At Odds with Omaha...

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Posted: 13-Oct-2009 13:38 by CheezYCheezY

Doesnt really matter how many r still in the deck..only diff is with NL Holdem you see 5 cards postflop so 47/52 r unknown...

with omaha you will see 7 that means 45/52 r unknown..

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Posted: 13-Oct-2009 13:39 by CheezYCheezY

2 hole cards 3 community cards for the peepz that will react on my 1st post..

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Posted: 13-Oct-2009 13:59 by gladdened

Errr.....thanks for your reply.
From your response, it seems you are saying that the probability of hitting a specified number of outs is unaffected by the number of cards in the deck...which doesn't fit with my understanding of probabilty!
/chuckles...and returns to google for more reading.

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Posted: 13-Oct-2009 14:19 by IneedUrChips

Errr.....thanks for your reply.
From your response, it seems you are saying that the probability of hitting a specified number of outs is unaffected by the number of cards in the deck...which doesn't fit with my understanding of probabilty!
/chuckles...and returns to google for more reading.

gladdened, 13/10/2009

It doesnt affect at al. Think of it this way: Normally the deck in holdem at the end of a hand F/T/R on a full ring + 3 burn cards leave 24 cards that arent used. These 24 dead + 3 burnt cards are never ever gonna influence probability Because they fysicly dont exist.

Thus odds and outs stay the same regardless.

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Posted: 13-Oct-2009 17:13 by Robbieweeza

As long as you don't know which cards are out, the unknown cards in the deck could be any of them.

Try looking at it this way: if you have all outs but one (i.o.w. your opponent has 1 out) and there's only 1 card left, this doesn't mean it's the 1 out he needs.

Any 2 unknown cards are interchangable statistically.

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Posted: 13-Oct-2009 20:34 by gpj77

please correct me if im wrong, but it doesnt make sense if there are the same number of outs but less cards in the deck for the odds of hitting the outs to remain constant.
Lets compare 2 flush draws 1 in omaha and 1 in holdem

in hold em if you are holding 4 cards to a flush and want to find out what the percentage of hitting a flush on the turn or river it is

(9/47 +9/46) *100= 38.7%
but in the same situation in omaha you know 2 more cards which are not your outs, so

(9/45+9/44)*100=40.5%

so you are slightly more likely to hit a flush draw in omaha than in holdem, i think.

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Posted: 13-Oct-2009 21:00 by gladdened

Thanks for the replies.
It seems that what is being worked out above are the odds of specified outs from a full deck less known cards, all other dead cards (other players cards and burn cards) are included in the calculation since they are unknown.
How realistic/accurate these odds are will vary from game to game, situation to situation. For example playing ten handed hold' em (with all the cards face up), would engender very different 'true' odds to heads up play (because it is likely that some or all of our outs are held by other players). Since we can't possibly know the information that 'face-up' would provide, probability is assumed...

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Posted: 13-Oct-2009 22:51 by Robbieweeza

It's a false logic many players use (no offense). People tend to think that in order to have outs, they need to be in the deck.

The cards in other players' hands or in the muck are of the same value as the ones on the bottom of the deck. After all, they're not being used. All that's relevant to your odds of winning are the cards that are on top of the deck. The rest of the deck you can put in the muck, in your pants or frisbee them around the room, they're not ever going to be of any relevance.

In the beginning of the game the deck is shuffled and the cards remain in the position they're in when the shuffling ends. Suppose 51 cards are dealt and 1 is left. Then what is relevant to you are the odds of one of your outs being in that position. Assuming the deck was shuffled perfectly, every card has a 1/52 chance of being in any given position in the deck. Remember, the cards' positions are fixed after the shuffle.

So if you have a flush draw (9 outs) and there's 1 card left, there's a 9/52 chance that one of your outs ended up there in the beginning of the hand. If one of them was in that position, then it doesn't matter what happened to the other cards, and the other way around.

When you speak of things like 'true odds', you're being results oriented. It doesn't matter to us whether or not someone holds our outs, what matters to us is the likelyhood of them NOT holding our outs. You don't know if the cat is dead until you open the box.

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Posted: 14-Oct-2009 00:26 by gladdened

Ahhhhhh...I see...said the blind man as he walked into the lamp post.
I understand the logic there robbie, thanks for taking the time to explain it.

edit 1 nice schrodingers cat reference, me and a good pal were talking about the observer effect/nature of duality in poker just the other day, and i rather smoothly (/coughs) dropped some schrodinger on him. He thought i was making it up i reck...

edit 2 absolutely no offence taken...i've been feeling ill ease with my conception of probability in poker, and have had a creeping certainty that i've thinking about it in either a different way (or, more likely, the wrong the way). I needed tellin!

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Posted: 14-Oct-2009 20:56 by gpj77

at robbie,
does this mean my previous post is wrong,
if so i dont get it

sorry if this sounds stupid but surely it would be 9/47 chance for the turn card to be the out, and then 9/46 on the river.

think of it this way if a 52 card deck is spread out in front of you and you had to pick one, say K of clubs. the odds of you picking that card is 1/52.
but
if you fail to pick that card there is 51 unknown cards left. when you try to pick the card again you have 1/51 chance of choosing correctly. and so on. so the odds of choosing correctly increases as there is less cards to choose from.

so back to the flush draw, in omaha you have got 2 more hole cards so you have "chose 2 more cards incorrectly" so there is less cards for the 9 outs to come from on the turn and river therefore increasing the odds of it hitting.

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Posted: 15-Oct-2009 12:27 by Chivalrousgent

Probabilities are calculated according to unknown cards, not cards in play/in deck.

So yes, the more cards become known (by being dealt to the flop, turn or being accidentally exposed), the more you reduce the denominator (outs/cards remaining).

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Posted: 04-Nov-2009 16:46 by datgsta

i should never have looked at this thread, im now so confused half of you are saying 1 thing and half are saying the opposite. who is right?

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Posted: 07-Nov-2009 11:53 by gladdened

Robbie's last post is answered my original query, not so sure what some of the other guys are sayin...maybe it's a bit like the offside rule, everyone thinks they know what it is, just can't explain very well!

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Posted: 07-Nov-2009 13:29 by Sinistar

Robbie's post gets a little confusing where he says:

So if you have a flush draw (9 outs) and there's 1 card left, there's a 9/52 chance that one of your outs ended up there in the beginning of the hand.

At this point you'd know where 8 cards are (4 in your hand + 4 on the board) so it's impossible for any of those to be dealt on the river and that changes the probabilities. Any specific other card can now be in 1 of 44 possible places so will appear on the river 1 time in 44. (43 times in 44 it will be elsewhere, whether that be dealt to another player or in the unused portion of the deck.) So the probability of hitting the flush on the river is 9/44, not 9/52 as Robbie suggests.

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Posted: 07-Nov-2009 14:37 by Robbieweeza

You're right, Now that I read it again it makes no sense. I meant that sentence in a different way but even in that way it's wrong.

Anyone else find that chance calculation skills are not like riding a bike and get rusty really fast?