p(BIA) - P(A I B)P(B) - what the hell ?

Hey :) . 

Jeg er igang med at læse en matematisk bog, og jeg fatter ikke helt de udregninger der bliver lavet.

Nogen der kan forklare mig hvordan ?

Her er et lille udsnit :

Now that we have an a priori distribution of win rates, we can apply Bayes' theorem to this
problem. For each win rate, we calculate:
A = the chance of a win rate of 1.15 being observed.
B = the chance that this particular win rate is the true one (a prion).
We cannot directly find the probability of a particular win rate being observed (because the
normal is a continuous distribution). We will instead substitute the probability of a win rate
between 1.14 and 1.16 being observed as a proxy for this value. Recall that the standard
deviation of a sample of this size was 1.61 bets. P(A I B) is calculated by simply calculating the
weighted mean of P(A j B) excluding the current row. From a probability chart:

En skala :

WinRate      : P(B)      : p(A IB)       : p(A IB )         : p(B)
------------------------------------------------------------------------------------------------------
-5  BB/100 : 0.25%    0.000004    0.00222555    99.75%
-4  BB/100 : 2%         0.000031    0.00226468    98%
-3  BB/100 : 8%         0.000182    0.00239720    92%
-2  BB/100 : 20%       0.000738    0.00259053    80%
-1  BB/100 : 39.5%    0.002037    0.00233943    60.5%
+0 BB/100 : 20%       0.003834    0.00181659    80%
+1 BB/100 : 8%         0.004918    0.00198539    92%
+2 BB/100 : 2%         0.004301    0.00217753    98%
+3 BB/100 : 0.25%    0.002564    0.00221914    99.75%
-----------------------------------------------------------------------------------------------------

Applying Bayes' theorem to each of these rows:

              P(A I B)P(B)
p(BIA) = --------------------------------------
             P(A I B)P(B) + P(A I B)P(B)