The chance someone has AA if you have KK at a 9 person table

[BruceZ]

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The chances that someone holds KK or AA when you have either or, is 44:1. No matter what

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Sounds like you're ready to make some side bets! I'll let you give me 40:1. [img]/images/graemlins/cool.gif[/img]

[binions]

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The chances that someone holds KK or AA when you have either or, is 44:1. No matter what

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Um, no. The chance of AA when you have KK is highly dependent on how many hands are dealt. The others have pegged it. It's ~24:1 at a 9 person table.

At a 2 person table, it would be rarer - 203:1 to be exact. After all, it's 216:1 to get AA before any cards are dealt.

At a 26 person table, AA over KK would be more common than 24:1.

[Gavagai]

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Actually no one gave the right answer since he asked about a 9 person table, though closer gave the correct answer (mine) for a 10 person table. For a 9 person table, the exact answer is 8*6/C(50,2) - C(8,2)/C(50,4) =~ 3.9% or 1 in 25.6 = 24.6:1.

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Could you explain this calculation? What is 'C'? I'm afraid I never took maths to a very advanced level.

Thanks,

Gavagai

[BruceZ]

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Actually no one gave the right answer since he asked about a 9 person table, though closer gave the correct answer (mine) for a 10 person table. For a 9 person table, the exact answer is 8*6/C(50,2) - C(8,2)/C(50,4) =~ 3.9% or 1 in 25.6 = 24.6:1.

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Could you explain this calculation? What is 'C'? I'm afraid I never took maths to a very advanced level.

Thanks,

Gavagai

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Try this post or one of a million others that discuss the same thing.

[J.Copperthite]

Of course dealing a flop in a 26 person game would be impossible... [img]/images/graemlins/smile.gif[/img] [img]/images/graemlins/smile.gif[/img]

[eOXevious]

[ QUOTE ]
[ QUOTE ]
Actually no one gave the right answer since he asked about a 9 person table, though closer gave the correct answer (mine) for a 10 person table. For a 9 person table, the exact answer is 8*6/C(50,2) - C(8,2)/C(50,4) =~ 3.9% or 1 in 25.6 = 24.6:1.

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Could you explain this calculation? What is 'C'? I'm afraid I never took maths to a very advanced level.

Thanks,

Gavagai

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C stands for combination, google combinations and permutations... its very interesting and easy stuff

[Cobra]

I did something similar. I don't know if this is what you are looking for.

http://forumserver.twoplustwo.com/sh...mp;o=&vc=1

Cobra

[etgryphon]

Haha

Yeah that was the orginial attempt at it. I kinda left it alone for a bit. Can you either PM or just post it here and explain all the terms?

Thanks,

-Gryph

[SrGuapo]

I follow the math on the AA vs. KK example that closer posted, and was thinking about the probabilty someone will have AA or KK if I have QQ on a ten person table. This math becomes another level more complicated, right?

[BruceZ]

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I follow the math on the AA vs. KK example that closer posted, and was thinking about the probabilty someone will have AA or KK if I have QQ on a ten person table. This math becomes another level more complicated, right?

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If you want an exact answer, then you need 4 terms since up to 4 people can have AA or KK, but as a practical matter, 2 terms will still get you to within 0.001%. Here is the exact calculation:

P(AA or KK vs. QQ) =

9*12/C(50,2) -
C(9,2)*(12*7)/C(50,2)/C(48,2) +
C(9,3)*12*(6*2 + 1*6)/C(50,2)/C(48,2)/C(46,2) -
C(9,4)*12*(6*2*1 + 1*6*1)/C(50,2)/C(48,2)/C(46,2)/C(44,2)

=~8.6% or 1 in 11.6 or 10.6-1.

As an explanation of the third term, notice that the first player has a choice of 12 hands, the second player has a choice of either 6 of one hand or 1 of the other, and if the second player chooses one of the 6, then the remaining player has a choice of 2 hands, otherwise he has a choice of 6 hands. As these problems get more complicated, you can draw a tree-like structure to describe the various possibilities.