Bayesian argument to believe in aliens?

johne1618

Active Member
Let A = Hypothesis that aliens are visiting Earth

Let Not[A] = Hypothesis that aliens are not visiting Earth

Let { CE_i } = A set of i independent Close Encounter events

Using Bayes' theorem we find that the (posterior) probability that aliens are visiting Earth given a set of i independent close encounters, P( A | { CE_i } ), is given by

P( A | { CE_i } ) = P( { CE_i } | A ) P( A ) / [ P( { CE_i } | A ) P( A ) + P( { CE_i } | Not[A] ) P( Not[A] ) ]

Now I argue that the probability of a set of i independent close encounters given the hypothesis that aliens are visiting Earth, P( { CE_i } | A ), can be taken to be one by definition. A believer in the alien hypothesis will not be surprised to learn of a set of close encounters with aliens.

I wish to argue backwards to find an expression for the prior probability for aliens P( A ) that gives us an appreciable posterior probability for aliens given reports of i close encounters, P( A | { CE_i } ) = 1/2.

Substituting P( { CE_i } | A ) = 1 and P( A | { CE_i } ) = 1/2 into Bayes' theorem we get

P( A ) / P( Not[A] ) = P( { CE_i } | Not[A] )

P( A ) / P( Not[A] ) = P( CE_1 | Not[A] ) * P( CE_2 | Not[A] ) * P( CE_3 | Not[A] ) * ...

Let us assume that aliens are not visiting Earth and that the witness was awake during the encounter (no bed-time accounts) and that the encounter was faithfully recorded soon after the event. I would say that the witness was either lying, hallucinating or the victim of a hoax.

For the sake of argument let us assume that we have a compelling case so that P( CE | Not[A] ) = 1 / 100

Let us assume that we have 10 good cases of close encounters. In order to end up with an appreciable posterior probability of aliens given those cases, i.e. P( A | { CE_i } ) = 1/2, the prior ratio of our belief to unbelief in aliens, P( A ) / P( Not[A] ) is given by

P( A ) / P( Not[A] ) = (1/100)*(1/100)*(1/100)*(1/100)*(1/100)*(1/100)*(1/100)*(1/100)*(1/100)*(1/100) = 10^-20

Thus if we combine data from compelling close encounter cases even the most hardened skeptic should begin to believe in the alien hypothesis.
 
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Landru

Moderator
Staff member
Let A = Hypothesis that aliens are visiting Earth

Let Not[A] = Hypothesis that aliens are not visiting Earth

Let { CE_i } = A set of i independent Close Encounter events

Using Bayes' theorem we find that the (posterior) probability that aliens are visiting Earth given a set of i independent close encounters, P( A | { CE_i } ), is given by

P( A | { CE_i } ) = P( { CE_i } | A ) P( A ) / [ P( { CE_i } | A ) P( A ) + P( { CE_i } | Not[A] ) P( Not[A] ) ]

Now I argue that the probability of a set of i independent close encounters given the hypothesis that aliens are visiting Earth, P( { CE_i } | A ), can be taken to be one by definition. A believer in the alien hypothesis will not be surprised to learn of a set of close encounters with aliens.

I wish to argue backwards to find an expression for the prior probability for aliens P( A ) that gives us an appreciable posterior probability for aliens given reports of i close encounters, P( A | { CE_i } ) = 1/2.

Substituting P( { CE_i } | A ) = 1 and P( A | { CE_i } ) = 1/2 into Bayes' theorem we get

P( A ) / P( Not[A] ) = P( { CE_i } | Not[A] )

P( A ) / P( Not[A] ) = P( CE_1 | Not[A] ) * P( CE_2 | Not[A] ) * P( CE_3 | Not[A] ) * ...

Let us assume that aliens are not visiting Earth and that the witness was awake during the encounter (no bed-time accounts) and that the encounter was faithfully recorded soon after the event. I would say that the witness was either lying, hallucinating or the victim of a hoax.

For the sake of argument let us assume that we have a compelling case so that P( CE | Not[A] ) = 1 / 100

Let us assume that we have 10 good cases of close encounters. In order to end up with an appreciable posterior probability of aliens given those cases, i.e. P( A | { CE_i } ) = 1/2, the prior ratio of our belief to unbelief in aliens, P( A ) / P( Not[A] ) is given by

P( A ) / P( Not[A] ) = (1/100)*(1/100)*(1/100)*(1/100)*(1/100)*(1/100)*(1/100)*(1/100)*(1/100)*(1/100) = 10^-20

Thus if we combine data from compelling close encounter cases even the most hardened skeptic should begin to believe in the alien hypothesis.
This has been done http://www.sci-news.com/astronomy/b...extraterrestrial-life-intelligence-08443.html

In a new study published this week in the Proceeding of the National Academy of Sciences, Dr. David Kipping of Columbia University and Flatiron Institute used a statistical technique called Bayesian inference to estimate the odds of complex life and intelligence emerging beyond Earth.
 

