Omaha Hi Low A2 question

Within how many hands can you expect to be dealt A2 with any two random cards


and

A3 with any two random cards


Thankyou very much

Michael - The answer for A3XX, or any two particular cards, is the same as for A2XX, about one time in every 15.6 deals.


For A2XX, I assumed you meant to include duplications of aces and deuces. If so, you must include AAA2, AA22, A222, AA2X, A22X and A2XX, where X is any of the 44 cards which are neither ace nor deuce.


Following is my math, in chart form.


AAA2 4*4 = 16

AA22 6*6 = 36

A222 4*4 = 16

AA2X 6*4*44 = 1056

A22X 6*4*44 = 1056

A2XX 4*4*44*43/2 = 15136


16 + 36 + 16 + 1056 + 1056 + 15136 = 17316


17316/270725 = 0.06396


1/0.06396 = 15.6


If you wanted any combination of A2 and/or A3, you'd want to include AAA2, AA22, A222, AA2X, A22X, A2XX, AAA3, AA33, A333, AA3X, A33X, A3XX, AA23, A223, A233, and A23X, where X is any of the 40 cards which are neither ace, deuce, nor trey.


scratchwork:

4*4 + 6*6 + 4*4 + 6*4*40 + 4*6*40 + 4*4*40*39/2 +

4*4 + 6*6 + 4*4 + 6*4*40 + 4*6*40 + 4*4*40*39/2 + 6*4*4 + 4*6*4 + 4*4*6 + 4*4*4*40 = 27808


27808/270725 = 0.103


or roughly once in every ten deals you'll have some combination of A2 and/or A3.


If there is something you don't understand, ask and I'll give a brief explanation.


Never a guarantee my math is correct but someone is usually kind enough to point out my errors.


Buzz

Thankyou very much for all of your help buzz, i really thought that the information you provided was useful.


One more question: I am pretty knew to omaha hi low, i mean i play better starting hands than many of the people in my regular 4-8 games but i am not sure whether i am too tight or too loose. Do you recommend any strategies for starting hands, is it like holdem does it depend on position or is there some sort of a system?


Thankyou very much in advance

Michael - There's too much to say to put it all in one post, but here are some pointers.


(1) Game selection is the key to winning in poker. Period. Find a game where the players are weaker than you.


(2) Buy Ray Zee's excellent split poker book and closely follow the advice in it. You'll obviously need more than the book to win, but Ray's book will help. Also read whatever else you can find about Omaha-8, in books, articles, and posts.


You also need to "play good poker," as Ray puts it. This involves figuring out what cards your opponents are playing and playing accordingly.


(3a) Play very tight, at least at the start.


(3b) Play tight. Play only hands with a good chance to scoop. If you play hands with high cards (hoping that a low is not possible), you want all your cards "working together." For example, K-Q-J-9 and K-Q-Q-T are playable but K-Q-T-3 is not.


(3c) Play tight. If you play hands with low cards, you want an ace. Even 2-3-4-5 is not a very good hand, although it is generally playable. Any A-2-X-X is playable, but without another wheel card, preferably a 3, is not very good. I also generally play many, but not all, A-3-X-X hands. (I play all A-3 hands with a suited ace). I also generally play most, but not all A-4-5-X hands if they have a suited ace.


(3d) Play tight. Note that you are not playing A-2-X-X because it has a good chance to win for low. You are playing it because it has a good chance to scoop.


(3e) Play tight. Sometimes I play slightly weaker hands on the button - and also one seat before the button when a very tight player has the button. However I hate to advise anyone to play weaker hands on the button without at least cautioning that doing so can be dangerous. Starting hand selection is very important in Omaha-8.


(3f) Play tight. Playing tight is so boring that I usually defend my big blind, just to keep from being harrassed by constant pre-flop raises from fools when I post the big blind. The same is not true of the little blind. Don't bother about the little blind.


(3g) Play tight. Although you like all your cards "working together," avoid hands with middle cards. For example, J-T-9-8 T-9-8-7 and J-J-T-9 are all generally unplayable, even though the cards are working together, unless you feel like losing your money. (As an aside these hands are probably reasonably good hands in high-only-Omaha).


(4a) Don't chase. If you are playing K-Q-J-9 and the flop is J-T-3, you like your hand. If you are playing K-Q-T-9 and the flop is J-5-3, you should fold rather than hoping for two favorable cards.


