Omaha odds-question

[petsei]

Could anyone tell me how to correctly calculate the odds of a 17outer against a made set on the flop.
According to my own calculations the percentage of making a straight with 17 outs on either turn or river amounts to approximately 62%, whereas it is about 35% to improve a set to a boat.
How would I fare with this possible straight, if I were calling a potsize bet (being fairly certain that my sole opponent has flopped a set) moving all-in at the same time?
Thanks in advance.

[Buzz]

Petsei - I like to think in terms of actual cards. Let's presume you hold

A/images/graemlins/spade.gifJ/images/graemlins/spade.gif9/images/graemlins/heart.gif7/images/graemlins/spade.gif, your opponent holds T/images/graemlins/spade.gifT/images/graemlins/heart.gif5/images/graemlins/diamond.gif2/images/graemlins/club.gif and the flop is

T/images/graemlins/club.gif8/images/graemlins/diamond.gif3/images/graemlins/heart.gif. In this case you have a 17 outer, but don't hold any of your opponent's outs (nor does your opponent hold any of your outs). But if you held the 3/images/graemlins/spade.gif instead of the A/images/graemlins/spade.gif, it would be a different matter. Although you'd still have 17 outs, you'd tie up one of your opponent's outs.

At any rate, I've contrived the hands and flop so that you have 17 outs to make a straight, your opponent has a set, and there is no chance of either of you making a flush.

If you take the bets one at a time, I think it's correct to compare the implied pot odds to the odds against making your hand on each successive betting round. However when you go all-in on the third betting round, you see both of the next two cards for the price of one.

You need to hit any one of your outs on the turn or river. But even when you hit one of your outs if the board pairs you still lose.

On the turn, there are 17 good cards for you, 7 good cards for your opponent, and 17 neutral cards. But then that changes on the river, depending on what the turn card is. In other words, the board can pair either by pairing one of the flop cards or with a back-door pair.

If you catch one of your outs on the turn, and the board subsequently pairs that out (and there are 6*2+3*3 ways that can happen), then you still lose.

O.K. thus far?

With all that in mind, you win
17*16/2-(6*2+3*3) + 17*17 ways (out of 990). Otherwise your opponent wins.

The probability you will win is 404/990.
The probability your opponent will win is 586/990. The odds against your 17 outer winning are 586 to 404 or roughly 3 to 2.

Note that all 17 outers don't have these exact odds. For example, if, as mentioned above, you held 3/images/graemlins/spade.gif instead of the A/images/graemlins/spade.gif, then there would be 17*16/2-(6*2+3*3) + 17*18 = 421 ways (out of 990) for you to win. That would make the odds against you 569 to 421.

Also note that considering flush draws will alter the odds significantly.

Buzz

[bigpooch]

Actually, if you hold 4 cards and your opponent holds 4 and
there are 3 on the flop, there are only 41 cards that can
show up on the turn and river, i.e., C(41,2) = 820 possible
combinations. In LHE, there would be only C(45,2)=990 such
combinations; or alternatively, just considering one hand,
there would be 990 combinations.

[Buzz]

Thanks, Big Pooch.

The probability a 17 outer will win is 404/820.
The probability the opponent who flopped the set will win is 416/820. The odds against the 17 outer winning are 416 to 404 or roughly even.

Note that all 17 outers don't have these exact odds. For example, if, as already mentioned, the player with the 17 outer held the 3/images/graemlins/spade.gif instead of the A/images/graemlins/spade.gif, then there would be 17*16/2-(6*2+3*3) + 17*18 = 421 ways (out of 820) for the 17 outer to win. That would make the odds 421 to 399 in favor of the player with the 17 outer (a bit better than even).

I didn't mention in the previous post that the 17 outer clearly has favorable odds to call the all-in-forcing bet on the third betting round.

Between all of us, I guess we'll eventually get the odds right. Thanks again for correcting the careless error on my part, Big Pooch.

Buzz

[petsei]

Thank you very much for going to the trouble of explaining at such great length. But I'm afraid it's casting pearls before swine. I understand only part of it. I gather that 990 are the possible card combinations. What I don't get is how you arrived at the formula:17*16/2-(6*2+3*3)+17*17.
I would have thought the combinations in my favour (without having subtracted the combinations that are helpng the opponent) should be 17*16/2 +1 7*28. The rest was over my head anyway. Thanks again.
petsei

[Buzz]

[ QUOTE ]
I understand only part of it.

