View Full Version : math on the flop
imported_Robert Andersson
04-26-2005, 06:15 PM
Ok you have J-9 off as starting hand.
How big is the probability that you flop either:
- one pair?
- two pair?
- an openended straight draw?
Any comments highly appreciated.
/Robert
Filip
04-26-2005, 08:19 PM
[ QUOTE ]
Ok you have J-9 off as starting hand.
How big is the probability that you flop either:
- one pair?
- two pair?
- an openended straight draw?
Any comments highly appreciated.
/Robert
[/ QUOTE ]
Odds of flopping:
one pair = 29%
I'll need a moment to do the other calcs.
Cobra
04-27-2005, 02:16 PM
If you are holding J 9 offsuit the following are the primary flops you are interested in. Remember with you holding two cards there are 50 unknown cards so the flop will come in (50c3) = 19600 ways.
Prob of four of a kind or a boat with J 9
= (6c3) = 20/19600 = .10%
Prob of a straight
You will get a straight if you flop TKQ, TQ8, T87 so:
= 3*4*4*4 = 192/19600 = .98%
Prob of three of a kind
= 2*(3c2)*44 = 264/19600 = 1.35%
Prob of two pair using J and 9
= (3c1)*(3c1)*44 = 396/19600 = 2.02%
Prob of eight out strt draw and J or 9
You can get an eight out straight draw and J or 9 with a QT/J or 9, or a T8/J or 9 so:
= 2*4*4*6 = 192/19600 = .98%
Clean J or 9
= 6*(44c2) = 5676 - 192 = 5484/19600 = 27.98% ( remember you must subtract out the J or 9 above that is included in the eight out computation)
Clean eight out straight
You can get an eight out straight with the following hands:
qtx were x is not qtk8j9
t8x were x is not t8q7j9
tk7
Lets take qtx and look at it. You eliminate the qt to avoid a double count, you eliminate the q7 that gives a straight, and you eliminate the J9 thats counted above. You then add back in the times when you get qtq, qtt.
= 2*4*4*28 + 2*(4c2)*4 + 4*4*4 =1056/19600 = 5.39%
Cobra
imported_Robert Andersson
04-27-2005, 04:47 PM
Thanks Cobra!
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