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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?

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

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    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%
    – BowlOfRed
    Dec 9 '15 at 21:56
  • This is the calc for AK facing AK
    – paparazzo
    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.
    – Duncan
    Feb 17 '16 at 23:45
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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

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