How your hole cards affect the flush chances in Omaha.

[Filip]

What got me thinking about this little problem was after being berated by railbird after i cracked his friends nut straight with a flush. I had 3 clubs in my hand and he thought that i sucked for playing the hand.

Using a Hypergeometric probability distribution, h(x,n,M,N), where

x = Sample: number of clubs, 3 to 5
n = Sample size: 5 (flop, turn, river)
M = Population: number of clubs in deck, 11 to 9
N = Population size: 48 (the cards left after my 4 hole cards)

I add the probabilites for x = 3..5 and of course alter M = 11..9 accordingly. The results for how the number of clubs in my hand affect my chances of getting a flush are as follows:

2 clubs, P(flush) = 0,071577243

3 clubs, P(flush) = 0,05407451

4 clubs, P(flush) = 0,039294424

Nothing chocking or unexptected there. But now to my question.

In a full table game (10 players) we have 40 cards out and i cant stop to reason that if i got 3 clubs in my hand the affect they will have on the flop,turn,river will be neglected by the other 36 cards out.

Since 40 cards out of 52 are being dealt out i say that these 40 cards have a discrete probability distribution of suits. So what if the 4 cards i choose to look at (my hand) out of the 40 happens to be 2,3 or 4 clubs.

Hence i come to the conclusion that the chances of me getting a flush are the same no matter what number of clubs i hold.

Thoughts?


/Filip

[gaming_mouse]

I'm not sure I understand the question.

Are you asking about the chance of flopping a flush given that you hold 4 of a suit? Hitting a flush by the river?

Obviously, if you hold 4 of one suit, you have the best chance of flopping a flush or hitting one by the river, and you don't need the hypergeomtric function to calculate it. Eg, given that you hold 4 clubs, the chance of flopping a flush is:

1 - ((39 choose 3)/(48 choose 3)) = .471

If you hold only 3 clubs, that chance drops down to:

((10 choose 2)*39 + (10 choose 3))/(48 choose 3) = .108

Again, please clarify if this is not what you were asking.

gm

[Filip]

In omaha you must use 2 cards and only 2 cards.

Sorry if my question was kinda blurry, what i boils down is that i say that on a full table it doesnt matter how many clubs u hold since it is being balanced by the other 36 cards. Hence the chances of a flush are the same if u have 2, 3 or 4 clubs. Its a statement/question.

[tylerdurden]

[ QUOTE ]
Hence the chances of a flush are the same if u have 2, 3 or 4 clubs. Its a statement/question.

[/ QUOTE ]

No.

OTOH, if you fix the cards in your hand at X clubs (where X is 2 3 or 4) then your chances of getting a flush are the same no matter how many opponents you have.

[gaming_mouse]

[ QUOTE ]
In omaha you must use 2 cards and only 2 cards.

Sorry if my question was kinda blurry, what i boils down is that i say that on a full table it doesnt matter how many clubs u hold since it is being balanced by the other 36 cards. Hence the chances of a flush are the same if u have 2, 3 or 4 clubs. Its a statement/question.

[/ QUOTE ]

Okay. Sorry about that. I've played Omaha once, like a year ago. The chance of flopping a flush has nothing to do with how many oppos you have. If you hold 4 of a suit, this will certainly decrease your chance of flopping a flush.

Eg, you hold 4 of a suit. Chance of flopping a flush:

(9 choose 3)/(48 choose 3) = 0.00485661425

If you hold only 2 of a suit:

(11 choose 3)/(48 choose 3) = 0.00953977798

So your chance is almost twice as good holding 2 of a suit.

gm

[pzhon]

Holding 3 or 4 of a suit gives you a lower chance to make a flush. There are fewer ways to flop a flush or flush draw, and if you flop a flush draw, you have 7 or 8 outs to the flush rather than 9. However, if you get the flush with 3 or 4 of that suit, there is a greatly reduced chance that you are up against another flush. That can help if you don't have the nut flush.

[Filip]

Ok, try and look at the problem in two different ways.

Looking at the whole 52 card deck the probabilities that i have a flush on the river are undisputed:

2 clubs, P(flush) = 0,071577243
3 clubs, P(flush) = 0,05407451
4 clubs, P(flush) = 0,039294424

This is must certinaly the correct solution.

BUT

What if i describe the problem like this:

You remove the top 40 cards from the deck.
Isnt it correct too say that both piles of 12 and 40 cards will each have a discrete probability distribution of suits?

I then choose 4 cards from the 40. How can the suits of these 4 cards change the probability distribution of the suits of the other pile with 12 cards (that will make the flop/turn/river).


Can you see my thought process now?

[gaming_mouse]

[ QUOTE ]

I then choose 4 cards from the 40. How can the suits of these 4 cards change the probability distribution of the suits of the other pile with 12 cards (that will make the flop/turn/river).


Can you see my thought process now?

[/ QUOTE ]

This is incorrect thinking. You are not taking into account the new information given to you by knowing that you hold, say, 4 clubs. This DOES change the distribution of the remaining cards. In Bayesian parlance, you are now interested in the posterior distribution of the clubs, given that you hold 4 clubs. The prior distribution does not change, but you are no longer interested in that.

To see why your logic does not hold, take your argument to an extreme: Say we are looking at the number of aces, rather than the number of clubs. We choose 4 cards from the 40. They are all aces. How can this change the distribution of aces in the remaining pile of 12 cards?

[Filip]

Thanks, i knew my reasoning was incorrect.