AA vs KK - Gambling and Probability - Probability Theory Forum

Two Plus Two Forums Other Topics Probability

AA vs KK

Post Reply Subscribe

...

07-08-2019 , 01:22 AM

hokum

newbie

Join Date: Jul 2009 Posts: 41

I am trying to wrap my head around this-
I read the chances of AA vs KK occurring in a 6 handed game are just under 3% (2.88).
Does that mean out of 100 hands KK will run into AA almost 3 times?
That does not seem right....
Where is my logic failing or is that actually the case, that can't be the case, can it?
Thanks for any help,

Cornfused

Quote

07-08-2019 , 02:00 AM

whosnext

Carpal \'Tunnel

Join Date: Mar 2009 Posts: 6,729

I think this is the chance of KK running up against AA vs 6 opponents.

That is, given that you have KK (this part is assumed), what is the chance one of your six opponents has AA.

Quote

07-08-2019 , 02:29 AM

hokum

newbie

Join Date: Jul 2009 Posts: 41

Quote:

Originally Posted by whosnext

I think this is the chance of KK running up against AA vs 6 opponents.

That is, given that you have KK (this part is assumed), what is the chance one of your six opponents has AA.

0.45

Ok, I see the light.

Thanks!

Quote

07-08-2019 , 12:19 PM

statmanhal

Pooh-Bah

Join Date: Jan 2009 Posts: 4,986

Quote:

Originally Posted by hokum

-
I read the chances of AA vs KK occurring in a 6 handed game are just under 3% (2.88).

Quote:
Originally Posted by whosnext View Post
I think this is the chance of KK running up against AA vs 6 opponents.

That is, given that you have KK (this part is assumed), what is the chance one of your six opponents has AA.
-------------------------
In the interest of absolute nerdiness, for facing at least one AA with hero holding KK, I got 2.45% against 5 opponents and 2.93% against 6.

Anyone care to do a check?

Quote

07-08-2019 , 01:56 PM

whosnext

Carpal \'Tunnel

Join Date: Mar 2009 Posts: 6,729

These type of probs are often rounded (poorly) on various websites. I think ESPN shows some probs on their WSOP coverage that are not quite correct, but close enough for their audience.

Using the simplest manifestation of the Principle of Inclusion-Exclusion (PIE), I get the respective formulas to be:

Vs 5 Opponents
= 5*C(4,2)/C(50,2) - C(5,2)*C(4,4)/C(50,4)
= 2.444637429%

Vs 6 Opponents
= 6*C(4,2)/C(50,2) - C(6,2)*C(4,4)/C(50,4)
= 2.932262267%

Quote

Post Reply Subscribe

...