Odds of 3 suited cards on the board after the river |
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Posted Thu Aug 11, 2005 1:33 pm GMT by ktw85
What are the odds of the board having 3 suited cards after the river? Seems to happen to me a lot but I generally don't have that suit...(!)
Thanks,
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Posted Thu Aug 11, 2005 1:35 pm GMT by Dave B
I am guessing 20%
Posted Thu Aug 11, 2005 2:07 pm GMT by supafrey
we don't need your silly guesses, math man!
wait, was your question of whether this would happen with someone holding two of those suit in their hand, or just 3 suited in general... which is pretty likely...
Posted Thu Aug 11, 2005 2:13 pm GMT by Dave B
my guess is based on remembering if you are suited, you have a 5% chance of hitting a flush. 4 suites=20% chance of any suite.
Posted Thu Aug 11, 2005 2:35 pm GMT by supafrey
yeah, i know. I was just being a jerk. But that's also why I asked if he meant whether someone had the two suited cards in their hand or not.
Posted Fri Aug 12, 2005 9:20 am GMT by ktw85
Thanks for your posts; I'm actually just talking about the board (exclusive of what you have in your hand). It's based on a question from my newbie brother, who asked, "How often will the board end up with 3 suited cards after the river? Seems to happen to me all the time (whether I actually have any of those cards or not).
Hope that helps, although I don't really know what he'll do with this information other than lose some more money.
Posted Fri Aug 12, 2005 2:06 pm GMT by Adam Marshall
I did some quick math and came up with 32.62%, assuming you didn't consider hole cards. That also assumes it is 3 to a flush exactly. 4 and 5-flushed boards were not considered.
Posted Fri Aug 12, 2005 2:12 pm GMT by ktw85
Thanks for figuring that out...can you share the math with me?
Ken
Posted Sat Aug 13, 2005 3:23 am GMT by Adam Marshall
Ug... 
Sure!
It's based mostly on the formula for good ole combinations...
n! / (n-r)! r!
With a board of 5 without considering pocket cards there are 2598960 possible combinations of cards...
52! / (52-5)! 5!
Of the total possible number of 3 card flushes that exists there are 1144...
13! / (13-3)! 3! X 4 different suits...
If you have a three card flush with a five card board there are two leftover cards, a total possible 741 different combinations that don't contain more flush cards...
39! / (39-2)! 2!
So ( 1144 X 741 OR 847704 ) would be the total possible number of 5 card boards that exist that have three card flushes.
So the probability of boards that have three card flushes vs. the possible combinations of all five card combinations of boards would be 847704 / 2598960 = 0.3261704681872749099639855942377
Multiply by 100 to make a percentage.
I'll say also that at first it didn't sound correct to me, but then I realized that you HAD to have a two card flush no matter what. There is no way that you couldn't have less than a two card flush. Made it sound a little more digestable to me.
I'm not a big math guy outside of poker, so I'd love to have the math geniuses in the forum tell me that I'm correct or an idiot.
Posted Sun Aug 14, 2005 10:40 am GMT by ORGrinder
32% is the correct number.
Posted Mon Aug 15, 2005 9:51 am GMT by Loonbat
Looks good to me ...
LOON'S STAMP of APPROVAL
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