NaClhv

Theology, philosophy, math, science, and random other things
                                                                                                                                                                                                                                                                  
2016-04-04
Importance: 

Bayesian evaluation for the likelihood of Christ's resurrection (Part 3)

The previously given probability value for the resurrection - 0.000 000 000 000 000 000 000 1 (which can also be written as 10^-22, or 1e-22) - is a prior probability. That is, it's the probability based on the background information, taking into consideration the fact that Jesus was human, and that humans don't rise from the dead.

However, it is just the starting point. It does not take into account any evidence we have specifically about Jesus's resurrection. Remember Bayes' rule: the final, posterior odds is the prior odds times the likelihood ratio. The number we have now is just the prior odds. We now need a numerical value for the likelihood ratio of the evidence, and then we can get our posterior odds.

But what kind of evidence is there for Christ's resurrection? And how could it possibly overcome a prior odds of 1 to 10 000 000 000 000 000 000 000 against it? Well, as for the evidence, we have the writings of the New Testament, where Jesus's resurrection and his follower's testimonies are documented. Okay, but is this "evidence" any good? How can we put numerical likelihood ratios to these things?

What we need is the numerical strength of a human testimony. As it turns out, we can actually get a not-unreasonable, order of magnitude estimate of this value. Remember your answers to the probability questions at the beginning? I hope you have them written down or otherwise recorded, because we will use them to calculate the likelihood ratio value that you would personally assign to a typical human testimony.

Let's use my personal answers, given below, as an example for how to do these calculations. These are my gut answers to the questions, before doing an actual probability calculations. Remember, these are the events that I'm willing to give even odds (50/50 chance) on, based solely on an earnest, personal testimony. It does not mean that I'm willing to believe 100%, and it does not mean that I'd stop looking for more evidence. It only points to how much I'm willing to adjust my beliefs based on someone saying "yes, I know it's unlikely, but it really happened".

For the shared birthday question, I would easily believe that my friend shared a birthday with me. I would also not have any real problem believing that our mothers also shared birthdays. At three people - myself, mother, and father - I would start becoming skeptical, but would probably give my friend the benefit of doubt. Starting with four shared birthdays in the family, I would start leaning more heavily towards skepticism.

On winning the lottery, I would not really doubt that my friend won the lottery. I would start doubting if he says that he won two consecutive lotteries.

On getting a royal flush, I think I could almost believe that my friend got two such hands in a very lucky night at the table. I feel like three would be entering the realm of the fantastical, and I would doubt my friend at around this number.

On pocket aces, I would be willing to believe that my friend had up to four or five pocket aces in a lucky night of Hold'em.

On the multiple births, I would not have any real problems believing that someone was a part of quadruplets. A claim to be in a quintuplet would start to cause a little bit of doubt to me, and a claim of sextuplets would need additional evidence.

On being struck by lightning, I actually had someone around me claim that this had recently happened to her. I had no problem believing it. Even if she had claimed two such accidents I don't think I would have really doubted her. If she had claimed three, I would start to be skeptical.

Now, calculating the numerical probability values for all these things is pretty straightforward:

The probability of sharing a single birthday is 1/365, or 1/3.65e2. The probability of sharing the three birthdays for your family is then simply this number cubed - 1 in 4.86e7.

The probability of winning the lottery varies by exactly which lottery you're talking about, but the odds for the jackpot are generally somewhere around 1 in 1e8.

The probability of getting a single royal flush is 1 in 6.5e5. The probability of getting two in two hands is therefore this number squared, 1 in 4.2e11. We can then take it down by a couple orders of magnitude, to account for the fact that there's dozens of hands played in a poker night. That gives us something like 1 in 1e9 for the odds.

The probability for getting pocket aces is 1 in 221. Getting five would then be 1 in 5.3e11. Taking it down again by several orders of magnitude to account for multiple hands, that brings us to something like 1 in 3e7.

The probability of quadruplets is about 1 in 1e6, and for quintuplets it's about 1 in 5.5e7. We'll split the difference here and call it 1e7.

The probability of getting struck by lightning in a given year is about 1 in 1e6. If we count "recently" as the last 5 years, that would bring it down to 1 in 2e5. Getting struck twice would then be 1 in 4e10, then maybe take off an order of magnitude for possible dependency factors to give us 1 in 4e9.

So, looking at the final numbers above - 1/4.9e7, 1/1e8, 1/1e9, 1/3e7, 1/1e7, 1/4e9 - we seem to be getting a reasonably consistent estimate for how I value the strength of an earnest, personal testimony. There are a lot of small details we can go over again (how many hands of poker did you play last night? Is your friend someone likely to play the lottery, or to be outdoors during a thunderstorm?), but these will largely be random, small, unknowable effects that will get washed out in this order-of-magnitude calculation.