DavidB66

Senior Member
Now I argue that the probability of a set of i independent close encounters given the hypothesis that aliens are visiting Earth, P( { CE_i } | A ), can be taken to be one by definition. A believer in the alien hypothesis will not be surprised to learn of a set of close encounters with aliens.

I think you need to explain your thinking on this a bit more. Unless your definition of 'alien visitor' includes 'frequent deliberate or reckless contactor of humans', I don't see how the probability of close encounters, given the occurrence of alien visits, can be high 'by definition'.

The meaning of 'independent' encounters also needs explaining. What we have are various reports of close encounters, but few if any of these reports are entirely independent of previous reports. Ever since the 1940s, at least, most people in the literate world will have been familiar with reports and fictional representations of flying saucers, etc. Even school children in Zimbabwe.

I suspect that using the same form of argument one could also prove the existence of ghosts, fairies and witches beyond a reasonable doubt!
 

Mendel

Senior Member.
How do you define a "close encounter event"?

Now I argue that the probability of a set of i independent close encounters given the hypothesis that aliens are visiting Earth, P( { CE_i } | A ), can be taken to be one by definition.
Below, you expand this set notation as the product of all individual P(CE_i | A), and that's certainly not 1 unless you want to argue that the presence of aliens results in no more hoaxes or false reports. The maths doesn't add up here.
 

Ann K

Senior Member.
Let { CE_i } = A set of i independent Close Encounter events.
It's my understanding that the only thing we have are unidentified events, that is, there are no verified "close encounter" events, let alone independent ones, and thus there is no connection with the concept of aliens. That being the case, isn't this just an idle math exercise?
 

NorCal Dave

Senior Member.
1. Let W= Hypothesis that under a full moon some people transform into wolves.

2. Let Not [W]= Hypothesis that under a full moon no people transform into wolves.

3. Let (WT i)= A set of i independent Wolf Transformations events.

Ergo, per your analysis, the most harden skeptic should begin to believe in were-wolves.

I know squat about Bayesian arguments, but in this case and I believe with yours above, it's line #3 that's the gist of the matter. As @Ann K pointed out above, a collection of anecdotes and stories isn't a valid sub-set.
 

FatPhil

Senior Member.
Let A = Hypothesis that aliens are visiting Earth

Let Not[A] = Hypothesis that aliens are not visiting Earth

Let { CE_i } = A set of i independent Close Encounter events

Using Bayes' theorem we find that the (posterior) probability that aliens are visiting Earth given a set of i independent close encounters, P( A | { CE_i } ), is given by

P( A | { CE_i } ) = P( { CE_i } | A ) P( A ) / [ P( { CE_i } | A ) P( A ) + P( { CE_i } | Not[A] ) P( Not[A] ) ]

Now I argue that the probability of a set of i independent close encounters given the hypothesis that aliens are visiting Earth, P( { CE_i } | A ), can be taken to be one by definition. A believer in the alien hypothesis will not be surprised to learn of a set of close encounters with aliens.

I propose that P(CE_i)=0 for most of the CEs.
 

Mendel

Senior Member.
Headline from your source:
Astronomer Uses Bayesian Statistics to Weigh Likelihood of Complex Life and Intelligence beyond Earth
Content from External Source
This doesn't actually deal with extraterrestrial visits to Earth, it's just computing the likelihood of intelligent life emerging on a planet, which is part of the input data to the drake equation:
Article:
The Drake equation is a probabilistic argument used to estimate the number of active, communicative extraterrestrial civilizations in the Milky Way Galaxy.

@johne1618, on the other hand, is trying a variant of "because we have so many CE reports, one of them must be real", dressed up with unmotivated ad-hoc formulas.
 
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johne1618

Active Member
How do you define a "close encounter event"?


Below, you expand this set notation as the product of all individual P(CE_i | A), and that's certainly not 1 unless you want to argue that the presence of aliens results in no more hoaxes or false reports. The maths doesn't add up here.

