Poker Odds question

Hitman26 · Jan 13, 2007

If someone has KK preflop at a table of 9, what are the odds that any other one player will have AA?

ECS · Jan 13, 2007

100 percent if im the one with pocket Kings

RobFunk · Jan 13, 2007

ECS said:
100 percent if im the one with pocket Kings

Pretty much.

goodcall · Jan 13, 2007

3.918%

CHOPTALK · Jan 13, 2007

The odds one person getting AA is 220 to 1.

Hitman26 · Jan 13, 2007

Thanks Caller.

ensign_lee · Jan 13, 2007

Same as if you had any other hand that didn't include an ace

4/50 *4/50 = .0064, therefore .64%

ensign_lee · Jan 13, 2007

goodcall said:
3.918%

This would be a horrible abberation if true.

RobFunk · Jan 13, 2007

ensign_lee said:
Same as if you had any other hand that didn't include an ace

4/50 *4/50 = .0064, therefore .64%

I knew this off the bat except :

I didnt know the proper math.
I wasnt sure if Hitman was being sarcastic or not.

goodcall · Jan 13, 2007

ensign_lee said:
Same as if you had any other hand that didn't include an ace

4/50 *4/50 = .0064, therefore .64%

4/50 * 3/49 * 8

Hitman26 · Jan 13, 2007

No I was truly wondering.

Hitman26 · Jan 13, 2007

4% seems awfuly high Caller.

goodcall · Jan 13, 2007

Hitman26 said:
4% seems awfuly high Caller.

It does to me too, but 8 opponents is a lot of chances. I'll stick to that number until someone points out the flaw in my math (there probably is one...it's late).

Hitman26 · Jan 13, 2007

Ok, I appreciate the input even if it's wrong.

bulldog77 · Jan 13, 2007

If its 220/1 anybody has it than its 220/9 that somebody else has it at a ten player ring game.

CHOPTALK · Jan 13, 2007

<TABLE class=odds cellSpacing=0 cellPadding=5 border=0><TBODY><TR bgColor=#ede085><TD vAlign=top>AA versus KK preflop (heads up)</TD><TD>.004</TD><TD>1 in 22560 - 22559:1</TD></TR></TBODY></TABLE>http://www.learn-texas-holdem.com/texas-holdem-odds-probabilities.htm

ensign_lee · Jan 13, 2007

goodcall said:
4/50 * 3/49 * 8

D'oh. Didn't think that through enough. Good call goodcall.

ensign_lee · Jan 13, 2007

Oh wait. Forgot to take into account that each person has that chance to have AA

So it would be

4/50 * 3/49 * 8 = about 4%

I'm wrong...

Woody0 · Jan 13, 2007

goodcall said:
It does to me too, but 8 opponents is a lot of chances. I'll stick to that number until someone points out the flaw in my math (there probably is one...it's late).

You can't answer this question using an algebraic method but have to use combinatorial mathematics. Unfortunate I can't handle combinatorials. However combinatorials allow for the chance that the four aces come out before the second card is dealt to any player while your method does not consider this possibility. Ask at 2 X 2.

ClipJoint · Jan 13, 2007

Online = 50/50

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