[/ QUOTE ]

Petsei - I'll try to make it clearer. If you really want to understand, I think it is essential for you to get a deck of cards, so that you can follow along.

First remove the four cards I have conjured up for your hand, the four cards for your opponent's hand, and the three cards for the flop. Put these eleven cards in three different spread-out piles. I could also have picked different cards, but the ones I have chosen will do.

Next take your 17 outs from the deck (Q Q Q Q J J J 9 9 9 7 7 7 6 6 6 6) and put them in a fourth pile. We'll call this the "friendly" pile.

Next take the 7 cards that will pair the flop from the deck and put them in a fifth pile. (T 8 8 8 3 3 3). We'll call this the "bad card" pile.

There should be 17 cards left in the deck. Put them in a sixth pile. We'll call this the "neutral" pile.

O.K.? The above may sound too simplistic to you, but it's not, in my humble opinion. I believe you have to see what is actually happening with all the cards.

[ QUOTE ]
I gather that 990 are the possible card combinations.

[/ QUOTE ]

Well... 990 should be corrected to 820. But yes, there are 820 possible two card combinations if you know the whereabouts of eleven cards.

[ QUOTE ]
What I don't get is how you arrived at the formula:17*16/2-(6*2+3*3)+17*17.

[/ QUOTE ]

"Formula" is not the word I would use for the expression I wrote, but no matter. Let me try to explain the mathematics.

Piles 4, 5, and 6 contain the cards that may possibly be dealt on the turn and river. To beat the set your opponent has already flopped, you need to do one of two things. You need to either
• (1) combine two of the cards in your friendly pile with each other (but not two that make a pair), or
• (2) combine a card from the friendly pile with a card from the neutral pile.

• (1) Here is the number of ways you can combine two cards from your friendly pile: 17*16/2. Big Pooch would write this as C(17,2). This step should not be a mystery. If you take one card away from the friendly pile, there are sixteen cards left with which it could combine. Make sure you see this is the truth before you continue. O.K.?

Now take that card and start a seventh pile with it. There should be sixteen cards left in the friendly pile. Take one of the sixteen cards and make sure you can see that there are fifteen cards you could combine it with. Then for the next card, there will be fourteen ways to combine it with a card with which it has not already been combined. Can you see it coming?

16+15+14+13+12+11+10+9+8+7+6+5+4+3+2+1.

That all adds up to 136.
C(17,2) also equals 136.
18*17/2 also equals 136.

C(17,2) and 18*17/2 are just short-cut ways to add those sixteen numbers.

But wait. Some of those combinations of two cards from the friendly pile make pairs. How many of them? Well....there are six ways you can make a pair from the four queens and also six ways to make a pair from the four sixes. There are three ways to make a pair from the three jacks, another three ways to make a pair from the three nines, and another way to make a pair from the three sevens. Thus there are (6*2+3*3) ways to make a pair from the cards in your friendly pile.

So you have to subtract all these ways of making pairs using two cards from you friendly pile from the total number of two-card combinations of your friendly pile.

18*17/2-(6*2+3*3) is how I wrote it.
18*17/2-(6*2+3*3) is a correct mathematical statement of that part of the procedure. Basically it says to subtract the total of the pairs from 136.

• (2) The other way for you to win the hand is to combine any card from the friendly pile with any card from the neutral pile. There are 17 cards in each pile - thus there are 17 times 17 ways to make a two card combination.

I think that's about it. Ask some more if there's something you still don't understand. Maybe I (or someone else here) can make it clear.

[ QUOTE ]
I would have thought the combinations in my favour (without having subtracted the combinations that are helpng the opponent) should be 17*16/2 +1 7*28.

[/ QUOTE ]

No. It's true that 17+28 = 45, but if you're going to insist we give your opponent a set, then I'm, going to insist we also give your opponent two other cards. We don't have to make them known cards (like the five and the deuce) - but two cards need to be taken out of the pack to complete your opponent's hand - and which cards they are affects your outs and also your opponent's outs (to improve).