Maybe I should be more careful. On being presented with a good close encounter case , a believer in aliens would assign the following likelihood:

P( CE | A ) = 99/100

where the missing 1% is the combined probability of the witness either lying, hallucinating or being the victim of a hoax.

Just assuming an appreciable posterior probability of P( A | { CE_i } ) = 1/2 and plugging into Bayes' theorem we get

P( A ) / P( Not[A] ) = P( { CE_i } | Not[A] ) / P( { CE_i } | A )

If we take P( { CE_i } | Not[A] ) = (1/100)^10 and P( { CE_i } | A ) = (99/100)^10 then

P( A ) / P( Not[A] ) is approximately 1.1 * 10^-20

Therefore not an appreciable difference.
 

FatPhil

Senior Member.
Maybe I should be more careful. On being presented with a good close encounter case , a believer in aliens would assign the following likelihood:

P( CE | A ) = 99/100

where the missing 1% is the combined probability of the witness either lying, hallucinating or being the victim of a hoax.

Just assuming an appreciable posterior probability of P( A | { CE_i } ) = 1/2 and plugging into Bayes' theorem we get

P( A ) / P( Not[A] ) = P( { CE_i } | Not[A] ) / P( { CE_i } | A )

If we take P( { CE_i } | Not[A] ) = (1/100)^10 and P( { CE_i } | A ) = (99/100)^10 then

P( A ) / P( Not[A] ) is approximately 1.1 * 10^-20

Therefore not an appreciable difference.

But this is just extrarectal numerology, and it completely ignores the fact that the events are not independent.

If you insist on re-running the numbers, please factor into the equations the fact that some skeptics, on being presented with a good close encounter case, would assign the following probability: P(CE) = 10^-10^100, and the fact that the skeptics have a much better history of being able to evaluate probabilities correctly.
 

Mendel

Senior Member.
Replying to the OP #1:
P( A | { CE_i } ) = P( { CE_i } | A ) P( A ) / [ P( { CE_i } | A ) P( A ) + P( { CE_i } | Not[A] ) P( Not[A] ) ]
Since you're applying simple Bayes, the denominator should be P( { CE_i } ). This means you are setting
P( { CE_i }) =? P( { CE_i } | A ) P( A ) + P( { CE_i } | Not[A] ) P( Not[A] )
Is that identity true, though?

We know that
P( { CE_i } ) = P( { CE_i } | A ) + P( { CE_i } | Not[A] )​
because each event occurs either with A or not[A]. Also,
P(A)+P(not[A]) = 1​

Taken together,
P( { CE_i } ) = P( { CE_i } ) × 1 = (P( { CE_i } | A ) + P( { CE_i } | Not[A] )) × (P(A)+P(not[A]))​
Multiplying the sums in the parentheses yields
P( { CE_i } ) = P( { CE_i } | A )×P(A) + P( { CE_i } | Not[A] )×P(A) + P( { CE_i } | A )×P(not[A]) + P( { CE_i } | Not[A] )×P(not[A])
I've bolded the parts that correspond to your assumption, which means the rest must be zero:
0 =? P( { CE_i } | Not[A] )×P(A) + P( { CE_i } | A )×P(not[A])​
Since we have no negative probabilities, and all of these partial probabilities ought to be non-zero (since otherwise we'd already know whether there are aliens or not), this holds:
0 < P( { CE_i } | Not[A] )×P(A) + P( { CE_i } | A )×P(not[A])​
And that means your denominator identity is mathematically false.

I'd try to correct it, but since you haven't defined the meaning CE_i and P(CE_i) well, I find this difficult.
 

Mauro

Senior Member
Let A = Hypothesis that aliens are visiting Earth

Let Not[A] = Hypothesis that aliens are not visiting Earth

Let { CE_i } = A set of i independent Close Encounter events

Using Bayes' theorem we find that the (posterior) probability that aliens are visiting Earth given a set of i independent close encounters, P( A | { CE_i } ), is given by

P( A | { CE_i } ) = P( { CE_i } | A ) P( A ) / [ P( { CE_i } | A ) P( A ) + P( { CE_i } | Not[A] ) P( Not[A] ) ]
Up to now everything is right. The formula you use is Bayes' theorem in which P(CE_i) at the denominator has been rewritten using the formula of total probability.


Now I argue that the probability of a set of i independent close encounters given the hypothesis that aliens are visiting Earth, P( { CE_i } | A ), can be taken to be one by definition. A believer in the alien hypothesis will not be surprised to learn of a set of close encounters with aliens.
Well, of course if you take by definition that there is a probability of one to observe close encounters if aliens are visiting Earth, then that is. You give no proof of your assertion, then I could as well argue that that probability is about zero (I would not be surprised of aliens being quite able and quite careful in avoiding closer encounters with us).

