On May 8, 12:00 pm, FFMG <spambuc...@myoddweb.com> wrote:
> On Tuesday, 8 May 2012 16:41:18 UTC+2, Jussi Piitulainen wrote:Neither. There is no requirement on the number of documents, and the
> > FFMG writes:
> > > > You have made an incorrect independence assumption. As both
> > > Sorry, that's not an assumption, that's the way the problem
> > > And they are independent variables, the presence of "naughty" is not
> > > The formula is P(C|F1...Fn) = P(C)P(F1|C)...P(Fn|C)
> > > So, given the problem in my original post, the result is not between
> > Probability theory only gives you
> > Then come the independence assumptions which allow you to expand
> > If "naughty" and "money" were exactly independent and probabilities
> > Since we don't want to accept 1/2 = 1 and we think that relative
> So, if I understand you correctly the 2 issues at hand are:
Bayes formula works. However, your original stab was certainly false,
since it contained "money" twice and did not contain "naughty" at all.
As such it is difficult even to figure out what you were trying to
> So, as the formula seem to be correct in my example, I guess my question would be, is there any way of binging the number back between 0 and 1? or can I simply assume that anything > 1 is in fact 1, (or almost 1).As Jussi explained, you have to use the data correctly, and you got to
think a little. What is Prob(spam|naughty)? What is Prob(spam|money)?
What then do you think of Prob(spam|naughty AND money)? Hint, in this
case you do not even need Bayes.
> Following on to that, I also see many examples where the denominator can be ignored as it can be regarded as constant. But then how can I calculate how close the probability of a number is to 1? (because without a denominator I have no idea how close the probability of a document is to be 'spam').You`ll have to explain better what you mean by this. As is, it make no
sense to me.
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