For what it's worth I have noticed that people who are familiar with Bayesian filtering seem to have a mental block when it comes to understanding this. People who know nothing about bayesian get it instantly. Here's the actual formula.

card(Test_message intersect Spam diff Ham) minus card(Test_message intersect 
Ham diff Spam)



On 08/17/16 09:16, Shawn Bakhtiar wrote:

On Aug 17, 2016, at 3:43 AM, Matus UHLAR - fantomas <uh...@fantomas.sk <mailto:uh...@fantomas.sk>> wrote:

On 16.08.16 20:06, Marc Perkel wrote:
What I'm doing is looking for fingerprints in email that intersect HAM and not in SPAM - which would be a HAM result.
If it matches SPAM and does NOT match HAM - then it's SPAM.

The magic is in the NOT matching on the other side.

so, if mail matches both hammy and spammy tokens (or token sets), you don't
classify at all?


I guess what is confusing me (and I imagine others, as alluded to by Matus) is the fact that you are describing a special condition of Bayes' probability theorem. You are testing two variables (match SPAM and match HAM) (not matching is simply the negation of matching) thus giving you four conditions:

1) SPAM  &&HAM
2) SPAM  &&~HAM
3) ~SPAM &&HAM
4) ~SPAM &&~HAM

Here is a great diagram to show the four probable conditions:
https://en.wikipedia.org/wiki/Bayes%27_theorem#/media/File:Bayes%27_Theorem_2D.svg

So (if I am correct) Matus is asking what if condition 1 is true? How are you classifying an email than? Which is often the state of most emails, and thus why the use of Naive Bayes spam filtering, which generates a probability based on Bayes' probability theorem and is the conventional methodology to date. A Rose by any other name....

Condition 4 is obvious it's nothing you have ever seen so classifying it anything other than HAM would be a huge mistake (IMHO), and fully covered by the aforementioned theorem as the probability of SPAM would (should) be 0. Same with Condition 3, obviously it never hits SPAM so wether it matches HAM or not you're going to treat it as HAM anyway same as condition 4.

That leaves condition 2. Which (if I'm not mistaken) is "... it matches SPAM and does NOT match HAM - then it's SPAM.". Which brings us back to Matus question, what if the email contains a single HAM token? Two HAM tokens? This is exactly what Bayes' probability theorem is designed for. All you are doing is defining a special condition in which the HAM probability is ZERO.

I think that's were I need to understand a bit more about what HAM means in this solution, does getting a hit on HAM somehow negate it being SPAM completely? In other words if the email contains some set of tokens that are SPAM, yet only one HAM token, that single HAM token makes it not SPAM? If so, you have a long way to go in convincing me that this is a good solution.

So if I say to you, "Let's get some lunch" that's ham because spammers never say that, but normal people do. So the way to test what "spammers never say" is to store what they do say and see if it's NOT in the list. (Thus the infinite set)


Actually I get SPAM with that very set of tokes in it. If somehow the HAM rating of it overrides the SPAM, I don't believe it would have a desirable effect.

I get plenty of:

"
Hay Shawn,

Hope you have time to do some lunch, click on this link and check out my new pictures!

Wannabe Phisher
"

Based on your example there's plenty of HAM and SPAM tokens in there, "Click on this link" high probability of SPAM-e-ness, would it get HAMed based on "hope you have time to do lunch". Or am I missing something?


Similarly, there's only so many ways to misspell viagra, and good email wouldn't have it spelled wrong.

Does that make sense?


Again, what you are saying makes sense in that it is special condition of the probability theory, What does not make sense is why would you not simply use the probability theory, that already encompasses that condition?

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