But the problem is actually worse: it seems you are confusing the set of events you actually have observed, { CE_1, CE_2 .... CE_i }, with some generic set of observations one could possibly make. Example: if CE_1 is a close encounter with one-meter tall, silver skinned aliens, then which is the probability of that particular encounter happening, if aliens are visiting Earth? This is the number which is needed and it cannot just be assumed to be one.

To say it another way: you can probably convince many people that 'if aliens are visiting Earth, then there must have been close encounters', this looks possible enough, and even if 'looking possible' does not mean 'it is true', so this will not help with hardline skeptics, you have at least gained some support. But this is not enough: you now must prove that the specific encounters you observed in your set are actually close encounters with aliens and this is going to be much more difficult.


I wish to argue backwards to find an expression for the prior probability for aliens P( A ) that gives us an appreciable posterior probability for aliens given reports of i close encounters, P( A | { CE_i } ) = 1/2.
Well, if you assume that what you observed has a 50% probability of being caused by aliens, then that it is, it was aliens with 50% probability.

(.... omissis, irrelevant to the core of the argument which follows)

For the sake of argument let us assume that we have a compelling case so that P( CE | Not[A] ) = 1 / 100

Let us assume that we have 10 good cases of close encounters. In order to end up with an appreciable posterior probability of aliens given those cases, i.e. P( A | { CE_i } ) = 1/2, the prior ratio of our belief to unbelief in aliens, P( A ) / P( Not[A] ) is given by

P( A ) / P( Not[A] ) = (1/100)*(1/100)*(1/100)*(1/100)*(1/100)*(1/100)*(1/100)*(1/100)*(1/100)*(1/100) = 10^-20

Thus if we combine data from compelling close encounter cases even the most hardened skeptic should begin to believe in the alien hypothesis.
And this is very much true, and it's true even without any need to make hypothesis on what P(CE_i|A) and P(A|CE_i) are! I'm a pretty hardened skeptic, but give me just one case where you can demonstrate the probability of it not being caused by aliens to be 1/100 and I assure you my skepticism will be badly shaken, give me ten independent cases like that and it'd be an absolute overkill. It's obvious that extraordinary evidence can validate even extraordinary claims, the problem is to get the evidence first.
 
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Mendel

Senior Member.
Maybe I should be more careful. On being presented with a good close encounter case , a believer in aliens would assign the following likelihood:

P( CE | A ) = 99/100

where the missing 1% is the combined probability of the witness either lying, hallucinating or being the victim of a hoax.
W=Witness saw an alien
^W=Witness did not see an alien (lie, hallucination, hoax, or no report)
CE=Witness reports a close encounter

We want to know the probability for a witness having seen an alien, given they reported one.
P(W|CE)=P(CE|W)×P(W)/P(CE)​
P(CE|W) is the probability that someone who sees an alien reports it.
P(W) is the probability that someone sees an alien.
P(CE) is the probability that someone makes a report (bunk or real).

Which of these probabilities do you know?
 

johne1618

Active Member
I now think a more fundamental problem with the argument is the following definition:

Let { CE_i } = A set of i independent Close Encounter events

The probability of an actual close encounter event given the hypothesis that aliens are not visiting Earth, P( CE | Not[A] ) must be zero by definition!

We cannot presume that we have actual close encounter events but rather only reports of close encounter events. The simplifying assumption that they are independent of each other seems ok.

Therefore the argument should be:

Let A = Hypothesis that aliens are visiting Earth

Let Not[A] = Hypothesis that aliens are not visiting Earth

Let { RCE_i } = A set of i independent reports of Close Encounter events

If we just assume an appreciable posterior probability of P( A | { RCE_i } ) = 1/2 and plug into Bayes' theorem we get

P( A ) / P( Not[A] ) = P( { RCE_i } | Not[A] ) / P( { RCE_i } | A )

Now we can set P( RCE | Not[A] ) = 1/100 without contradiction. A non-believer in aliens can look at a particular report of a close encounter by a witness and assign it a non-zero probability. The more reputable the witness the lower the probability that a non-believer will assign to that witness coming up with a story about a close encounter with aliens.

