High Hand Odds

My local card room runs a 'High Hand' promo hourly. You must have aces full or better to qualify. In Holdem, both hole cards must play. Omaha 08 is also eligible for this promotion but must qualify using only the flop.

My question is: Which game has the better percentage of hitting a qualifying hand based on the above information (not considering number of hands played per hour or any other factors)?

Edit: All the games are all fixed limit or 2-100 spread limit. I'm looking for someone that can is skilled enough to give me an actual percentage (like 51.5% omaha vs. 48.5% holdem), we've discussed the topic to death at the tables and believe its close but would like know the numbers.

Quote: Choirboy

My local card room runs a 'High Hand' promo hourly. You must have aces full or better to qualify. In Holdem, both hole cards must play. Omaha 08 is also eligible for this promotion but must qualify using only the flop.

My question is: Which game has the better percentage of hitting a qualifying hand based on the above information (not considering number of hands played per hour or any other factors)?

If I am understanding this correctly, all else equal, you get seven cards either way but one game requires two to be part of the hand and the other game requires three specific cards to be part of the hand.

Also THE and minimum aces full using your hole cards. You wouldn’t generally fold pairs, and especially not aces…so that leaves matching a pair and single card on the board with your hand containing Ax, which is relatively low in terms of folding percentage…

I guess there’s the matter of quads with your hand ALSO containing a better kicker than anything on the board, as well as the card that matches the trips on the board, but I’d think the probability of that is negligible and would involve a lot of aces besides.

What percentage of hands do you see in Omaha 8? Does it have a high rate of limps? I assume most pairs in your hand wouldn’t fold, but could be wrong. (Actually, are low pairs considered bad?)

Do they have a Limit THE game there? Does that game qualify, if so. Depending on the number of players, you could just load a table with confederates and agree to limp to flops with all justifiable hands if your hourly expectation would justify it compared to the rake.

DISCLAIMER: This post contains opinion, speculation and theory. While I have played poker cash games, I have never personally done so in pursuit of a particular jackpot.

Maybe this is naive, but simplifying to playing all the same rate of seeing flops in Omaha to rivers in Holdem, then I think its easier in Hold’em because:

Omaha you can choose 2 from 4 from your hand to play so 6 ways. Holdem you can choose 3 from 5 on the board to play so 10 ways.

At Mission, playing smart you should see less Omaha flops than Holdem flops. But the comparison is really flop to River. And I’m not sure how much it will matter because you are likely starting with cards more likely to see flops and rivers ri get Aces full or better

Without doing any math, I am certain the Omaha 8 game has higher odds of winning the 'Aces Full or Better High Hand bonus.'

In Omaha 8, you are dealt 4 hole cards and must make a hand with exactly 2 hole cards and 3 common (Or board) cards.

In Holdem, this HIgh Hand bonus requires you to make a hand with two (Both) hole cards and three common cards.

So, on basic principles, you are 6x times likely to make a full house or better at Omaha 8 then in Hold'em (with both hole cards.) That is because 4 hole cards in hold'em are equivalent to 6 different combinations of 2 hole cards in Hold'em.

Quote: gordonm888

Without doing any math, I am certain the Omaha 8 game has higher odds of winning the 'Aces Full or Better High Hand bonus.'

In Omaha 8, you are dealt 4 hole cards and must make a hand with exactly 2 hole cards and 3 common (Or board) cards.

In Holdem, this HIgh Hand bonus requires you to make a hand with two (Both) hole cards and three common cards.

So, on basic principles, you are 6x times likely to make a full house or better at Omaha 8 then in Hold'em (with both hole cards.) That is because 4 hole cards in hold'em are equivalent to 6 different combinations of 2 hole cards in Hold'em.

Did you read the part where you have to flop it in Omaha??

Quote: unJon

Did you read the part where you have to flop it in Omaha??

No, my bad. Thanks for pointing that out to me.