Therefore the 28 part of 17+28 = 45 has become 4+7+17. The 28 has become four cards for your opponent's hand, 7 outs to pair the flop, and 17 neutral cards. The 28 has become your opponent's hand pile, the "bad card" pile, and the neutral pile.

[ QUOTE ]
The rest was over my head anyway. Thanks again.

[/ QUOTE ]

I hope I have made it clearer for you. I honestly don't think the math is really over your head. You just need someone who can explain it to you. (I'm not sure I'm that person, but ask if you have questions and I'll respond if I can).

Buzz

[Big Dave D]

Download PokerCalculator or use http://www.twodimes.net/poker/

If you must do it by hand, and you dont have Buzz's combinatrics (?) skills, then build a simple spread sheet. Calculate how many unknown cards there are on the flop and then multiply that number by itself, -1, to get the total possible number of card combinations on the turn and river. We'll call this Z. Then put how many cards win for you on the turn on the table, and next to it how many safe cards there are are on the river. Multiply this out, lets call this A, for the number of "hit the turn, safe on the river" combinations there are. In reality, it may be easier to break this down into several entries if you have several winning combinations. Then calculate how many safe cards on the turn where you can still improve, but havent yet, and next to it the number of turn outs. Multiply this out as before, lets call this B. Lastly, include your runner-runner outs by just putting the individual numbers for turn and river next to each other and again multiplying out (C).

Now your winning % is simply (A+B+C)/Z. This method, described in more detail, can be found in several of Bob Ciaffone's books.

gl

dd

[petsei]

Thank you, Buzz, I think I’ve got it.
There is just one detail that I still don't understand. When I assumed that the opponent held a set, only two of his cards were "known" to me. So this would account for 9 cards, leaving 901 combinations. Aren’t the other two ("the not assumed ones")of the opponents cards interchangeable with the rest of the stack? What I mean is this: When you have flopped, for example, at Holdem a fourflush, it's 9 out of 47 unknown cards, regardless of what your opponents might hold. I hope, I have made myself understood. You were of great help anyway.

And thanks to Big Dave D too for his contribution. The Pokercalculator is a great gadget

Regards
petsei

[Buzz]

[ QUOTE ]
Thank you, Buzz, I think I’ve got it.

[/ QUOTE ]

Petsei - You're welcome.

[ QUOTE ]
There is just one detail that I still don't understand. When I assumed that the opponent held a set, only two of his cards were "known" to me. So this would account for 9 cards, leaving 901 combinations. Aren’t the other two ("the not assumed ones")of the opponents cards interchangeable with the rest of the stack?

[/ QUOTE ]

Yes. They are interchangeable with the rest of the pack.

There are different ways to think about it. Maybe this is a self-indulgence, but what works really well for me is thinking in terms of four card hands when thinking about Omaha or Omaha-8. (In Chowaha, which is two card multi flop Omaha, I think in terms of two card hands, and if I were playing in a game of five card Omaha, then I'd try to think in terms of five card hands).

In my favored way of thinking, whatever those two other cards are, they are still missing from the rest of the stack and located in your opponent's hand. Thus in my way of thinking, opponents don't just have two cards; they always have four cards.

If you play coordinated hands and if your opponent also plays coordinated hands, and if you both have a fit with the flop, very often at least one of those other two cards in your opponent's hand is one of your own outs.

[ QUOTE ]
What I mean is this: When you have flopped, for example, at Holdem a fourflush, it's 9 out of 47 unknown cards, regardless of what your opponents might hold. I hope, I have made myself understood.

[/ QUOTE ]

Yes. I do understand. But I humbly disagree. If you put specific cards in your opponent's hands, then it is no longer 9 out of 47 unknown cards.

It's very similar in Omaha. If you don't put cards into your opponent's hand, and if you have two hearts and if there are two hearts on the flop, then it's 9 out of 45 unknown cards.

I recognize that there are those out there who think differently and who disagree with me on this issue. You're welcome to disagree yourself. In terms of the calculations, I'll admit I could have done them differently.

However, I'm very comfortable with my way of always thinking of an opponent's Omaha or Omaha-8 hand as a four card hand. It's a model that works well for me - and probably would work well for you too.

Take care.

Buzz