Also since we are only talking about reports of close encounter events we can take the probability of obtaining a set of reports of close encounter events, given the hypothesis that aliens are visiting Earth, P( { RCE_i } | A ), to equal one. A believer in aliens will not be surprised to hear a report of a close encounter regardless of whether the witness is reputable or not.

Thus if we have 10 reports of close encounters by seemingly good witnesses we can deduce that even if our prior odds that aliens are visiting Earth is 100^10 to 1 against we will still end up with posterior even odds that the alien hypothesis is true.
 
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Mendel

Senior Member.
If we just assume an appreciable posterior probability of P( A | { RCE_i } ) = 1/2 and plug into Bayes' theorem we get

P( A ) / P( Not[A] ) = P( { RCE_i } | Not[A] ) / P( { RCE_i } | A )
That's equivalent to
P( { RCE_i } | A )×P(A) = P( { RCE_i } | Not[A] )×P(not[A])​
Applying Bayes on each side results in
P( A|{ RCE_i })×P( { RCE_i }) = P(not A | { RCE_i })×P( { RCE_i })​

Divide by P( { RCE_i }), and you get
P( A|{ RCE_i }) = P(not A | { RCE_i })​
and since
P( A|{ RCE_i }) + P(not A | { RCE_i }) =1​
your "If we just assume an appreciable posterior probability of P( A | { RCE_i } ) = 1/2" becomes a necessity.

But why would you assume that, and what happens to your maths when you assume something else?
You also wrote, "P( { RCE_i } | A ), to equal one" and I think you were arguing P( { RCE_i } | Not[A] ) should be 0.01^10 = 10^-20 ?

Let's plug this in:
P( { RCE_i } | A )×P(A) = P( { RCE_i } | Not[A] )×P(not[A])​
1×P(A) = 10^-20 × P(not[A])​
and with
P(not[A]) < 1​
we get
P(A) < 10^-20​

Result: P(A) is very small.

Your argument shows mathematically that we probably don't have alien visitors.
 

Ann K

Senior Member.
@Mendel
"The Drake equation is a probabilistic argument used to estimate the number of active, communicative extraterrestrial civilizations in the Milky Way Galaxy."

Every term in the Drake equation contains an unknown, therefore it can only be used to speculate, not to estimate.
 

JMartJr

Senior Member
Every term in the Drake equation contains an unknown, therefore it can only be used to speculate, not to estimate.
I'd argue that we have a decent idea of the first two variables, and are starting to get a rough idea of the third. The remaining four are totally unknown, other than they cannot be 0, given that we exist. The more closely any of them approach 0, the more likely we are functionally alone.

As a more general observation I am put in mind of the old saw (attributed to multiple sources), "Logic is a system for going wrong with confidence."
 

Mauro

Senior Member
........ Now we can set P( RCE | Not[A] ) = 1/100........... Thus if we have 10 reports of close encounters by seemingly good witnesses we can deduce that even if our prior belief in the hypothesis that aliens are visiting Earth is 100^10 to 1 against we will still end up with an appreciable posterior belief in the alien hypothesis.
Having a "report from a seemingly good witness" is not the same thing as "it was demonstrated it was aliens with 99% probability".
 

Mendel

Senior Member.
Having a "report from a seemingly good witness" is not the same thing as "it was demonstrated it was aliens with 99% probability".
@johne1618 's argument boils down to "if a UFO believer sees 10 reports that he's 99% confident in, he's going to believe with 50% confidence that we actually have alien visitors, unless the probability of aliens visiting us is actually smaller than 1:100,000,000,000,000,000,000".

One problem is that here on Metabunk, no debunker has ever had that level of confidence in an alien report.

The other problem is that the UFO community promotes reports as credible that say "aliens are small grey people with big heads" as well as "aliens are sexy females with big eyes that come to visit when the wife's not home", and if you believe in both with 99% confidence, you need to face the fact we're being visited by two alien civilisation, which means P(A) is really P(A_1)×P(A_2); so, worst-case, these 10 great reports correspond to 10 alien visits which now have to have a 1:100 chance each for the whole thing to be believable. If that's the case, somebody out there is marketing Earth as a tourist destination right now.

The third problem is that a lot of "close encounter" reports could really be time travel reports, and who's to say time travel is less likely than interstellar/intergalactic travel? So inherently the confidence in most CE reports should be 50% at best.
 

Ann K

Senior Member.
I'd argue that we have a decent idea of the first two variables, and are starting to get a rough idea of the third. The remaining four are totally unknown, other than they cannot be 0, given that we exist. The more closely any of them approach 0, the more likely we are functionally alone.
if even just one of the terms of the Drake equation is "totally unknown", that leaves the product totally unknown, doesn't it?
 
