Odds of facing AA/KK when you have A/K in Hold\'em
 

In a fullring texas hold them game, what are the odds that someone else has AA or KK when you hold AK? Thx.

Re: Odds of facing AA/KK when you have A/K in Hold\'em
 

N=number of other players. 3 combos each=6/1326=0.5%
If N=8->8*0.5%=4%
More precisely 1-0.995^8
Wanna go all-in with AK?

[img]/images/graemlins/wink.gif[/img]

Re: Odds of facing AA/KK when you have A/K in Hold\'em
 

[ QUOTE ]
In a fullring texas hold them game, what are the odds that someone else has AA or KK when you hold AK? Thx.

[/ QUOTE ]

~1/20

Re: Odds of facing AA/KK when you have A/K in Hold\'em
 

50*49/2=1225 (not 52*51/2=1326)
6/1225=0.5%
9 other players->5%

Re: Odds of facing AA/KK when you have A/K in Hold\'em
 

[ QUOTE ]
In a fullring texas hold them game, what are the odds that someone else has AA or KK when you hold AK? Thx.

[/ QUOTE ]

With 9 opponents:

P(AA) = P(KK) = 9*3/C(50,2) = 27/1225 EXACTLY.

P(AA or KK) = P(AA) + P(KK) - P(AA and KK)

= 27/1225 + 27/1225 - C(9,2)*6/C(50,2)*3/C(48,2) EXACTLY,

=~ 21.9-to-1

Re: Odds of facing AA/KK when you have A/K in Hold\'em
 

if you are worried about being called pf by AA/KK when holding AK, reads on your opponent will be far more helpful than these numbers.

Re: Odds of facing AA/KK when you have A/K in Hold\'em
 

[ QUOTE ]
In a fullring texas hold them game, what are the odds that someone else has AA or KK when you hold AK? Thx.

[/ QUOTE ]

They are exactly the same as the odds of someone having AA when you have KK. Do you see why?

Re: Odds of facing AA/KK when you have A/K in Hold\'em
 

[ QUOTE ]
[ QUOTE ]
In a fullring texas hold them game, what are the odds that someone else has AA or KK when you hold AK? Thx.

[/ QUOTE ]

They are exactly the same as the odds of someone having AA when you have KK. Do you see why?

[/ QUOTE ]

They are not exactly the same. The probability of AA or KK when you hold AK is:

9*6/C(50,2) - C(9,2)*6/C(50,2)*3/C(48,2) =~ 4.36%


The probability of AA when you hold KK is:

9*6/C(50,2) - C(9,2)*6/C(50,2)*1/C(48,2) =~ 4.39%


The slight difference arises from the case where 2 opponents have the hands in question. In the first case, there are still 3 remaining ways for the 2nd opponent to have AA or KK once the first opponent has KK or AA respectively. In the second case, there is only 1 remaining AA when the first opponent has AA.

EDIT: I had final numbers reversed.