Odds of 3 suited cards on the board after the river
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Odds of 3 suited cards on the board after the river
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,
Thanks,
- ktw85
- Posts: 3
- Joined: Thu Aug 11, 2005 2:29 pm GMT
Some Clarification
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.
Hope that helps, although I don't really know what he'll do with this information other than lose some more money.
- ktw85
- Posts: 3
- Joined: Thu Aug 11, 2005 2:29 pm GMT
Can you share the math you used to reach that?
Thanks for figuring that out...can you share the math with me?
Ken
Ken
- ktw85
- Posts: 3
- Joined: Thu Aug 11, 2005 2:29 pm GMT
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.
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.
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