Just looked at a few formulas for holdem and trying to apply to omaha or really any game.

(1 - 4x4/48)*(1 - 4x4/47)*(1 - 4x4/46) x (1-4x4/45)

This formula I made after looking at some holdem fomulas (I changed the 2 to a 4 (first 4 after the 1-) for 4 card omaha). I also added the first part to it and changed it to 48 instead of 50 on the first bit of the equation since its omaha and not holdem. It's to workout in a 5 handed game what are the % of time your opponents won't have a ace or any card for that matter you decide. I don't have a ace.

Working from this one of my four opponents will have a ace 81.6% of the time. or not have one 18.4% of the time working it out from this formula.

I'm not sure if this formula is correct to work out this stuff since i don't know any math besides elementary school mathematics and failed high school. I have no clue what the -1 is or why its used and the brackets I have no clue either. took me about 10 mins just mucking around on my smart caculator that they gave me for high school that i hardly ever used to figure out you mutiple between the two fours in the equation (number of cards 4 x number of opponents 4 ).

Any Oxford or harvard grads or just math nerds wanna tell me I'm wrong or right so i'm on the right path?

The 4 villains combined have 16 cards of the 48 unknown cards. The chance at least one of them has at least one ace is 100% minus the chance that all the aces are in the deck pile rather than their hole cards. There are 32 cards not belonging to them.

If you think of the A, there's a 32/48 chance it's in the deck pile. Then for the next ace you consider, there's a 31/47 chance, and so on.

P(someone has an ace) = 1 - (32/48)(31/47)(30/46)(29/45) = 1 - (32×31×30×29)/(48×47×46×45) = 81.52%

Or with combinations, there are C(32,4) ways to place the aces in the deck pile, out of C(48,4) ways to place them anywhere other than hero's hole cards, so 1 - C(32,4)/C(48,4)

The work you showed equals 90.76%, so I don't know where your 81.6% came from or whether it's a coincidence that it's close to 81.5%.

As is almost always true, there are a number of ways to solve problems like this. Here is another.

Given you don't have an ace, there are 44 non-aces in the deck of 48 cards before dealing to opponents. With 4 opponents, for all to not have an ace, the probability is C(44,16)/C(48,16) = 18.5%. Then the probability at least one ace is dealt is 100% -18.5% = 81.5%.

It actually equals 0.184

100-18.4 = 81.6

Sorry, I made a calculator typo yesterday. Yes, what you did works. It gets .1848 and 1-that=.8152

Now I have to figure out why it works...

4×4/48 is the overestimated chance one player has an ace, overcounting the chance of multiple aces.
1-that = underestimated chance they lack an ace

Then you did the same thing for the other 3 players, then multiplied the 4 underestimated chances of each player lacking aces. And for some reason you only decreased the denominators by 1, even though ostensibly 4 cards had been removed each time.

But somehow the combination of those wonky things equals a working formula, and it beats the hell out of me. I love when this happens! Such a routine problem and yet I'm about to learn something.

It looks like if we expand the terms, an inclusion-exclusion formula will emerge.

1 - 16(1/45 + 1/46 + 1/47 + 1/48) +
256[1/(46*45) + 1/(47*45) + 1/(48*45) + 1/(47*46) + 1/(48*46) + 1/(48*47)] -
4096[1/(47*46*45) + 1/(48*46*45) + 1/(48*47*45) + 1/(48P3)] +
2^16 / 48P4

Sanity check: there should be 2^4 terms in the alternating sum and there are.

Great but that's not helping me much right now. I'll come back to this later.

OP, I'm glad you posted this! If you have a reason that this works then I'm all ears!

Ok, each 16/n is the chance of a particular ace being somewhere in the 16 collective hole cards. OP's product is [1 - P(4 people lack the Ad)] * [1 - P(4 people lack 4h | lack 4d)] * ... which is sound.

(And of course, 1-16/48 = 32/48, and so on, so the terms of the products match my terms.)

The above is one conceptual lens through which to view what's happening, and it makes perfect sense through that lens.

My struggle was attempting to view that equation through a different lens. It was a valid lens, but one through which the above math would be difficult (if not impossible?) to make sense of on a conceptual level (rather than just algebraic).

For instance, what statmanhal showed is a different lens. It algebraically reduces to my 1-C(32,4)/C(48,4), but conceptually our two solutions are very different.

Right, but what you have is a multi-lens Hasselblad-equivalent camera while most of us try to get by with our single lens Brownie-equivalent.

Lol! That's only because I've been OCD about collecting these lenses, spending hours to do so if need be, for like 10 years now. It's actually sad how many poker hours I sacrificed during the glory days of poker for math time. I was a way worse player back then but I made more $/hr because of how amazing the games were. Yet most of my time wasn't spent playing poker. If I could go back in time and physically beat some sense into my former self...except he would win a fight too! In summary, instead of getting rich like y'all were probably doing, I know how to...checks notes...solve some math problems more than one way lol. Saaaad! Other than that legendary Nordic guy who loses thousands per month at 2nl or whatever, there aren't many poker players more pathetic than me.

Anyway, yesterday I said: all of which is still true. Through one lens, 16/48 is the chance of a specific ace being in the 16 combined hole cards. Through another lens, it's the overestimated chance of one player having any of the aces. The next however many nights while going to sleep, I'll try to think of some mind-bending way for the 2nd lens to make sense in that context. It's probably a fool's errand, which is why I won't spend waking time on it.