Hey! Assuming a HU match where the button opens and calls a 3-bet with A2o and the Big Blind has 3-bet AK, what are the chances an ace will hit the flop given both players HOLD an ace?

Let's say the flop is A95r.

I am just trying to see how often domination occurs when calling a 3-bet. Also, if someone could show me how to do the math on this that would be great A little rusty! Thanks!

Given each of two players have 1 ace, there are 2 aces left in the 48 card deck. The number of flops with exactly one ace is C(2, 1)*C(46,2). The number of flops with 2 aces is C(2,2)*C(46,1). The total number of flops is C(48,3). Therefore, the probability of 1 or 2 aces on the flop if two players each have an ace is

Pr = (C(2,1)*C(46,2)+C(2,2)*C(46,1)) / C(48,3) = (2*23*44+1*46)/(8*47*46) = 12.2%

For just one flop ace, delete the second term in the numerator.

An easier way is to calculate the probability of no aces and subtract It from 1.0

Pr = 1 – C(46,3)/C(48,3) = 1- 23*15*44/(8*47*46) = 12.2%

There are 2 aces left in the deck of 52 - 4 = 48 unknown cards (we know what our opponent holds).

If you want to know the percentage of flops with at least 1 ace you can find the number of flops without an ace and subtract from 1.

Flops without an Ace:

We know that there are 52-4 cards = 48 left in the deck. We also know that of those 48-2 = 46 are not aces. That means that we just need to choose 3 cards from the 46 remaining cards that are not aces:

46c3 = 46!/((46!-3!)*3!) = (46*45*44)/6 = 15,180

Now since there are 48 cards left in the deck the total number of ways to deal all flops (not just those without aces) is 48c3

48c3 = 48!/((48!-3!)*3!) = (48*47*46)/6 = 17,296

So the percentages of flops without aces is 15,180/17,296 = .8778 * 100 = 87.78%.

So that means the flops with at least one ace are 1-.8778 = .1222 * 100 = 12.22%

But that is assuming you know your opponents cards exactly and that you know they only contain 1 ace.

Since we never know our opponent's hands exactly the answer to the true domination effects are different as you would have to determine how often villain holds Ax or AA in their range and calculate based on that.

It happens at roughly the same frequency that you flop a set with a pocket pair, since it requires the hitting of a 2-outer, but with three opportunities to do so. Very approximately, it's (2/50)*3 = 12%
I'm not great at probabilities, but I think to get the more precise number you have to consider the probability that the first flop card is not an ace (46/48, because there are 48 possible cards, 46 of which are not aces), the second card is also not an ace (45/47), and the third card is also not an ace (44/46), and then multiply these probabilities and subtract from 1 to find the probability that at one of the cards is an ace.
1 - (46/48)*(45/47)*(44/46) = 1 - 0.878 = 0.122 = ~12.2%

Someone with better math skills than me while no doubt post a better answer, but it's something like that.

EDIT: I'm a slow typer as well as a poor mathematician.

Another way to evaluate domination with Ax is to calculate the probability a heads up opponent has an Ace with a higher kicker. If you have Ax, where x is rank R, then

Pr(domination) = (C(3,1)*C(4*(13-R),1) + C(3,2))/1225.

The first numerator term is the number of villain hands with 1 ace and a kicker higher than R. The second numerator term is for vilian having two aces. 1225 is the number of possible hands for villain.

Example:Given you have AT, the probability villain is dealt a hand that dominates is

Pr= (3* 4*3 + 3)/1225 =39/1225 = 3.2%

My normal public-service announcement in these types of threads.

2+2 has a Probability Forum for these types of questions.

Obviously, there is a fair amount of overlap between "Poker Theory" and "Poker Probability", but my view has always been that questions pertaining specifically to probabilities probably belong in the Probability Forum.

Why do you ask the question? I wonder because it doesn't seem very useful beyond curiosity.

just_grindin gives the correct answer with a good explanation, except for a typo. (46!-3!) should be (46-3)! and the next calculation has a similar error. His math works out as if this were written correctly, though.

Arty gets an identical answer using a slightly different method if you like that way better.