If you must flop the High Hand in Omaha to win the bonus, you are trying to make an Aces-up FH or better from 7 cards (4 hole cards and 3 board cards) in which the three flop cards must be used in the hand. That is a more difficult requirement than the Hold'em version, which only requires the two hole cards to be used out of a total of seven cards.

Therefore, the hold'em game version of the High Hand Bonus offers you a higher probability of winning.

The hold'em game is more likely than the Omaha game to make the High Hand Bonus by approximately 1.667X; that is by a ratio that is = c(5,3)/c(4,2).

Agree. Matches my reasoning.

Quote: unJon

Agree. Matches my reasoning.

We’re unanimous so far.

I know nothing of Omaha 8 strategy, so the only variable for me is if there was an inordinate amount of limping to flops.

Quote: Mission146

We’re unanimous so far.

I know nothing of Omaha 8 strategy, so the only variable for me is if there was an inordinate amount of limping to flops.

Omaha 8 is a hi-lo game where the low hand must be 8-high or lower to split the pot. In my experience, many players do tend to limp preflop, because you can win two ways (high or low) and your hand prospects are ambiguous until you see a flop. High stakes games may be different however.

Quote: gordonm888

No, my bad. Thanks for pointing that out to me.

If you must flop the High Hand in Omaha to win the bonus, you are trying to make an Aces-up FH or better from 7 cards (4 hole cards and 3 board cards) in which the three flop cards must be used in the hand. That is a more difficult requirement than the Hold'em version, which only requires the two hole cards to be used out of a total of seven cards.

Therefore, the hold'em game version of the High Hand Bonus offers you a higher probability of winning.

The hold'em game is more likely than the Omaha game to make the High Hand Bonus by approximately 1.667X; that is by a ratio that is = c(5,3)/c(4,2).

This is the kind of mathematical answer that I am looking for however it diifers from thoughts on how the math should be done. Please show me where I'm going wrong.

In both games you're playing the best 5 card hand out of 7 cards, Hold'em 2 cards in your hand and 5 cards on the board, Omaha 4 cards in hand and 3 on the flop. I would think that running a simulation or a computer program you would get an equal percentage of hands that were aces full or better. At that point the games would be equal in the possibility of qualifying for the high hand promotion.

Now, add in to it the fact that both cards hole cards must play in Hold'em, a simulation would have to be run using only 6 cards to make an aces full or better hand to figure it using only one hole card and that number deducted from a 7 card aces full or better count to determine the percentage of advantage that actually goes to the Omaha game. Or, possibly the number of games in which one or both hole cards do not play would be slightly more than 1 out of 7 or 14.28%.

Now, I've never considered myself very good at this so I welcome any correction.

Quote: Choirboy

This is the kind of mathematical answer that I am looking for however it diifers from thoughts on how the math should be done. Please show me where I'm going wrong.

In both games you're playing the best 5 card hand out of 7 cards, Hold'em 2 cards in your hand and 5 cards on the board, Omaha 4 cards in hand and 3 on the flop. I would think that running a simulation or a computer program you would get an equal percentage of hands that were aces full or better. At that point the games would be equal in the possibility of qualifying for the high hand promotion.

Now, add in to it the fact that both cards hole cards must play in Hold'em, a simulation would have to be run using only 6 cards to make an aces full or better hand to figure it using only one hole card and that number deducted from a 7 card aces full or better count to determine the percentage of advantage that actually goes to the Omaha game. Or, possibly the number of games in which one or both hole cards do not play would be slightly more than 1 out of 7 or 14.28%.

Now, I've never considered myself very good at this so I welcome any correction.

As an initial matter you are ignoring that for Omaha you must use exactly two of the four cards in your hand.

Quote: Choirboy

This is the kind of mathematical answer that I am looking for however it diifers from thoughts on how the math should be done. Please show me where I'm going wrong.

In both games you're playing the best 5 card hand out of 7 cards, Hold'em 2 cards in your hand and 5 cards on the board, Omaha 4 cards in hand and 3 on the flop. I would think that running a simulation or a computer program you would get an equal percentage of hands that were aces full or better. At that point the games would be equal in the possibility of qualifying for the high hand promotion.

