over-flush odds

[cowl]

If I'm holding K [img]/images/graemlins/diamond.gif[/img]Q [img]/images/graemlins/diamond.gif[/img] and the flop comes all [img]/images/graemlins/diamond.gif[/img]'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 [img]/images/graemlins/diamond.gif[/img]'s on board.

[gaming_mouse]

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]

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. [img]/images/graemlins/smile.gif[/img]

[gamble4pro]

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]

[ 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 [img]/images/graemlins/tongue.gif[/img].

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. [img]/images/graemlins/blush.gif[/img]

[gaming_mouse]

[ 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]

"Gaming mouse"'s explanations from his reply are entirely correct.

[Reef]

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]

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]

[ 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