(4b) Don't chase. Fold to a bet if you are playing A-2-4-6 and the flop is 2-4-K. Here you have flopped bottom two pair, which is unplayable unless you feel like throwing your money away. You still have a low here, but it is not very good and low may not even be possible. (Should low become possible, say with an 8 on the turn, your low would be beaten by A-3-X-X, A-5-X-X or 3-5-X-X).


Hope this helps.


Buzz

Thanks Buzz your a Champ !!!


I will keep an eye out for all your posts lol


Take Care

Here's a rough way to do the calculation to good accuracy.


First consider hands that have A2 plus any other 2 cards 4 through K. There are 16 distinct A2 combinations and 40*39/2 distinct combinations of cards in the range 4 through K (40 choices for the first card and 39 choices for the second card, but the order doesn't count [e.g. 4h Kd is tha same as Kd 4h]). Hence the number of hands with A2 and 2 cards in the range 4-K is


16*40*39/2 = 12480.


Similarly, the number of hands with A3 and 2 cards in the range 4-K is 12480.


The largest number of remaining hands that contain at least A2 or A3 consists of those with A23 and a card in the range 4-K. The number of distinct combinations of A23 is 4*4*4 = 64. Hence the number of hands with A23 and a card in the range 4-K is


64*40 = 2560.


In a first approximation, we will neglect the relatively small number of hands that contain A23 along with another A, 2, or 3.


Hence, to a good approximation, the number of distinct 4-card hands containing at least A2 or A3 is


2*12480 + 2560 = 27520.


The total number of distinct 4-card hands is


52*51*50*49/(4*3*2) = 270725


(52 choices for the first card, 51 for the second, etc., but the order doesn't matter, so we divide by 24, the number of ways of ordering 4 objects).


Finally, the probability P that a 4-card hand will contain at least A2 or A3 is approximately given by


P = 27520/270725 = 0.102 .


I hope this is useful.

JI - Thanks. Your approximation is useful, at least to me, but not exactly in the way you might have thought it would be. I couldn’t understand how you could get within one percent of my result while only using three out of the sixteen terms included in my set-up.


I originally wrote


"4*4 + 6*6 + 4*4 + 6*4*40 + 4*6*40 + 4*4*40*39/2 +

4*4 + 6*6 + 4*4 + 6*4*40 + 4*6*40 + 4*4*40*39/2 +

6*4*4 + 4*6*4 + 4*4*6 + 4*4*4*40 = 27808"


Upon re-adding the sixteen terms, I found the total shown above is incorrect. The correct sum is 31784.


16+36+16+960+960+12480+16+36+16+960+960+12480+96+9 6+96+2560 = 31784.


And, in that case, 31784/270725 = 0.1174,


meaning you are dealt some combination of ace plus two and/or three about once in every 8.5 deals, roughly an average of once a round at a full L.A. Omaha-8 table (nine players).


Mea culpa for my error. Thanks for your post which led me to find it.


Buzz

Buzz,

Thanks for your most recent post correcting your number. Your new number is exactly correct.


Here's how I would do it.


The combinations left out of my original approximation are those with at least 3 cards in the range A-3, with at least 2 of these paired. These consist of the following


Group I

1 pair with the fourth card X in the range 4-K: AA2X, A22X, AA3X, A33X


4*(6*4*40) = 3840.


4 ways of choosing the ranks of the pair and the remaining card in the range A-3

*

6 choices for the cards that make up the pair

*

4 choices for the remaining card in range A-3

*

40 choices for X.


Group II

All four cards in the range A-3 with only one pair:

AA23, A223, A233


3*(6*16) = 288


3 choices for the rank of the pair * 6 choices for the cards that make up the pair * 16 choices for the cards of the other 2 ranks.


Group III

All four cards in the range A-3 with 2 pair:

AA22, AA33


6*2*6 = 72


6 choices for AA * 2 choices for the remaining rank * 6 choices for a pair of that rank.


Group IV

All four cards in the range A-3 with trips:

AAA2, AAA3, 222A, 333A


4*8 + 2*4*4 = 64


4 choices for AAA * 8 choices for a 2 or 3

+

2 choices for the rank of trips in the range 2-3

* 4 choices for the cards making up the trips

* 4 choices for an A.


These are to be added to the 27520 hands of the form A2XX, A3XX, A23X, as calculated in my post of 10 December 2001.


Hence total number of hands with at least A2 or A2

is given by


27520 + 3840 +288 +72 +64 = 31784


and the probability of such a hand is


P = 31784/270725 = 0.1174.


These are exactly your numbers.


I should mention that your most recent post helped me to realize that some of the combinations that I left out of my original approximation are just as important as A23X, which I included there.


Thanks Buzz.