Now, add in to it the fact that both cards hole cards must play in Hold'em, a simulation would have to be run using only 6 cards to make an aces full or better hand to figure it using only one hole card and that number deducted from a 7 card aces full or better count to determine the percentage of advantage that actually goes to the Omaha game. Or, possibly the number of games in which one or both hole cards do not play would be slightly more than 1 out of 7 or 14.28%.

Now, I've never considered myself very good at this so I welcome any correction.

You wouldn't. Why do you even care about analyzing this promotion if you are not willing to read the answers until you understand them?

THE YOUR HAND: 1, 2

COMMUNITY CARDS: 3, 4, 5, 6, 7

OMAHA 8 YOUR HAND: 1, 2, 3, 4

COMMUNITY FLOP: 5, 6, 7

Hands: 1, 2, 3, 4, 5, 6, 7 OR 1, 2, 3, 4, 5, 6, 7

A priori, these are seven random cards. WHERE THEY ARE DOESN'T MATTER INITIALLY!!! They are seven random cards. It is a random seven-card hand.

Now, we will look at a Full House. Trips and a pair. The probability of a dealt full house in ANY SEVEN CARDS is roughly 2.6% in frequency. That is for all Full Houses, not just the aces, but it's the same concept.

In one instance, I have two bolded cards that must be in the hand. In the other, I have three bolded cards that must be part of the Full House (or other).

That means, in both instances, two cards are NOT part of the hand.

THEREFORE, five cards out of seven are part of the completed hand, and two are not.

nCr(3,3)*nCr(4,2)/nCr(7,5) = 0.2857142857142857

nCr(2,2)*nCr(5,3)/nCr(7,5) = 0.4761904761904762

Therefore, ASSUMING a completed Full House of any kind, the probability in Texas Hold 'Em that two of the cards in your pocket cards, all else being equal, will be part of that Full House is 47.62%, rounded. In the meantime, the probability that three specific cards will be part of the Full House, all else being equal, is 28.57%, rounded.***

THUS, the question then becomes one of opportunities to see more such hands. What are opportunities? Opportunities are hands that you do not fold.

THAT YOU HAVE FOUR CARDS IN YOUR POCKET COMPARED TO TWO MEANS NOTHING!!!! IT'S STILL SEVEN RANDOM CARDS!!!

So, how many flops do you see? Do you see so many more flops in Omaha8 that it offsets the fact that you need three cards to be part of the hand rather than two?

HERE IS THE MATH: Three is a bigger number than two. That's the math.

***This isn't exactly correct because there could be a Full House on the board AND a player could be holding a pocket pair in the same hand. Also, board such as AA699 with an Ace and Six in the players hand is a worse Full House than AAA99, unless it still plays for the purposes of this promotion.

Without knowing how many flops you see in one game compared to another, what the hell do you expect a simulation to accomplish?

This is why I can NEVER be concise, in case anyone wondered. It's never good enough.

I spent SO MUCH more time on the more recent post than I did in my initial response, in which the general concept of the problem occurred to me INSTANTLY, to produce a post that, in substance, is no better than my first one.

Some would argue, in this world of 280 characters, that my more recent post is WORSE---because it is longer.

One way or another, someone's going to complain, right? Right. That wasn't actually a question.

I'm not a good poker player. I'm a bad poker player. I didn't even know what Omaha 8 was. And I knew the general solution to the problem presented instantly. The time spend was because that first post was from my phone.

But, not good enough. I give math, my posts are too long. I give a simple answer and I should have given math.

Three is a greater number than two.

Quote: Mission146

So, how many flops do you see? Do you see so many more flops in Omaha8 that it offsets the fact that you need three cards to be part of the hand rather than two?

It’s not that simple. In O8, a high proportion of good starting hands have Aces and wheel cards and suited. Seeing the flop with those cards is less likely to lead to a qualifying high hand (but lots of nut lows and potential scoops). I would insta-fold JJ66, losing out on some possible quads high hands.

Your analysis is plenty good enough.

