In a 6-max game, the percentage for running AA into KK is around 3%, if I did my calculations right. Does anyone have the probability of this event for games with 2, 6 & 9 players?

  • I didn't do the math (which is otherwise not complicated) but off the top of my head I remember reading it's about 1 in every 25 times you have KK someone will have AA.
    – DPM
    Nov 5, 2015 at 22:26

2 Answers 2


The chance that any one person has AA when you have no aces (and given no other information) is 6 / (50 * 49 / 2) or ~ .5% (note: you know your two cards, so it's not 52 * 51). This means that there's a 99.5% chance that they don't. When doing probabilities the probability that any one of n things with probability p_n is going to happen is 1 minus the the product of 1 minus the p_n's: i.e. the chance that any one of them happens is 1 minus the chance of all of them not happening, the correct answer is 1 - (1 - .005)^(number of seats not counting you). Rounding to .1%:

1: .5% (~1 in 200 heads up)
2: 1%
3: 1.5%
4: 2%  (1 in 50)
5: 2.5% (1 in 40 six-handed)
6: 3%
7: 3.5%
8: 4% (1 in 25 full 9-player table)

Have a look here, Poker Maths and Probability and this single page is good to have beside you. One Page Probability It has a load of these sort of answers.

Basically, you having KK doesn't matter as far as the odds of someone else having AA since you don't have an Ace. Therefore the odds of another person having AA are 0.45% or 220:1

  • Actually because you don't have an Ace, the odds someone else has AA goes up ever so slightly, but not enough to change the odds considerably.
    – Kenshin
    Dec 27, 2015 at 8:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.