that the example isn't likely to happen ?? 

Then what constitutes a defeater ?  Does it need to be something that makes the premise more probably false ?  Or less plausible than it's negation ?

[kurros]

I don't see why it would be a defeater. It's enough to say that the premise is probably true. Or that it's more plausible than it's negation.

So showing a live example where the premise is false is not a defeater for the premise, provided that the example isn't likely to happen ?? 

Then what constitutes a defeater ?  Does it need to be something that makes the premise more probably false ?  Or less plausible than it's negation ?

"I am in my car" is not permanently false just because I am demonstrably not always in my car.

[Johan Biemans (jbiemans)]

Granted, but that is where context comes into play.  Given that, I don't think that you can say that it is true that "I am in my car", but you can say "I am currently in my car", or "I was in my car", etc.  I guess I am having trouble with using the label "true" for contingent statements; but I suppose, we do it all the time, don't we ?

Maybe it is just all the years on this site talking about logic that this seems overly pedantic, but isn't it technically correct ?

[Moot]

Granted, but that is where context comes into play.  Given that, I don't think that you can say that it is true that "I am in my car", but you can say "I am currently in my car", or "I was in my car", etc.  I guess I am having trouble with using the label "true" for contingent statements; but I suppose, we do it all the time, don't we ?

Maybe it is just all the years on this site talking about logic that this seems overly pedantic, but isn't it technically correct ?

"I am in my car" means "I am currently in my car". I can't see that you are technically correct.

[pat1911]

The problem is simple. You are doing multiplication when you should be doing addition.

Mmmm... so the conclusion is 120% true?

If your premises are numerical values then yes. How does one go about assigning numerical values to premises of a logical argument anyway, unless the values themselves are pertinent to the premise?

[Zbigge1031]

So this is something that's been addressed quite a lot on this forum but I've never understood the objections to it. Long ago SolElQoy tried explaining this to me and others but I came away none the wiser. So here's another attempt at getting to the bottom of this.

The idea is that the probability of the premises become multiplied in a deductive argument. This means that if an argument has two premises that have a .6 probability of being true, the conclusion only has a .36 probability of being true.

Let's take the newer version of the KCA (forgive me if I'm not formulating it exactly right):

1. If the Universe began to exist, it had a cause.
2. The Universe began to exist.
C. Therefore, the Universe had a cause.

Now let's say that we give each premise a .6 confidence level. How likely is the conclusion to be true? It seems pretty obvious to me that the answer is .36. There are three possibilities and here's how I'd put the probabilities:

1. The Universe began to exist and had a cause = .36
2. The Universe began to exist without a cause = .24
3. The Universe did not begin to exist = .4

This seems right to me, but Dr. Craig (and others I'm sure) has a standard that says a good argument simply needs to have premises that are more plausible than their negation. But that seems to contradict the above.  Somethings gotta give! It's been suggested that the answer lies in the difference between probability and plausibility but I can't understand why that would be. Even if plausibility and probability are different things, I can't see why a plausibility estimate couldn't be roughly translated into a probability estimate.

I do think probabilities cause odd occurrences when applied to deduction, and it seems to me that's part of why we shouldn't use them.  In contrast to your statement, if one believes with a 60% certainty in each of the premises of the KCA, then while the conclusion may only have a 36% likelihood, it can be the only rational position for such a person.    This is why Dr. Craig only argues for more plausible than not; the rules of deduction can compel someone to accept a conclusion based upon their acceptance of the premises.  I believe that to be rational, one must accept any belief formed from a conjunction of other accepted beliefs.

[Moot]

The problem is simple. You are doing multiplication when you should be doing addition.

Mmmm... so the conclusion is 120% true?

If your premises are numerical values then yes. How does one go about assigning numerical values to premises of a logical argument anyway, unless the values themselves are pertinent to the premise?

Hmm... I don't know what to say here... it's just not how it works.

[Moot]

So this is something that's been addressed quite a lot on this forum but I've never understood the objections to it. Long ago SolElQoy tried explaining this to me and others but I came away none the wiser. So here's another attempt at getting to the bottom of this.