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JMartJr

Senior Member
if any one of the terms of the Drake equation is "totally unknown", that leaves the product totally unknown, doesn't it?
Yeah. With what we know now, it can't be "solved." My disagreement was with the claim that ALL of the variables are unknown. We have a decent idea of a few of them now.
But I don't want to derail the thread into a Drake Equation sidetrack...
 

Domzh

Active Member
im not a mathematician or physicist but to me this looks like utter nonsense.

let me explain why.

the big issue here is that a "close encounter report" is an experience from a human dealing with something that only exist as an arbitrary idea.

i believe this model could work to estimate if there is a bird flying around with the size of a 747. if enough people report it independently this could suggest some truth.

a bird is proven to be real. so is the size of a 747. so we test something highly unlikely that is combined with two very real factors.

while an "alien" has zero proven describing factors. (scientifically proven beyond reasonable doubt).

i bet if we would jump back in time, visiting galileo, then this model would suggest that the earth is most likely flat.

its nonsense imho
 

NorCal Dave

Senior Member.
The simplifying assumption that they are independent of each other seems ok.
I don't think so, they are not independent.

The assumption that these reports and claims, good bad or otherwise, are independent seems to ignore the interrelations of CE/abduction stories. Each claim feeds and is built upon the previous.

The Hill's are one of the first popular CE/abductions claims and they introduced many of the standard items that would be in most claims from then on, such as but not limited to, "missing time" and "medical/reproductive exams".

By the '90s CE/abductions were part of the zeitgeist, formulaic and very popular. The X Files central theme revolved around Molder's sister being abducted. The leading researchers were claiming in bestselling books that tens of thousands of people were experiencing them (bold by me).

With Hopkins, Jacobs and Mack, accounts of alien abduction became a prominent aspect of ufology. There had been earlier abduction reports (the Hills being the best known), but they were believed to be few and far between and saw rather little attention from ufology (and even less attention from mainstream professionals or academics). Jacobs and Hopkins argued that alien abduction was far more common than earlier suspected; they estimate that tens of thousands (or more) North Americans had been taken by unexplained beings.[20]
Content from External Source
en.wikipedia.org/wiki/Alien_abduction

So how does one isolate "good independent accounts of CE/abduction" from all the hopelessly intermingled and interrelated stories out there?
 

johne1618

Active Member
That's equivalent to
P( { RCE_i } | A )×P(A) = P( { RCE_i } | Not[A] )×P(not[A])​
Applying Bayes on each side results in
P( A|{ RCE_i })×P( { RCE_i }) = P(not A | { RCE_i })×P( { RCE_i })​

Divide by P( { RCE_i }), and you get
P( A|{ RCE_i }) = P(not A | { RCE_i })​
and since
P( A|{ RCE_i }) + P(not A | { RCE_i }) =1​
your "If we just assume an appreciable posterior probability of P( A | { RCE_i } ) = 1/2" becomes a necessity.

But why would you assume that, and what happens to your maths when you assume something else?
You also wrote, "P( { RCE_i } | A ), to equal one" and I think you were arguing P( { RCE_i } | Not[A] ) should be 0.01^10 = 10^-20 ?

Let's plug this in:
P( { RCE_i } | A )×P(A) = P( { RCE_i } | Not[A] )×P(not[A])​
1×P(A) = 10^-20 × P(not[A])​
and with
P(not[A]) < 1​
we get
P(A) < 10^-20​

Result: P(A) is very small.

Your argument shows mathematically that we probably don't have alien visitors.

My argument is that even if the skeptic starts out with a very, very small prior belief in alien visitation, P( A ) / P( Not[A] ) = 10^-20, after learning of 10 reports of different close encounter events by good witnesses each with P( RCE | Not[A] ) = 1/100, his/her posterior probability of alien visitation, P( A | { RCE_10 } ) = 50%.
 
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Mauro

Senior Member
My argument is that even if the skeptic starts out with a very, very small prior belief in alien visitation, P( A ) / P( Not[A] ) = 10^-20, after learning of 10 reports of different close encounter events by good witnesses each with P( RCE | Not[A] ) = 1/100, his/her posterior probability of alien visitation, P( A | { RCE_10 } ) = 50%.
Yeah, but you are stating the obvious: a sufficiently strong evidence can overcome any low prior probability. This holds for aliens as for angels, ghosts, Zeus, the Matrix, anything one can imagine.

