basic odds question

i've been practicing how to figure odds for various poker situations. i'm more concerned with how to figure this out than what the answer is.


what are the odds of being dealt AA23 double suited (any two suits) in omaha/8?


any A (4/52) x the suited 2 (1/51) x a remaining A (3/50) x the suited 3 (1/49) = 12/6497400


= 1/541450 or 541449:1 against


correct? if not where did i screw up?


thanks

There are 12 ways to get AA23 double suited since there are 6 ways to get AA, and there are 2 ways to match 23 for each one. The number of possible hands is 52*51*50*49/4! = 270725. So 12/270725 or 22560:1 against.


We divide by 4! = 24 to get total number of unique hands independent of order. You only counted Ax2xAy2y in exactly that order, so you missed alot of possible orderings. You have to be consistent in numerator and denominator as to whether or not you care about order. Sometimes one is easier than the other.

notAmathGUY - You asked, "what are the odds of being dealt AA23 double suited (any two suits) in omaha/8?"


Six ways to get two aces.*

Then two ways to get a deuce in the same suit as one of the aces.

Then one way to get the trey in the same suit as the other ace.


6*2*1 = 12


Here they all are:


AsAh2s3h,

AsAh2h3s,

AsAd2s3d,

AsAd2d3s,

AsAc2s3c,

AsAc2c3s,

AhAd2h3d,

AhAd2d3h,

AhAc2h3c,

AhAc2c3h,

AdAc2d3c,

AdAc2c3d.


Since 12/270725 = 0.0000443, you should see it about once in every 22,560 hands or so, making the odds about 22,559 to 1 against.


Not sure quite how your reasoning went, so that it's difficult to show you where you went wrong. Looks like you missed the total possible number of Omaha-8 hands. The total possible number of four card Omaha-8 hands is 270,725 (52 choose 4).


*That there are 6 ways to get a pair of aces is a bit of a shortcut (used because it occurs so often in these problems). Here there are four aces and you want to choose two of them. Sometimes this is written as "four choose two," or C(4,2). This means 4!/2!, which computes as (4*3)/(1*2) = 6, where "!" means "factorial."


Buzz


P.S. I wrote my answer before reading BruceZ's, or I would not have written it at all. Then I decided to read his response before posting. There are a variety of ways to think about these problems. Coincidentally BruceZ and I used about the same reasoning. Since I have already written the above response, and since it expands a bit on BruceZ's response, maybe you can get something worthwhile out of it.


Buzz

There are 12 distict AA23ds hands (as many as there are 32o HE hands). The probability to get one is thus


p = 12/C(52,4) = 12/270725 = 0.00443%


or 1:22559.4 against.


cu


Ignatius

Forum members frequently post inqueries of this nature. It would really be useful and informative for everbody if you would add a section on poker mathematics and probabilities. On his website in the library section, Mike Caro has posted all of the poker math tables from Doyle Brunson's "Super System". You could also write a short section on how to do these computations, perhaps similiar to the chapter "How to Figure to Odd" in "Poker Strategy and Winning Play" by A.D. Livingston.

The first section in David Sklansky's book Getting The Best of It is called "Probability: The Mathematics of Gambling." Everything you need to know is covered there.


Best wishes,

Mason