PDA

View Full Version : over-flush odds

cowl
01-05-2005, 01:59 AM
If I'm holding K /images/graemlins/diamond.gifQ /images/graemlins/diamond.gif and the flop comes all /images/graemlins/diamond.gif's, how do I calculate the odds that my opponent, (who on fifth street showed heavy aggression) is holding an over-flush?

By the way no more /images/graemlins/diamond.gif's on board.

gaming_mouse
01-05-2005, 03:24 AM
cowl,

It's not clear to me exactly what your asking. In particular, I'm confused about why you would be interested in the odds after the 5th street action. Were you trying to decide whether or not to call a raise?

cowl
01-05-2005, 03:44 AM
I bet it and my opponent called on the flop, I bet and my opponent called on the turn, on the river I bet and my opponent raised. I am going to make the call here regardless, but I would like to know how to calculate the odds of my opponent flopping an over-flush.

My mention of the board on the river was to discourage people from thinking that a fourth diamond had fallen on board.

My question. If I flop a flush with a King of that suit in my hand, how are the odds calculated of an oppenent flopping a flush that could beat me.

My oppologies for being unclear, somewhat new to the forum and very new to the probablility pages. I know that this will sound very stupid but if you could explain how to read your explanation I would appreciate that too. /images/graemlins/smile.gif

gamble4pro
01-05-2005, 04:29 AM
If I get it correctly, you want the probability for your opponent to get an over-flush in the moment when all community cards are dealt (7 cards seen). In this situation, the only possible over-flush is a diamond suit with an ace.
So, 45 unseen cards, 8 diamonds left with one ace.
If you want the probability for one opponent (a specific one) to get that over-flush, this is 7/C(45,2) = 7/990 = 0.70%.
If you want the probability for at least one opponent to get this, the formula is: 7*n/990, where n is the number of your opponents (maximum one opponent could hold Ax of diamond)

cowl
01-05-2005, 05:41 AM
[ QUOTE ]
If I get it correctly, you want the probability for your opponent to get an over-flush in the moment when all community cards are dealt (7 cards seen). In this situation, the only possible over-flush is a diamond suit with an ace.
So, 45 unseen cards, 8 diamonds left with one ace.
If you want the probability for one opponent (a specific one) to get that over-flush, this is 7/C(45,2) = 7/990 = 0.70%.
If you want the probability for at least one opponent to get this, the formula is: 7*n/990, where n is the number of your opponents (maximum one opponent could hold Ax of diamond)

[/ QUOTE ]

Ok, bear with me here, (still trying to learn the lingo). You answered my the right question but now I have questions about your answere /images/graemlins/tongue.gif.

What is 7/C(45,2)(where do these numbers come from?)
45 is the number of unseen cards i believe. 2 is the number of cards that they are holding maybe? And 7/C I just have no clue. And what do you do with (45,2) to turn it into 7/990 = 0.70%.

thanks for any input. /images/graemlins/blush.gif

gaming_mouse
01-05-2005, 07:04 AM
[ QUOTE ]
What is 7/C(45,2)(where do these numbers come from?)
45 is the number of unseen cards i believe. 2 is the number of cards that they are holding maybe?

[/ QUOTE ]

Yes, 2 is the # of cards in each person's hand.

[ QUOTE ]
And 7/C I just have no clue. And what do you do with (45,2) to turn it into 7/990 = 0.70%.

[/ QUOTE ]

The expression is "C(45,2)", short for "Choose(45,2)" or "45 choose 2". It stands for the number of ways to choose 2 things from a total of 45 things. Since there are 45 cards left, this is the total number of possible hands people can hold.

Since there are a total of 8 diamonds left, and only 1 ace, there are 7 suited diamond hands that contain the ace of diamonds -- the ace of diamonds together with each of the other diamonds.

For more on "45 choose 2" (which, by the way, google will calculate for you -- try pasting it in) see the response by Zapped in the thread "open message to gaming mouse." He gives an excellent explanation of the concept and derives the formula for solving problems with it.

HTH,
gm

gamble4pro
01-05-2005, 09:31 AM
"Gaming mouse"'s explanations from his reply are entirely correct.

Reef
01-05-2005, 09:52 AM
Ok, how about Axs vs. Kxs vs. another random flush. All 3 hands flop a flush, no more to that suit on turn or river.

Odds?
(I've had this happen to me in a tourny once- thank baby Jesus that I don't play K4s)

gamble4pro
01-05-2005, 12:14 PM
You have to be more specific in enouncing the problem. What you hold, what is on the board and the event whose probability you are looking for. I do not understand what you meant by "Axs vs. Kxs vs. another random flush".

Reef
01-06-2005, 05:41 AM
[ QUOTE ]
You have to be more specific in enouncing the problem. What you hold, what is on the board and the event whose probability you are looking for. I do not understand what you meant by "Axs vs. Kxs vs. another random flush".

[/ QUOTE ]

ok, As3s vs. Ks4s vs. 6s7s with Js2sTs flop

Cobra
01-06-2005, 10:44 AM
This information is derived from an article I read written by Brian Alspach. He has some great poker probability articles. Google Alspach + poker and you will get his web site. Specifically the following probabilities are that the board comes three to a suit, you have two of that suit, and you are wondering if anyone else was dealt two to that suit with a card higher than your high card.

If you have K obviously there can only be one higher card.

Probability 9 Opponents-.064, 6 opponents-.042, 4 oppon-.021

If you have a 6,7 and the board has a JT you have 5 cards left higher than your 7.

Prob 9 Opponents-.214, 6 Oppenents-.146, 4 opponents-.075

Cobra

elitegimp
01-06-2005, 04:39 PM
Here's another way to think about it:

If you are 100% sure your opponent has the flush (he's fairly passive and wouldn't raise a board with 3 /images/graemlins/diamond.gifs without two of his own), then there are C(8,2) = 28 possible hands [There are 8 diamonds left in the deck, and 28 ways to put 2 of them together in a 2-card hand]. As mentioned previously, there are 7 hands that contain A /images/graemlins/diamond.gif x /images/graemlins/diamond.gif.

Conclusion: IN THIS SCENARIO, where you are 100% sure your opponent has the flush as well, he beats you 7 times out of 28, or 25% of the time. Since you are a 3:1 favorite, this is a clear 3-bet /images/graemlins/smile.gif