Quote: unJon

It’s not that simple. In O8, a high proportion of good starting hands have Aces and wheel cards and suited. Seeing the flop with those cards is less likely to lead to a qualifying high hand (but lots of nut lows and potential scoops). I would insta-fold JJ66, losing out on some possible quads high hands.

Your analysis is plenty good enough.

I know it's not that simple. I agree 100% with your analysis that it is not that simple. But, how are we going to even hope to come together to form a comprehensive analysis if we can never get past the initial boundary condition that three is a greater number than two?

But, I agree completely that the first step would be to try to figure out how many overall flops we are making it to followed by how many flops are we making it to---and with what in our hand---that is more or less likely to result in a relevant hand.

That's where a simulation becomes useful. Simulations can't be programmed if we haven't accepted that three is more than two already.

In both Holdem and Omaha 8:

1. You will always see a flop with AA in your hand (great start to Aces-full)
2. You will usually see a flop with a single ace in your hand (good start to Aces-full)
3. You usually/often see a flop with any pair in your hand. (which is a good start to 4oak)

Omaha usually takes longer to deal and more time to complete a hand. More cards dealt to players takes more time and the complexity of 4 hole cards makes players take time thinking through their decisions.

So, there is no reason to believe that there is a decisive difference that will cause Omaha to produce more qualifying high hands than Hold'Em given the math.

Quote: gordonm888

In both Holdem and Omaha 8:

1. You will always see a flop with AA in your hand (good start to Aces-full)
2. You will usually see a flop with a single ace in your hand (good start to Aces-full)
3. You usually/often see a flop with any pair in your hand. (which is a good start to 4oak)

Omaha usually takes longer to deal and more time to complete a hand. More cards dealt to players takes more time and the complexity of 4 hole cards makes players take time thinking through their decisions.

So, there is no reason to believe that there is a decisive difference that will cause Omaha to produce more qualifying high hands than Hold'Em given the math.

3. Is interesting. I think you are materially less likely to see a flop with a pair in O8 than Holdem. On the other hand, you get dealt many times the number of pairs in O8.

I don’t disagree with your overall conclusion.

Quote: Mission146

You wouldn't. Why do you even care about analyzing this promotion if you are not willing to read the answers until you understand them?

THE YOUR HAND: 1, 2

COMMUNITY CARDS: 3, 4, 5, 6, 7

OMAHA 8 YOUR HAND: 1, 2, 3, 4

COMMUNITY FLOP: 5, 6, 7

Hands: 1, 2, 3, 4, 5, 6, 7 OR 1, 2, 3, 4, 5, 6, 7

A priori, these are seven random cards. WHERE THEY ARE DOESN'T MATTER INITIALLY!!! They are seven random cards. It is a random seven-card hand.

Now, we will look at a Full House. Trips and a pair. The probability of a dealt full house in ANY SEVEN CARDS is roughly 2.6% in frequency. That is for all Full Houses, not just the aces, but it's the same concept.

In one instance, I have two bolded cards that must be in the hand. In the other, I have three bolded cards that must be part of the Full House (or other).

That means, in both instances, two cards are NOT part of the hand.

THEREFORE, five cards out of seven are part of the completed hand, and two are not.

nCr(3,3)*nCr(4,2)/nCr(7,5) = 0.2857142857142857

nCr(2,2)*nCr(5,3)/nCr(7,5) = 0.4761904761904762

Therefore, ASSUMING a completed Full House of any kind, the probability in Texas Hold 'Em that two of the cards in your pocket cards, all else being equal, will be part of that Full House is 47.62%, rounded. In the meantime, the probability that three specific cards will be part of the Full House, all else being equal, is 28.57%, rounded.***

THUS, the question then becomes one of opportunities to see more such hands. What are opportunities? Opportunities are hands that you do not fold.

THAT YOU HAVE FOUR CARDS IN YOUR POCKET COMPARED TO TWO MEANS NOTHING!!!! IT'S STILL SEVEN RANDOM CARDS!!!