The idea is that the probability of the premises become multiplied in a deductive argument. This means that if an argument has two premises that have a .6 probability of being true, the conclusion only has a .36 probability of being true.

Let's take the newer version of the KCA (forgive me if I'm not formulating it exactly right):

1. If the Universe began to exist, it had a cause.
2. The Universe began to exist.
C. Therefore, the Universe had a cause.

Now let's say that we give each premise a .6 confidence level. How likely is the conclusion to be true? It seems pretty obvious to me that the answer is .36. There are three possibilities and here's how I'd put the probabilities:

1. The Universe began to exist and had a cause = .36
2. The Universe began to exist without a cause = .24
3. The Universe did not begin to exist = .4

This seems right to me, but Dr. Craig (and others I'm sure) has a standard that says a good argument simply needs to have premises that are more plausible than their negation. But that seems to contradict the above.  Somethings gotta give! It's been suggested that the answer lies in the difference between probability and plausibility but I can't understand why that would be. Even if plausibility and probability are different things, I can't see why a plausibility estimate couldn't be roughly translated into a probability estimate.

I do think probabilities cause odd occurrences when applied to deduction, and it seems to me that's part of why we shouldn't use them.  In contrast to your statement, if one believes with a 60% certainty in each of the premises of the KCA, then while the conclusion may only have a 36% likelihood, it can be the only rational position for such a person.    This is why Dr. Craig only argues for more plausible than not; the rules of deduction can compel someone to accept a conclusion based upon their acceptance of the premises.  I believe that to be rational, one must accept any belief formed from a conjunction of other accepted beliefs.

I don't see why. Would you say the same thing about my dog example? If no, what is the relevant difference?

[Emuse]

If we think that something is only likely to be true sometimes and not always in otherwise identical circumstances then ...

P1. If the dog is wearing a sweater then it is in the park.

Will be more plausible than ...

P1. If the dog is in the park then it is wearing a sweater.

The latter is claiming that the dog always wears a sweater in the park.
The former says that the dog only wears a sweater in the park and only requires the dog to do this sometimes.

Or am I missing something?

P1. Everything that has a cause begins to exist.

Seems more modest than ...

P1. Everything that begins to exist has a cause.

[Johan Biemans (jbiemans)]

If the latter says that the dog always wears a sweater in the park, then doesn't the former say that if the dog is in the park then it is always wearing a sweater while it is in the park ?  I don't see why one implies always and the other implies sometimes.

[Moot]

My premise was: "If my neighbour's dog is in the park, it's wearing a dog-sweater." This could be certain (if for some reason it would be logically impossible for the dog to be in the park without his sweater) or it can be probably true, as in "the dog usually wears a sweater in the park".

I added that the dog only wears the sweater in the park to make the structure of the argument identical to the KCA (the version i presented).

[Emuse]

If the latter says that the dog always wears a sweater in the park, then doesn't the former say that if the dog is in the park then it is always wearing a sweater while it is in the park ?  I don't see why one implies always and the other implies sometimes.

This is why I'm querying too.  However ..

P1. If the dog is wearing a sweater then it is in the park.
P2. The dog is in the park.
C. Therefore, the dog is wearing a sweater.

... affirms the consequent.  So do we have to apply logic to a conditional before assessing its probability?  Isn't that what logic is for?

[Moot]

If the latter says that the dog always wears a sweater in the park, then doesn't the former say that if the dog is in the park then it is always wearing a sweater while it is in the park ?  I don't see why one implies always and the other implies sometimes.

This is why I'm querying too.  However ..

P1. If the dog is wearing a sweater then it is in the park.
P2. The dog is in the park.
C. Therefore, the dog is wearing a sweater.

... affirms the consequent.  So do we have to apply logic to a conditional before assessing its probability?  Isn't that what logic is for?

I don't see why this example would be different. We take the probability that P1 is true, namely the probability that it's true that the dog is in the park if it's wearing a sweater, and we multiply it with the probability of P2.

Maybe I'm missing something?