Sunday 31 March 2013

A Return to Sweet Probability

A while back, in WLC8 – Sweet Probability, I looked into Craig’s use and abuse of the Bayes Theorem, which is really the unaccredited use and abuse of Richard Swinburne’s use and abuse of the Bayes Theorem.

In brief, I looked at the probabilistic thinking that underpins the standard Bayes Theorem, the one that I can find in every single statistics text book I’ve looked at, including those which are very much focussed on Bayesian probability.  The Bayes Theorem as expressed by this equation is (adapted from a handy set of University of Washington lecture notes) using the assumption that there are two mutually exclusive events, A1 and A2):

Pr(A1|B) =

Pr(A1) . Pr(B|A1)
-----------------------------
( Pr(A1) . Pr(B|A1) + Pr(A2) . Pr(B|A2) )

If we replace A1 with R, A2 with not-R and B with B&E, this then becomes:

Pr(R|(B&E)) =

Pr(R) . Pr((B&E)|R)
-----------------------------
Pr(R) . Pr((B&E)|R) + Pr(not-R) . Pr((B&E)|not-R)

Craig’s variation of the Bayes Theorem, however, looks like this (with variations marked in bold red):

Pr(R|(B&E)) =

Pr(R|B) . Pr(E|B&R)
-----------------------------
Pr(R|B) . Pr(E|B&R) + Pr(not-R|B) . Pr(E|B&not-R)

Both these equations seem pretty terrifying, even the standard one.

In order to clarify things a little, we will shortly return to the idea of jellybeans.  In Sweet Probability, we had a jar with 100 jellybeans, 60 that contain Red material, 60 that contain yEllow material and 60 that contain Blue material – with some overlap, obviously.  The corresponding probabilities would be (see the article for my working):

Pr(R)=Pr(B)=Pr(E)=0.6

Pr(R∩B)=Pr(B∩E)=Pr(E∩R)=0.2

Pr(R∩B∩E)=0.1

I have been told that the jellybean analogy totally misrepresents Craig’s argument, or at least stretches the analogy until it was very close to breaking.  Clearly the use of the analogy alone isn’t going to wash.

My response to this accusation has been to demand that my interlocutor produce a derivation of Craig’s form of the Bayes Theorem, a demand that I have made repeatedly and rather forcefully … a demand that has, at least so far, been met with a ringing silence.

Therefore, it seems, we have to do the work ourselves.

In the diagram below I show the derivation of both the standard Bayes Theorem and the version used by Craig (click on it for a larger version):



The first three steps are common, and the fourth could be common, but I see no benefit in replacing B with B&E until the last step – unless your intention is to confuse.

The only tricky thing that is done in this working is to use the very first relationship Pr(X|Y)=Pr(X&Y)/Pr(Y) over and over again, sometimes forwards and sometimes backwards.  You end up with a complex looking equation, but really what it gives you is no more complex than what we have in the second line, the point in which we could replace B with B&E to get:


In our analogy, we already know that Pr(R&B&E)=0.1 so we just need to work out what the value of Pr(^R&B&E) is.  Putting this into words, we wish to work out the probability of selecting jellybean that contains both Blue and yEllow jellybean material but not Red jellybean material.  This eliminates all the red (20), orange (10), purple (10) and white (10) jellybeans along with those that contain only Blue and only yEllow jellybean material (20 and 20), leaving only the green jellybeans, of which there are 10.  So Pr(^R&B&E)=0.1 which coincidentally is the same value as Pr(R&B&E).  Therefore Pr(R|B&E)=0.1/(0.1+0.1)=0.5 which is less than the inherent probability of R, Pr(R), before we started.

But this is, of course an unfair analogy.  The value of Pr(R) that Craig uses is the probability that a specific person was resurrected.  That’s a miraculous thing and, by definition, miracles don’t occur with a probability of six in ten (0.6) – they are much more likely to have a probability of one in a million (0.000001) or even one in a billion (0.000000001).  A resurrection is even less likely than that, judging from the number of resurrected people we have today.

Note that I am not talking about someone who almost dies, who is revived and might come back with some interesting reports of an out of body experience – something which is a little more common.  I am talking about someone who is well and truly dead, who is locked into a crypt and who rises from the dead three days later.  Each year we have more than 50 million people who die.  For the sake of the argument, let’s say that each year we get 50 bona fide resurrections, which means that there is a one in a million chance of getting resurrected and Pr(R)=0.000001.  This is equivalent to having a jar with a million jellybeans in it, of which only one contains Red jellybean material.

Turning to B, we find that we have a disagreement.  What does this mean, exactly?  Craig tells us that B is “Background knowledge”.  Now when I hear that, I think of the knowledge that we as a species have access to if we choose to seek it out and, in this specific context, it refers to the fact that there are no reliable reports of people being resurrected.  However, I’ve been told that this is unfair because I am assuming naturalism whereas, from what I can tell from the complainant’s argument, B “should” be construed as theism.

