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#1
07-10-2005, 05:41 PM
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AA v KK
i'm dealt KK.
what're the chances that playing 8 handed that my opponent has AA? this seems to have been happening rather a lot recently ! edit : just realised this should probably have gone in the Probability section...sorry 
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#3
07-10-2005, 07:39 PM
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Re: AA v KK
[ QUOTE ]
i'm dealt KK. what're the chances that playing 8 handed that my opponent has AA? this seems to have been happening rather a lot recently ! edit : just realised this should probably have gone in the Probability section...sorry [/ QUOTE ] 10 handed it's 24.5 or so to 1. 8 handed i'm guessing maybe 26-27 to 1.
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#4
07-10-2005, 08:20 PM
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Re: AA v KK
6 handed its 44/1 so, i dont think its 25/1 8 handed....
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#5
07-10-2005, 11:05 PM
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Re: AA v KK
[ QUOTE ]
6 handed its 44/1 so, i dont think its 25/1 8 handed.... [/ QUOTE ] Never know; odds are exponential. But I'll let someone with a calculator and fully-functioning brain work this'un out. Quick thought: do you figure the odds of another person getting rockets and multiply it by # of players? That doesn't make sense, but how do you account for multipule players? Oh, the anguish of a ill mind.
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#6
07-11-2005, 01:50 AM
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Re: AA v KK
for each player you can just do (4/50)*(3/49) instead of (4/52)*(3/51) to figure out the % chance. It changes from 221 to 1 and becomes 204 to 1 per player. With 7 opponents, that changes the chance to 29 to 1 that someone has AA. Although the chances are only some thousand to one that there is both KK and AA in a hand, you've already completed your part and that bears no significance to whether or not someone has AA except that you've taken out 2 non-ace cards from the deck.
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#7
07-11-2005, 02:10 AM
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Re: AA v KK
[ QUOTE ]
Quick thought: do you figure the odds of another person getting rockets and multiply it by # of players? That doesn't make sense, but how do you account for multipule players? Oh, the anguish of a ill mind. [/ QUOTE ] That will get you very close, but it double counts the deals where 2 opponents have AA, so for an exact answer we then subtract off the probability that 2 opponents have AA. The probability that 2 specific opponents have AA is 1/C(50,4). To get the probability that any 2 opponents have AA, this is C(7,2) times the probability that 2 specific opponents have it, and this is exact because no more than 2 opponents can have it, so the pairs of opponents are mutually exclusive. 7*6/C(50,2) - C(7,2)/C(50,4) =~ 28.2-to-1. This is the inclusion-exclusion principle.
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#8
07-11-2005, 02:21 AM
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Re: AA v KK
[ QUOTE ]
6 handed its 44/1 so, i dont think its 25/1 8 handed.... [/ QUOTE ] 6 handed is 39.9-to-1, not 44-to-1. It looks like you are the latest victim of bad WPT televised odds. I covered this extensively on the probability forum. 6 handed: 5*6/C(50,2) - C(5,2)/C(50,4) = 39.9-to-1. 8 handed: 7*6/C(50,2) - C(7,2)/C(50,4) = 28.2-to-1.
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