So, how many flops do you see? Do you see so many more flops in Omaha8 that it offsets the fact that you need three cards to be part of the hand rather than two?

HERE IS THE MATH: Three is a bigger number than two. That's the math.

***This isn't exactly correct because there could be a Full House on the board AND a player could be holding a pocket pair in the same hand. Also, board such as AA699 with an Ace and Six in the players hand is a worse Full House than AAA99, unless it still plays for the purposes of this promotion.

NO I don't mind any correction. Nothing in the previous posts explained anything well enough to change my paradign of the game. After reading this post I understand the point you are saying about 3 vs.2 and it makes perfect sense and that initially Hold'em has a huge advantage over Omaha. However, I would still like to know how much the fact that both hole cards must play in Hold'em offsets that advantage since aces full or better can be made without one or both cards playing.. In Omaha, two cards in the hand always play for purposes of the promotion because for promotion purposes the hand is complete after the flop.

In my OP I also stated to exclude other factors or variables. So to put my question another way, If there are only two table in a card room, Table A is playing Holdem and Table B is playing Omaha. Both have nine players and see the same number of flop with the same number of players seeing the flop and all other variables being equal..
What percentage will Table A win the High Hand and what percentage wiill Table B win the promotion?

Quote: Choirboy

NO I don't mind any correction. Nothing in the previous posts explained anything well enough to change my paradign of the game. After reading this post I understand the point you are saying about 3 vs.2 and it makes perfect sense and that initially Hold'em has a huge advantage over Omaha. However, I would still like to know how much the fact that both hole cards must play in Hold'em offsets that advantage since aces full or better can be made without one or both cards playing.. In Omaha, two cards in the hand always play for purposes of the promotion because for promotion purposes the hand is complete after the flop.

In my OP I also stated to exclude other factors or variables. So to put my question another way, If there are only two table in a card room, Table A is playing Holdem and Table B is playing Omaha. Both have nine players and see the same number of flop with the same number of players seeing the flop and all other variables being equal..
What percentage will Table A win the High Hand and what percentage wiill Table B win the promotion?

1.) The fact that both hole cards must play doesn't offset anything. The underlying proposition is to create a five-card hand out of seven particular cards of which either two or three must be in a specific place.

Think of it like a hypothetical Royal Flush Bonus. There is a 100,000 credit jackpot if the, "Fifth card," in the hand is the Ace of Clubs OR King of Clubs for the Royal Flush. Alternatively, a different game requires that the, "Fifth Card," be an Ace or King of Clubs and the, "Fourth Card," be the Ace or King of Clubs...everything else being equal, which game is such that you are more likely to win the jackpot?

The first one. Only one card position must be correct. One is a smaller number than two.

I think the problem is that you're not seeming to realize that we are discussing five random cards out of seven random cards, at least, in terms of initial boundary conditions. How likely is Aces full (or better) in seven cards? The probability of that event, initially, does not change.

2.) "All other variables being equal," my post that you quoted has already, by estimate, more or less answered the question. Just take the bigger number and divide by the smaller number and that, VERY, VERY, ROUGHLY is how much more likely it is.

Actually, doing that equation seems to agree exactly with GordonM888, so that gives me tremendous confidence in my response. It is roughly 1.667x more likely in the Texas Hold 'Em game.

Quote: Mission146

Actually, doing that equation seems to agree exactly with GordonM888, so that gives me tremendous confidence in my response. It is roughly 1.667x more likely in the Texas Hold 'Em game.

This is correct and we have answered this OP three times now. Having to use both hole cards in Holdem is way more than offset by having to use exactly two hole cards in Omaha.

Here’s an example of Holdem where it’s the Omaha equivalent: you have to use all three flop cards and exactly two of your hole cards, turn and River.

One interesting aspect of this is: in Omaha, when you flop A-A-A there are likely to be several winners of the High Hand bonus -anyone with a pair in their hole cards or the 4th ace -and they would have to split it, I imagine.

The probability of flopping an A-A-A is 0.000181.

