Quote:
Originally Posted by mutanNZ
Do I understand correctly that formula for:
a) hands with 4 number of successfful combos (f.e. 54s, 98s) is 4*C(46,2)/C(50,5)
b) hands with 3 number of successfful combos (f.e. 64s, QTs) is 3*C(46,2)/C(50,5)
c) hands with 2 number of successfful combos (f.e. 52s, K9s) is 2*C(46,2)/C(50,5)
d) hands with 1 number of successfful combos (f.e. A2s, 73s) is 1*C(46,2)/C(50,5)?
I mean probability by the river
Yes, I think this is correct.
Quote:
Originally Posted by statmanhal
You have to avoid double counting. For 54s, the successful outcomes are as follows: (x = excluding)
123 45 x6 Combos = 1*C(46,2)
x1 23 45 6 x7 Combos = 1*C(45,.2)
x2 3 45 67 x8 Combos = 1*C(45,.2)
x3 45 678 x9 Combos = 1*C(45,.2)
I don't understand what this is saying.
It seems to me that the formulas given above by mutanNZ are correct. Am I totally misunderstanding something?
Sorry for such a late reply. This is the first time I have taken a look at this thread.
