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#1
11-20-2007, 08:38 PM
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Holdem AK running into AA 9-handed
Ok, this must have been asked like 10000 times before and I apologize should I have failed more than average when using the forum search engine.
The game is Texas Holdem and itīs a nine handed table. How many % of the time does some villain have AA if Hero has AK? Thank you very much in advance. 
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#2
11-20-2007, 08:52 PM
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Re: Holdem AK running into AA 9-handed
[ QUOTE ]
Ok, this must have been asked like 10000 times before and I apologize should I have failed more than average when using the forum search engine. The game is Texas Holdem and itīs a nine handed table. How many % of the time does some villain have AA if Hero has AK? Thank you very much in advance. [/ QUOTE ] You can do this one in your head since at most 1 opponent can have AA when you have AK. Since the probability that a particular opponent has it is 3/C(50,2) = 3/1225, the probability that one of 8 opponents has it is exactly 8*3/1225 or about 2.0%. EDIT: Originally computed for 9 opponents rather than 8.
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#3
11-21-2007, 06:03 AM
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Re: Holdem AK running into AA 9-handed
[ QUOTE ]
You can do this one in your head since at most 1 opponent can have AA when you have AK. [/ QUOTE ] Yes, but... if the first villain does not have AA, then the second villain is less likely to have AA than the first, because the first does sometimes have a hand with one ace, so thereīs two aces left rather than three?
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#4
11-21-2007, 11:48 AM
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Re: Holdem AK running into AA 9-handed
[ QUOTE ]
[ QUOTE ] You can do this one in your head since at most 1 opponent can have AA when you have AK. [/ QUOTE ] Yes, but... if the first villain does not have AA, then the second villain is less likely to have AA than the first, because the first does sometimes have a hand with one ace, so thereīs two aces left rather than three? [/ QUOTE ] It would actually make it slightly more likely for the next villian to have AA. We don't know whether the first villian has an A or not; all we know is that he hasn't got two of them. However, it doesn't really matter anyway. It's impossible for two other opponents to both hold AA. So, they're mutually exclusive. Using basic probability theory, the total probability is just the sum of the individual probabilities of each opponent.
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