Any Omaha math wiz kid out there?

[Insistent_]

If you deal 4 random cards from a deck... how likeley is it that you are dealt:

1. AAxx
2. AAxx single-suited
3. AAxx double-suited

4. three of a kind
5. four of a kind

Thanks so much! [img]/images/graemlins/smile.gif[/img]

[TimberBee]

Try the probablility forum

[Insistent_]

Okay, I'm waiting for a moderator to move this thread.
If someone happen to know please let me know what the figures are.
Thanks.

[illph]

I once did the numbers for AAxx and it's 1:39.

[Buzz]

[ QUOTE ]
If you deal 4 random cards from a deck... how likely is it that you are dealt:

1. AAxx

[/ QUOTE ]Hi Insistent_ -

First, there are C(52,4) ways to choose four random cards from a deck of 52 cards.

That translates to 52!/4!/48! and =
52*51*50*49/2/3/4 = 270,725 different possible four card poker hands.

Second, there are 6 ways to choose two aces. And then after choosing the two aces, there are 48*47/2= 1128 ways to choose the other two cards (not aces).

So the probability of four cards being AAXX is 6*1128/270725 = 0.02499954.

That's very close to 0.025, or about one out of forty.

So we say the chance (or probability) of being dealt a four card Omaha hand with two aces (and only two aces) is about one in forty. Another way of saying that is the odds are about 39 to 1 against it.
-------
If you want one of the aces to be suited, then there are still 6 ways to choose two aces, but now we have to be more specific about the other two cards.

And you haven't been in your question.

Did you mean only one card in the same suit as one of the aces, or either one or two cards in that same suit? There's a difference, and there's also a difference in how each of those hands plays. A hand with only one card of the same suit as one of the aces has a better chance to find three cards of that suit on the board at the river. Thus there is a difference in the drawing probabilities of the two hands.

I'll answer for just one card in the same suit as an ace. In that case, one of the other two cards has to be one of 12 remaining cards of the same suit of one of the aces. Since there are two different aces and either of them can be suited, that amounts to 24 cards. And the fourth card in the hand has to be of one of the two suits not covered by the aces, another 24 cards.

Thus 6*24*24/270725 will give us the probability we seek.
6*24*24/270725 = 0.01276572
-----
Want double suited? Take a guess before looking below.

6*12*12/270725 = 0.00319143
-----
Want to know about three of a kind? (You don't want this).
There thirteen ways to choose any rank, then four ways to exclude one card of that rank, then forty eight ways to choose a fourth card of any other rank.

Get it? 13*4*48/270725 is the probability. Comes to be 0.00921969, or very roughly one time in a hundred or so. (Actually 1/0.00921969 = ~108).
-----
Four of a kind?
13 ranks. Only one way to choose all four cards of that rank. 13/270725, or about one time in twenty thousand (20825 is the closest whole number).

Now that I've given you the numbers, I have to add they're not worth memorizing because you probably won't ever use any of them - but it's nice for them not to be an unknown stumbling block for you.

There are other posters on this forum who can do the math and there are also posters on the probability forum who can do the math. You can ask a question about Omaha probabilities here or there.

But you have to be very specific about the question you ask. And just one question per post is better.

Buzz

[wazz]

I used to boast that I could give HU PLO flop equity matchups accurate to +- 3%, I was hoping this thread would be about that.