Bear with me here, I will start simple, but it begins to get complex.. which is where I'm having trouble.

The probability of having an Ace after drawing 1 card from a deck is:
A(1) = 4/52 = 7.69%

The probability of having an Ace after drawing 2 cards:
A(2) = (4/52) + (4/51) = 15.5%

Having AA after drawing 2 cards:
AA(2) = (4/52) * (3/51) = 0.45%

AAA(3) = (4/52) * (3/51) * (2/50) = 0.018%

AK(2) = (4/52) * (4/51) * 2 = 1.21%

But that's as far as I have gotten. I'm having trouble with:

Having AA after drawing 3 cards:
AA(3) = ?

AKQ(3) = ?

AAK(3) = ?

AA(3) = ?

There's 3 ways to get AA out of 3 cards all of which are equally likely:
AAx, AxA, xAA
Probability of having AAx: 4/52 * 3/51
thus AA(3) = 3(4/52 * 3/51)

Similarly there is 3! = 3 * 2 * 1 = 6 ways to get AKQ:
AKQ, AQK, KAQ, KQA, QAK, QKA
so AKQ(3) = 6(4/52 * 4/51 * 4/50)

AAK is the same as AAx in that you have 3 ways to get it:
AAK, AKA, KAA
so AAK(3) = 3(4/52 * 3/51 * 4/50)

Is that what you need?

brilliant!

thanks a lot.. exactly what i was looking for.

@db6, thanks for pointing me in the right direction.
after further investigation, i have adjusted your formula to be more accurate:

your approximation method:
AA(3) = 3(4/52 * 3/51) = 1.36%

more exact method:
AA(3) = p(AAx) + p(AxA) + p(xAA)
AA(3) = (4/52 * 3/51) + (4/52 * 3/50) + (4/51 * 3/50) = 1.38%

AKQ(4) is a monster calculation!!!!

Um... I'm pretty sure P(AAx) = P(AxA) = P(xAA) as long as you define x = not A which apparently is not what you're doing here.

P(AAx) = 4/52 * 3/51 * 48/50
P(AxA) = 4/52 * 48/51 * 3/50 = P(AAx)
P(xAA) = 48/52 * 4/51 * 3/30 = P(AAx)

Using this you can just add them together since there's no overlap. If you count AAA as part of your probability then you just add

P(AAA) = 4/52 * 3/51 * 2/50 to the whole thing.

Yea this is yet another answer what am I doing wrong?

P(AAX) = 3P(AAx) + P(AAA) where x is not an A
= 3 *(4/52 * 3/51 * 48/50) + (4/52 * 3/51 * 2/50)
= 1.32126697% according to google

I also found this cool link which you might want to check out: http://en.wikipedia.org/wiki/Poker_probability_(Omaha)

oh whoops i screwed that up. i guess you were right the first time.

of all the permutations, this first section is throwing me off because the formula is inconsistent!

p(A,1) = (4/52)
p(A,2) = (4/52) + (4/51)
p(A,3) = (4/52) + (4/51) + (4/50)
p(A,4) = (4/52) + (4/51) + (4/50) + (4/49)

p(AA,2) = (4/52) * (3/51) * 1
p(AA,3) = (4/52) * (3/51) * 3
p(AA,4) = (4/52) * (3/51) * 6

p(AK,2) = (4/52) * (4/51) * 2
p(AK,3) = (4/52) * (4/51) * 6
p(AK,4) = (4/52) * (4/51) * 12

p(AAA,3) = (4/52) * (3/51) * (2/50) * 1
p(AAA,4) = (4/52) * (3/51) * (2/50) * 4

p(AAK,3) = (4/52) * (3/51) * (4/50) * 3
p(AAK,4) = (4/52) * (3/51) * (4/50) * 12

p(AKQ,3) = (4/52) * (4/51) * (4/50) * 6
p(AKQ,4) = (4/52) * (4/51) * (4/50) * 24

after many of my own incorrect posts, i am going to redeem myself with a proof!

where A != x,

AAA = 4/52 * 3/51 * 2/50
AAx = 4/52 * 3/51 * 48/50
AxA = 4/52 * 48/51 * 3/50
xAA = 48/52 * 4/51 * 3/50
Axx = 4/52 * 48/51 * 47/50
xAx = 48/52 * 4/51 * 47/50
xxA = 48/52 * 47/51 * 4/50
xxx = 48/52 * 47/51 * 46/50

AAA + 3(AAx) + 3(Axx) + xxx = 100%

i think we can close the books on this one

you got it!