Yes, now I understand. I think what made it hard for me conceptually was that
P(Rj) can be rewritten n different ways for each feature ci 1 <= i <= n. I had
this problem with your example too. I first thought that the lines with the
factors were arbitarary, but then I realized that each line goes with one way
of rewriting P(Rj) it all became clear.
Quoting Rémi Coulom <[EMAIL PROTECTED]>:
to Magnus: If you consider the example of section 2.2: 1,2,3 wins
against 4,2 and 1,5,6,7. The probability is
P=c1c2c3/(c1c2c3+c4c2+c1c5c6c7). For this example:
N1j=c2c3,B1j=0,M1j=c2c3+c5c6c7,A1j=c4c2
N2j=c1c3,B2j=0
N3j=c1c2,B3j=0
N4j=0,B4j=c1c2c3
I will add this example to the paper if it makes things clearer.
Regarding the definition of Wi, |{Nij|Nij!=0}| means "number of
elements of the set of all the Nij that verify Nij!=0". It is
incorrect. It should be |{j|Nij!=0}|. I'll fix it in the final
version. As I wrote in the next sentence, it is equal to the number
of wins of i.
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