PDA

View Full Version : How often will the board pair


drawoh
05-01-2005, 05:57 PM
In Hold'em how ofter (%) will the board pair?

Mr Mojo Risin
05-01-2005, 07:00 PM
There is a 19.1% chance that it will pair on the turn, and a 26.1% chance that it will pair on the river.

gamble4pro
05-02-2005, 07:22 AM
I will show you the calculation of odds for the flop being unpaired. Then, you will do (1- calculated odds) for the opposite event (flop to be paired).
-From outside view (no pocket cards seen):
The favorable combinations are (ABC), with A, B and C mutually different (as value), in number of
C(13,3) *4*4*4 (each 3 of 13 values, 4 cards each), from possible.
The probability is C(13,3)*4*4*4 / C(52,3) = 18304/22100 = 82.823% .
-When taking into account the own pocket cards:
If you hold a pair let it be (AA) the favorable combinations of flop are (BCD), with B, C and D mutually different (B, C or D could be even A). These can be differentiated as follows:
(BCD), with B, C, D different from A, in number of C(13,2)*4*4*4 and
(ABC), with B, C different from A, in number of 2*C(12,2) *4*4 .
Totally, we have 9008 favorable combinations from 22100 possible.
The probability is 9008/22100 = 40.760% .
If you hold two unpaired cards let them be (AB) the favorable combinations of the flop are (CDE), with C, D and E mutually different. These can be differentiated as follows:
(CDE), with C, D, E different from A and B, in number of C(11,3)*4*4*4
(ACD), with C, D different from A and B, in number of 3* C(11,2)*4*4
(BCD), with C, D different from A and B, in number of 3*C(11,2) *4*4
(ABC), with C different from A and B, in number of 3*3*11*4 .
Totally, we have 16104 favorable combinations from 22100 possible.
The probability is 16104/22100 = 72.868% .