What does the math look like to calculate the probability of an all-in pre flop situation AA v KK? I've never seen the actual math broken down. Google is impossible with all the bs you have to wade through.

Three different questions right?

Two. I'm not interested in any sort of ev calc.

It isnt so much "math", it's just plain old counting. The only way to determine poker equity is to enumerate all the ways the hand can play out (all boards) and count the wins, losses, and ties. Equity = (wins + 1/2_ties) / total. Where total is the sum of all possible deals to complete the hand.

Certainly one of the many different types of mathematics must be able to handle this problem more elegantly than counting ??

Expanding on NOG’s answer and written somewhat in jest:

Let N be the total number of outcomes of AA vs KK assuming all hands go to showdown.

Let Na be the number of outcomes where AA wins
Let Nk be the number of outcomes where KK wins
Let Nt be the number of outcomes which tie
Na + Nk + Nt = N

Then P(AA "wins") = (Na + Nt/2)/N

Nope

If you think there is some magical way of calculating equity of AA vs. KK, what about QsTs vs. Kh2s (or any other "weird" heads-up pairing)?

You might be willing to entertain the possibility that there is simply no "math" that can determine the equities of QsTs vs. Kh2s. The situation is just too "complex" (straights, flushes, counterfeits, etc.).

It seems reasonable to me that there is no other way than to count all the boards for which QsTs wins, all the boards for which Kh2s wins, and all the boards for which they tie. Especially since those counts are not easy to come by, especially before computers could be programmed to run through all possible boards.

Doyle Brunson talks about in the early days of his Texas Road Gambling days sitting in a hotel room and dealing zillions of boards to try to get a sense of hold-em equities. I think he and Sailor Roberts dealt boards for hours trying to answer if AKs had more or less equity than QQ.

Again I have to reluctantly chide one of the nicest and usually one of the smartest people on this forum. Even if you are 90% sure that you remembered this anecdote correctly it is far more likely that the ten percent alternative that it was two deuces was the actual story given how quickly they would have come to a conclusion against two queens.

It seems silly to try to find answers through hand trials. The answer can be found exactly without a computer by grouping and counting types of boards each hand wins on.

It also might be an interesting problem to try to come up with a relatively simple approximate hand vs hand equity formula’s that minimizes the average error.

In hold'em poker we know approximately how one overcard comes out on average, how an overpair comes out on average, how much equity is added by being suited or lost by being dominated, and many other rules of thumb. There aren't that many "types" of matchups to be familiar with.

Google for 'holdem calculator' or somesuch. I use the site below for a quick probability estimate
https://www.cardplayer.com/poker-too...r/texas-holdem

Or program your own brute force algorithm (which I did just as an excercise to check if the above site is OK for checking probabilities... it's more user friendly than my own code....and I'm too lazy to code a good UI).
It's pretty easy to simulate 100k hands in a very short time, and the results are extremely likely to converge to the true mathematical value.