(23/208)*10 = 1.105769231
(23/104) * 3 = 0.663461538
(29/117) * 9 = 2.230769231

Add those up and you get 4....

But, in answer to the second part...

What I am doing is finding the probability of the area in question and then taking the "weight" of probability per square in the area.

The probability of the area C+A is 23/16, but yet there are only 13 squares...

The 16 is number of squares around 1 mine + number of squares around the other mine.

The division by 13 determines the weight per square... if the sum of the uncovered square it is touching is 1. If the sum is greater, the multiplier affect would go up.

I don't know if this clears up what I was thinking or not... let me know.