You rightly say that having 10 alien visitation reports, each with a confirmed probability of 99% of being aliens, would be compelling evidence for aliens: noone disputes this, as noone would dispute that, having 10 reports, each of which ascertained to be caused by Zeus with 99% probability, would be compelling evidence for the existence of Zeus.

But the problem is actually having those 99%-sure reports. You seem to imply that a report from a 'good witness' is enough to give you that 99% confidence, but this is far from true. Do you have any example of any such 99%-confirmed-to-be-aliens reports?
 

FatPhil

Senior Member.
However, the probability of ever getting 10 reports of different close encounter events by good witnesses is just as vanishingly small as the probability that there are alien visitations. You've not moved me from my original stance at all.
 

Mendel

Senior Member.
@johne1618 's argument boils down to "if a UFO believer sees 10 reports that he's 99% confident in, he's going to believe with 50% confidence that we actually have alien visitors, unless the probability of aliens visiting us is actually smaller than 1:100,000,000,000,000,000,000".

My argument is that even if the skeptic starts out with a very, very small prior belief in alien visitation, P( A ) / P( Not[A] ) = 10^-20, after learning of 10 reports of different close encounter events by good witnesses each with P( RCE | Not[A] ) = 1/100, his/her posterior probability of alien visitation, P( A | { RCE_10 } ) = 50%.
That's about the same, right?

So how do you know how big P(A) is? There are 10^24 stars in the universe, so if there are 10,000 alien civilisations out there, that's exactly that 1:10^20 chance that one of them visits us.

And what do you say to the three concerns I raised in that post (#20)?
 

Mendel

Senior Member.
You rightly say that having 10 alien visitation reports, each with a confirmed probability of 99% of being aliens, would be compelling evidence for aliens: noone disputes this, as noone would dispute that, having 10 reports, each of which ascertained to be caused by Zeus with 99% probability, would be compelling evidence for the existence of Zeus.
If they're all of different aliens, your analogy fails.
 

Ann K

Senior Member.
So how do you know how big P(A) is? There are 10^24 stars in the universe, so if there are 10,000 alien civilisations out there, that's exactly that 1:10^20 chance that one of them visits us.
Well....not really. If there are 10,000 out there, then there are 10,000 OUT THERE. That says nothing about their ability or desire to visit us. :)
 
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Mendel

Senior Member.
Well....not really. If there are 10,000 out there, then there are 10,000 OUT THERE. That says nothing about their ability or desire to visit us. :)
I was hoping to have @johne1618 figure that out.

My point is that we're very small compared to the universe, and his dismissing that unknown chance as being too small is just argument from incredulity.
 

Rory

Senior Member.
My theorem: introducing a "Bayesian argument" generally signifies the argument is either lost or the arguer is clutching at straws.

At least, that's been my experience thus far (based on a sample size of three).
 

Ann K

Senior Member.
I think the probability of a natural explanation for the Winchester 1976 case (lying, hallucinating or victims of a hoax), P( CE | Not[A] ), is less than 1/10. I would only need 10 such cases to amplify a prior P(A) / P(Not[A]) < 10^-10 up to a posterior probability of 50%.
Lying, hallucinating, victims of a hoax, OR simply mistaken. An experimental craft, a home-made contraption, a sci-fi movie being made, or an advertising stunt are all possibilities. I can imagine something like the driver being sufficiently startled, jerking the wheel to end up on the grass, and the car stalling as the gear shift goes into reverse. She says herself that the rash she suffered may just have been nerves. You're very generous with the possibilities you assign, but that's not a thing we have any real way to ascertain.
 

NorCal Dave

Senior Member.
For example see the following Winchester,UK encounter of 1976 with witnesses Joyce Bowles and Ted Pratt
She saw all kinds of things all the time and is not reliable. According to none other than you:

I must admit the number of her repeat experiences are astonishing. After her experience with Mr Pratt on 14 Nov 1976 they had a second experience together on 30 Dec 1976 when they claim that the whole car was abducted by the aliens! She also had a third experience in the car with a Mrs Strickland when an alien stopped her and gave her a message of a religious nature which it seems that she never divulged to anyone!
Content from External Source
-johne1618
www.metabunk.org/threads/winchester-close-encounter-1976.11957/ Post #32
 

johne1618

Active Member
She saw all kinds of things all the time and is not reliable. According to none other than you:

I must admit the number of her repeat experiences are astonishing. After her experience with Mr Pratt on 14 Nov 1976 they had a second experience together on 30 Dec 1976 when they claim that the whole car was abducted by the aliens! She also had a third experience in the car with a Mrs Strickland when an alien stopped her and gave her a message of a religious nature which it seems that she never divulged to anyone!
Content from External Source
-johne1618
www.metabunk.org/threads/winchester-close-encounter-1976.11957/ Post #32
Many people who report close encounters or abductions have repeated experiences. I don’t think that makes them unreliable per se. Also Mr Pratt seemed to be a reasonable fellow. He might have been coerced into making his statement but I don’t believe that was the case.
 
