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#1
03-12-2005, 07:57 PM
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Was the WPT correct?
You are dealt K,K in a six handed game. What are the odds that one of your opponents has A,A? This question comes from the WPT Legends of Poker Tournament at the Bicycle club. They gave the answer 44 to 1. I believe they are %10 too high. Well?
Vince 
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#2
03-12-2005, 08:43 PM
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Re: Was the WPT correct?
If I followed BruceZ's method from this thread correctly, the probability is
5*6/C(50,2) - C(5,2)/C(50,4)= .024446 which is odds of about 39.9 to 1. So yeah, about 10% too high. Good catch.
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#3
03-13-2005, 12:58 AM
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Re: Was the WPT correct?
I thought so. Thanks,
Vince
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#5
03-13-2005, 02:54 AM
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Re: Was the WPT correct?
[ QUOTE ]
220 to 1 or 1/221 for one player to have AA, 5 * 1/221 for any of other 5 to have it or 5/221 = .022624. [/ QUOTE ] Your calculation is wrong. The earlier calculation on this thread was correct. Here are two problems with your calculation: [img]/images/graemlins/diamond.gif[/img] You did not use the fact that the player has two cards that are not aces. The chance that a particular opponent has AA is 6/(50 choose 2) rather than 6/(52 choose 2). This correction is significant. I believe the same mistake started this discussion. [img]/images/graemlins/diamond.gif[/img] You assumed that the times your opponents have AA are disjoint. It is possible that two of your opponents have AA, so the probability of AA is lower than five times the probability that a particular opponent has AA. This correction is quite small.
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#7
03-13-2005, 02:14 PM
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Re: Was the WPT correct?
[ QUOTE ]
Thank you for the correction you are absolutely right. Help me understabd the second point. If I am asking the probability that one other opponent has AA do I subtract the probability that two have it? [/ QUOTE ] When you take 5 times the probability of a single opponent having it, that will double count all the times that 2 opponents have it. If you want the probability that at least one opponent has it, then you must subtract off the probability that 2 have it, so that you will be counting these cases only once. This is the inclusion-exclusion principle. If you want the probability that exactly one opponent has it, then you subtract off 2 times the probability that 2 have it, so that you will be counting these cases 0 times. Since no more than 2 can have it, the C(5,2) cases when 2 players have it are disjoint, so you can get the probability that 2 have it by simply multiplying C(5,2) by the probability that 2 specific opponents have it. In this case this is only 0.0043%.
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#9
03-14-2005, 02:29 PM
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Re: Was the WPT correct?
[ QUOTE ]
I will trust the WPT'S figures on the subject 44 to 1. .0227% still real low good enough for me [/ QUOTE ] 1st: 44 to 1 is 1/45, not 1/44 (0.0222, not 0.0227) 2nd: There is no percent sign in that (2.22%, not 0.0222%) 3rd: Why would you trust the WPT's numbers when the poster you responded to walked you through the math he used to get his answer? If you think the WPT is correct, come up with some numbers to back it up.
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#10
03-15-2005, 03:48 AM
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Re: Was the WPT correct?
[ QUOTE ]
[ QUOTE ] I will trust the WPT'S figures on the subject 44 to 1. .0227% still real low good enough for me [/ QUOTE ] 1st: 44 to 1 is 1/45, not 1/44 (0.0222, not 0.0227) 2nd: There is no percent sign in that (2.22%, not 0.0222%) 3rd: Why would you trust the WPT's numbers when the poster you responded to walked you through the math he used to get his answer? If you think the WPT is correct, come up with some numbers to back it up. [/ QUOTE ] At one time the WPT came here looking for answers to this kind of question. Looks like they ought to come back.
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