Ok – but if that is the case, then it can’t be used in any argument proving the existence of god.  You cannot arrive at a valid proof for the existence of anything if you’ve assumed it from the start.  This is why I give Craig the benefit of the doubt and assume that B is just what we know, and therefore P(B)=1.0 meaning that the likelihood of knowing what we know is one hundred per cent.  This is equivalent to B being the likelihood of selecting a jellybean from a jar containing one million of them and finding that it is a jellyBean.  Thus P(R|B)=0.000001 and our “Background knowledge” has had no effect on the likelihood of a resurrection.

Now finally, we have E.  What does Craig actually mean by this?  He says it means “Specific evidence (empty tomb, post-mortem appearances, etc.)”  But how can we turn this into a likelihood?  Normally, when the Bayes Theorem is applied, we talk about different events, or aspects of an event, rather than “evidence”.  In my example, I link E to the likelihood of a selected jellybean containing yEllow jellybean material.  This is an “event” that is linked to the jellybean being a jellyBean (B), but not to the jellybean containing Red (R).

I suspect, however, that we should look more at the likelihood of the “Specific evidence” arising.  What are the chances that a book that is getting on for two millennia old contains the following reports:

·         On the Sunday after his crucifixion, Jesus’ tomb was found empty by a group of his women followers”,

·         On separate occasions different individuals and groups saw appearances of Jesus alive after his death”, and

·         The original disciples suddenly came to believe in the resurrection of Jesus despite having every predisposition to the contrary”?

I’m going to follow Plantinga’s example here and say that there are two options:

·         The reports are true, or

·         The reports are false (either due to outright fabrication, delusion on the part of those making the reports or some mix of muddled reporting that led to a false conclusions which were then reported incredulously).

Since there are two options, the probability of each is 0.5 (see Planting a Tiger for an explanation of Plantigerant probability).  Therefore, P(E)=0.5 which is equivalent to half the jellybeans containing yEllow jellybean material.

So, once we have modified the analogy accordingly, we have a jar containing one million jellybeans, one of which contains Red jellybean material – P(R)=0.000001, one million of which are jellyBeans – P(B)=1.0 and five hundred thousand of which contain yEllow jellybean material – P(E)=0.5 – but what we don’t know is the distribution of yEllow and Red jellybean material.

Fortunately, we can rely on Plantigerant probability again.  If a jellybean contains Red jellybean material then it will, again, either contain yEllow jellybean material or it won’t.  Therefore P(E|R)=0.5.

We know that likelihood of a jellybean containing Red jellybean material (R) and being a jellyBean (B) is the same as the likelihood of a jellybean containing Red jellybean material, so P(R&B)=0.000001.  We know that the likelihood of a jellybean not containing Red jellybean material is 999,999 to one, so P(^R)=0.999999 and jellybeans which don’t contain Red jellybean material are nevertheless jellybeans so P(^R&B)=0.999999.

Appealing to Plantigerant probability once more, if a jellybean does not contain Red jellybean material then it will either contain yEllow jellybean material or it won’t.  Therefore P(E|^R)=0.5

Using the same logic as above, we know that P(E|^R&B)=0.5 and we now have all the numbers we need to plug into Craig’s equation to get a result.


Which means that P(R|B&E)=0.000001=P(R).

The only way you can get this equation to change in Craig’s favour is to assume that E has a likelihood that is higher than 0.5, which you would only do if you already assume that the resurrection story is reflective of an actual resurrection.

In other words, a calculation of P(R|B&E) which results in a value that approaches unity tells us nothing about the likelihood of the resurrection story being true and everything about the assumptions of the person performing that calculation.

---------------------------------------------

Something that Craig says in his debate with Krauss now makes more sense:

So the question “Is There Evidence for God?” isn’t really very debatable. Rather the really interesting question is whether God’s existence is more probable than not. That is, is

Pr (G | E & B) > 0.5 ?

Now I’ll leave it up to you to assess that probability. My purpose in tonight’s debate is more modest: to share with you five pieces of evidence each of which makes God’s existence more probable than it would have been without it. Each of them is therefore evidence for God. Together they provide powerful, cumulative evidence for theism.

Craig is using exactly the same sort of argument that Richard Swinburne made, using precisely the same sort of equation (although in the Krauss debate he just threw it out there and made no effort whatsoever to explain what it was supposed to mean).

When you have worked it all through you begin to see that, in fact, it was pretty cunning of Craig to leave it up to the listener to assess the probability, since you’d end up with Pr(G|E&B) = Pr(G) which is a totally subjective assessment based on whether or not you already believe in the god in question.

1 comment:

  1. I found myself trying to figure out how to deal with conditional probabilities when the conditional is uncertain, and ended up posting the question on the math stackexchange site. Nobody has given a proper answer yet and it occurred to me that this seems to be the kind of thing that you would probably be interested in tackling.

    ReplyDelete

Feel free to comment, but play nicely!

Sadly, the unremitting attention of a spambot means you may have to verify your humanity.