# What are the odds of facing AA or KK when holding AK? And odds of facing AK when holding AK

What are the odds of being crushed by kings or aces when already holding AK and what re the odds of facing AK when already holding AK?

• Against how many players?
– Paul
Dec 8 '15 at 0:07

I'm not 100% sure what you're looking for in the first part of the question, but I'll assume you mean what is your equity when you have A♥K♦ vs A♣A♠ or K♥K♣ (suits are just for the card graphic). Either way checkout this odds calculator and play around with it.

As for the situation of A♥K♦ vs A♣K♥ the odds of this occurring is as follows.

Player 1 has a choice of 8 cards for his first card either one of the 4 aces or 4 kings, likewise for his second card he has a choice of 4 cards, being which ever he did not get for his first card.

(8 / 52) * (4 / 51) = 32 / 2652 or 8 / 663 which in percentage is 1.2066%

Player 2 has a choice of 6 cards first card either one of the remaining 3 aces or 3 kings, likewise for his second card he has a choice of 3 cards, being which ever he did not get for his first card.

(6 / 50) * (3 / 49) = 18 / 2450 or 8 / 1225 which in percentage is .6531%

Finally then we do 1.2066 * .6531 = .788%

• You have tried to calculate the chances of 2 players getting AK, but you know that one player does already, so this is irrelevant. Only calculate the chance of the second player getting it. And the math is wrong: 1.2% * 0.65% = 0.0078% Dec 9 '15 at 21:56
• This is the calc for AK facing AK Feb 12 '16 at 19:06
• The point @BowlOfRed was trying to make (I think) is that Grinch91 is calculating the probability that two players get dealt Ace-King from scratch. The post appears to be asking the question - "Given that I have Ace-King. What is the chance that a second player has Ace-King also?" That's a different calculation than what Grinch91 did. Feb 17 '16 at 23:45

This is a single player
Multiple players is a much more complex

One way to approach this is binomial coefficient / combination

You have two cards
50 cards remaining
The number of 2 card combination in 50 is 1,225
(50/2) = 1,225

Let's say you did not have a blocker aces (e.g. you had QQ)
The number of ways to make AA(2) from 4 is
(4/2) = 6 (sc, sh, sd, ch, cd, dh)
The odd are 6 / 1,225 = 0.0049 = 203 : 1

Let's say you have AK so you have block on each
The number of ways to make AA from 3 is
(3/2) = 3
The odd are 3 / 1,225 = 0.00245 = 407 : 1

So with a single blocker you cut the odds of facing AA by 2

AA or KK is just additive as those hands are mutually exclusive
The change of AA or KK is
0.0049 = 203 : 1

There are 9 ways to make AK from 3 aces and 3 king
9 / = 0.007347 = 135 : 1

AA, KK, or AK
For stuff like this is why you need to use combinations
(6/2) = 15
15 / = 0.01224 = 80.7 : 1

I know not the question but interesting
What is the chance of facing AA, KK, or AK
Holding QQ then
(8/2) = 28
The odd are 28 / 1,225 = 0.0229 = 42.75 : 1