Quote: gordonm888

One interesting aspect of this is: in Omaha, when you flop A-A-A there are likely to be several winners of the High Hand bonus -anyone with a pair in their hole cards or the 4th ace -and they would have to split it, I imagine.

The probability of flopping an A-A-A is 0.000181.

Wouldn't it have to be the same under pair for the hour? Wouldn't AAA66, for instance, be a high hand over AAA55?

Quote: Mission146

Wouldn't it have to be the same under pair for the hour? Wouldn't AAA66, for instance, be a high hand over AAA55?

Unless I missed something, nothing in the statement of the rules in the first post said that this was an hourly event and that only the highest High Hand within a certain timeframe would win the bonus cash.

Quote: gordonm888

Unless I missed something, nothing in the statement of the rules in the first post said that this was an hourly event and that only the highest High Hand within a certain timeframe would win the bonus cash.

first two sentences of OP.

Quote: gordonm888

Unless I missed something, nothing in the statement of the rules in the first post said that this was an hourly event and that only the highest High Hand within a certain timeframe would win the bonus cash.

It definitely said that it’s an hourly promo. The other part I guess I’d have to ask for clarification.

Quote: Choirboy

In my OP I also stated to exclude other factors or variables. So to put my question another way, If there are only two table in a card room, Table A is playing Holdem and Table B is playing Omaha. Both have nine players and see the same number of flop with the same number of players seeing the flop and all other variables being equal..
What percentage will Table A win the High Hand and what percentage wiill Table B win the promotion?

We've already answered this question.

The Hold'em table will have 1.67/2.67 of the winners. The Omaha 8 table will have 1/2.67 of the winners.

Quote: gordonm888

We've already answered this question.

The Hold'em table will have 1.67/2.67 of the winners. The Omaha 8 table will have 1/2.67 of the winners.

Maybe it’s me, but I find it easier to use the fractions 3/8 and 5/8 for the ratios. :-P

Quote: unJon

Maybe it’s me, but I find it easier to use the fractions 3/8 and 5/8 for the ratios. :-P

In the long run, you'll see approximately 62.5% of the winners on the Hold 'Em table, compared to 37.5% on the Omaha 8 Table, all else being equal.

Quote: Mission146

In the long run, you'll see approximately 62.5% of the winners on the Hold 'Em table, compared to 37.5% on the Omaha 8 Table, all else being equal.

Thank you!

Quote: Mission146

In the long run, you'll see approximately 62.5% of the winners on the Hold 'Em table, compared to 37.5% on the Omaha 8 Table, all else being equal.

I guess %s are even easier to understand than reduced, whole number fractions!

Quote: unJon

I guess %s are even easier to understand than reduced, whole number fractions!

If I'm honest, I agree with that, but like decimals the best. Decimals are the easiest for me to mentally reverse engineer into approximate 1 in xxx type odds, unless we are talking about whole percentages, of course.

Looks like I am late to the party here, but has anyone considered how many hands per hour are being dealt at each table? Offhand, it's been my experience that Hold 'Em is dealt roughly twice as fast as Omaha 8. Omaha dealing is slower (4 cards vs 2), player decisions seem to be slower, showdown/dealer reading hands is a much slower process in Omaha than in Hold 'Em. Plus when there is a Low, it takes a long time for the dealer to chop up the pot.

Even without considering the excellent mathematical analyses presented above, Hold 'Em is a far better table to win an hourly high hand than Omaha, just based on hands per hour.

One other thing, OP mentioned a 2-100 spread limit. If that is not a typo, that is pretty darn close to no limit. There will be far fewer hands go to showdown at a 2-100 spread limit table than at a 2-4 limit table. Plus, I'd imagine you'd fold more hands like Ax and low pairs preflop if there is a lot of betting. You have a much better chance of hitting a high hand at a fixed limit table.

At my local cardroom during a HH promotion, there are typically (at least pre-pandemic, I haven't been there since before Covid) maybe 4 to 5 limit tables going, 6 to10 NL tables, 1 Stud table, and 1 Omaha table. The vast majority of high hands are won at one of the limit HE tables.