View Full Version : # of Omaha Starting hands

04-11-2005, 05:56 AM
Not exactly probability but,

how many starting hands are possible in omaha if suits are not taken into account (like how they are listed in PTOmaha)?

So that A /images/graemlins/spade.gif A /images/graemlins/diamond.gif 2 /images/graemlins/spade.gif 3 /images/graemlins/diamond.gif

A /images/graemlins/heart.gif A /images/graemlins/club.gif 2 /images/graemlins/heart.gif 3 /images/graemlins/club.gif

are counted as the same hand.

Im not familiar with this kind of math, but my initial guesses were:




but i really am not sure.

If you do know how to figure this out, could you please explain it for me.

Thanks in advance


04-11-2005, 09:44 AM
not for nothing, but i believe your example hand IS counted as the same hand in PTO, but i could be wrong

04-11-2005, 10:23 AM
The correct answer is

13^4 = 28,561

I think the best way to explain why is to use a HE example. We all know there are 169 different starting HE examples (again assuming K /images/graemlins/diamond.gifQ /images/graemlins/diamond.gif = K /images/graemlins/spade.gifQ /images/graemlins/spade.gif = any KQ suited, but is different than KQo). Your first card can be one of the 13 ranks. Your second card can be one of the 13 ranks (if we said twelve here then you could NEVER have a pocket pair). So,

13^2 = 169

Likewise for Omaha. If it were 13*12*11*10 then a hand like four aces would not be possible. In fact, your hand would always have four different ranks of cards.

In case you're interested, the number of Omaha starting hands if suits matter is

C(52,4) = 270,725

04-11-2005, 02:21 PM
The answer is 16432. You can do a search here or RGP for more details. This assumes that As2sAh3h is the same hand as Ad2dAc3c but a different hand from As2hAc3d. Your example was confusing about what "suits are not taken into account" means.


Edit: fixed suit typo

04-11-2005, 02:35 PM
Your post contains several errors.

It is entirely coincidence that 13 x 13 = 169 = number of starting hands in hold'em.

The proper way to count hold'em starting hands is

Pairs = 13
Unpaired = 12+11+10+9+8+7+6+5+4+3+2+1 = 78
(this represents Ace + 12 lower ranked cards. King + 11 lower ranked cards . . .)

Unpaired hands can be suited or offsuit so total hands is equal to 78 + 78 + 13 = 169.

In order to count Omaha hands you would need to count separately quads, trips, two pair, one pair and no pair hands. Your method double (or more) counts many hands. For example, AAA3 would be counted four times (AAA3,AA3A, A3AA, and 3AAA).