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johne1618

Active Member
@johne1618 's argument boils down to "if a UFO believer sees 10 reports that he's 99% confident in, he's going to believe with 50% confidence that we actually have alien visitors, unless the probability of aliens visiting us is actually smaller than 1:100,000,000,000,000,000,000".

One problem is that here on Metabunk, no debunker has ever had that level of confidence in an alien report.

The other problem is that the UFO community promotes reports as credible that say "aliens are small grey people with big heads" as well as "aliens are sexy females with big eyes that come to visit when the wife's not home", and if you believe in both with 99% confidence, you need to face the fact we're being visited by two alien civilisation, which means P(A) is really P(A_1)×P(A_2); so, worst-case, these 10 great reports correspond to 10 alien visits which now have to have a 1:100 chance each for the whole thing to be believable. If that's the case, somebody out there is marketing Earth as a tourist destination right now.

The third problem is that a lot of "close encounter" reports could really be time travel reports, and who's to say time travel is less likely than interstellar/intergalactic travel? So inherently the confidence in most CE reports should be 50% at best.

I think a better way to present the argument is in terms of supernatural vs natural rather than specifically aliens vs non-aliens. After receiving reports from 10 good cases of the supernatural that can’t be explained by natural causes, with say 90% confidence, one should believe in the supernatural with 50% confidence unless one’s prior belief in the supernatural is less than 1:10,000,000,000. The supernatural cases could consist of any phenomena that can’t be explained by our current understanding of science.
 

johne1618

Active Member
Lying, hallucinating, victims of a hoax, OR simply mistaken. An experimental craft, a home-made contraption, a sci-fi movie being made, or an advertising stunt are all possibilities. I can imagine something like the driver being sufficiently startled, jerking the wheel to end up on the grass, and the car stalling as the gear shift goes into reverse. She says herself that the rash she suffered may just have been nerves. You're very generous with the possibilities you assign, but that's not a thing we have any real way to ascertain.

An experimental craft, a home-made contraption, a sci-fi movie being made, or an advertising stunt seem very unlikely to me.

Is it your position that this event must have a “natural” cause? If so then aren’t you begging the question from the start?
 

Mendel

Senior Member.
After receiving reports from 10 good cases of the supernatural that can’t be explained by natural causes,
the first problem is that your example doesn't clear the "can't be explained" hurdle for the report (in a sense of "no imaginable explanation according to established science", not in the "we know precisely what it was" sense)

the second problem is that your percentage isn't anchored anywhere, it's a "sounds good" number, and with these, Rory's observation applies in full.

If I claimed Mrs Bowles's report was only 75% trustworthy, could you defend your choice of 90%? What if I claimed 95%?
 

Ann K

Senior Member.
An experimental craft, a home-made contraption, a sci-fi movie being made, or an advertising stunt seem very unlikely to me.

Is it your position that this event must have a “natural” cause? If so then aren’t you begging the question from the start?
It's my position that ALL known events have a natural cause. i understood your original point to be a discussion of the probability of that natural cause to be extraterrestrial. Given the known difficulties of space travel and the enormous extent of space, it seems prudent to do a thorough investigation of earthly possibilities. All of them must be considered to be more likely than life on another planet having evolved to resemble humans, making multiple excursions to earth over many years from locations which are years-to-centuries away, and individuals of different descriptions using craft which span a wide variety of shapes and behaviors.

And since your argument depends upon assessing the reliability of witness accounts, we are right to investigate their sincerity, accuracy, mental state, propensity to fantasize, motivation to exaggerate or lie, and sobriety as well.

Now are you sure you really want to throw the "supernatural" into the mix?

I see you define it as "The supernatural cases could consist of any phenomena that can’t be explained by our current understanding of science." Would that mean that thunder was supernatural until the causes were discovered and suddenly became natural after that point? No, it was merely natural but not yet understood. Ditto fo EVERY discovery of science. Please don't throw around the word "supernatural" by using your idiosyncratic definition.
 
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