So, we'll take the geometric mean of the above values(1/7e9), and then conservatively knock off a couple orders of magnitude, to get 1/1e8 as their "average" probability. In other words, even if an event had only a 1/1e8 prior chance of happening, I would be willing to give even odds on that event having occurred based on someone's earnest, personal testimony.

At such small probability values, "probability" is nearly synonymous with "odds". Therefore, I can re-state the above as saying that an earnest, personal testimony will shift the odds from 1/1e8 to 1/1. Or, to put it yet another way: the typical Baye's factor for an earnest, personal testimony is around 1e8. That is my numerical value for the strength of a human testimony.

It is important to note that this number is not something that I just made up. The math that gives this value is described above in its entirity. What answer did you get when you plugged in the numbers? That is the number that you, personally, must be willing to assign to the strength of a personal testimony, in order to be consistent. I believe that most reasonable people will be within a couple of orders of magnitude of my answer.

Furthermore, this number is something that you can check for yourself, based on a thought experiment that you can perform on yourself. Imagine a future where you yourself are telling someone, "I just hit the jackpot in the lottery". You are being earnest and sincere. Now, what is the probability that you're telling the truth in your own hypothetical future?

Given that the odds of winning the lottery is about 1/1e8, if you agree with my assessment that personal testimony should be valued at a Bayes' factor of around 1e8, then you are about equally likely to be telling the truth or lying in this scenario. However, if you disagree with that assessment - for example, if you think that personal testimony should only be valued at 1e6 - then you're saying that the posterior odds of you having won the lottery is still only 1/100, and so you're 99% likely to be lying in that scenario. Which is it?

In fact, this thought experiment suggests a way to empirically verify this value. Simply investigate a random sample of the people who claimed to have won the lottery. Remember, we're only counting earnest, personal claims to the jackpot. What fraction of them are telling the truth? How many of them are actual lottery winners? If you say "maybe around half?", then you're agreeing with my Bayes' factor of 1e8. If you want the Bayes' factor to be 1e6 instead, then you need 99% of these people to be liars.

Do you still doubt that you can assign a numerical value to the strength of a personal, human testimony? Or maybe worry that the correct value is far from 1e8? Well, fortunately for us, this "lottery liars" experiment has actually been naturally conducted, and we can compare its result with my numbers.

On January 13, 2016, the Powerball lottery produced the largest jackpot in history to date: 1.6 billion dollars. This jackpot ended up being split three ways. But - were there people who lied about having won this jackpot? As a matter of fact, there were. Several people on social media claimed to be a winner, presumably in an attempt at some quick, cheap fame. How many such people were there?

I couldn't get an exact number for the number of Powerball jackpot liars, but we can still get a sense, an order-of-magnitude estimate. Snopes, for example, mentions two people by name, and "several" or "numerous" others. Another report claims "a number" of similar hoaxes. So - it sounds like maybe ten people lied about winning the jackpot? It's certainly not in the hundreds or thousands.

How does that compare with the estimates from my probability calculation? Well, the odds of hitting the jackpot in Powerball are about 1/3e8. However, people may buy multiple tickets - which many people certainly did on such a well-publicized jackpot. In the end, there were 3 actual winners, out of the total American population of 3e8 people. So the prior odds for a specific person in the United States being a winner was 3/3e8, or 1/1e8.

Now, if the Bayes' factor for an earnest personal testimony is 1e8, then the posterior odds is just the product of 1/1e8 and 1e8, which is 1. That translates into 1 actual winner for every liar. So, given that there were 3 actual winners to the jackpot, we should expect around 3 liars - and that is roughly what we actually appear to have, within an order of magnitude.

You can again nitpick at this example (the great publicity of this jackpot, the people who made an earnest claim offline, the relative certainty of a short-lived notoriety for lying, etc.) But as an order-of magnitude estimate, the results of this natural experiment are about as good as I can possibly hope for. So, the proper Bayes' factor for an earnest, personal testimony is typically about 1e8, and this has now been validated through multiple lines of thought. It is certainly not several orders of magnitude less than that.

Next week, we will therefore continue the rest of the calculation using 1e8 as the Bayes' factor.


You may next want to read:
Sherlock Bayes, logical detective: a murder mystery game
How to think about the future (Part 1)
Another post, from the table of contents

Show/hide comments(No Comments)

Leave a Reply

Copyright