Was the WPT correct?

[Vince Lepore]

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

[uuDevil]

If I followed BruceZ's method from this thread (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Number=1771837&page=&view=&s b=5&o=&vc=1) 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.

[Vince Lepore]

I thought so. Thanks,

Vince

[Mark1808]

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.

[pzhon]

[ 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:
/images/graemlins/diamond.gif 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.
/images/graemlins/diamond.gif 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.

[Mark1808]

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?

[BruceZ]

[ 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 (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Board=probability&Number=417 383&Forum=probability&Words=inclusion-exclusion&Match=Entire% 20Phrase&Searchpage=0&Limit=25&Old=1year&Main=4169 81&Search=true#Post417383). 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%.

[TEKEE]

I will trust the WPT'S figures on the subject 44 to 1. .0227% still real low good enough for me

[elitegimp]

[ 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.

[BruceZ]

[ 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 (http://archiveserver.twoplustwo.com/showthreaded.php?Cat=&Number=212797&page=&view=&sb =5&o=) looking for answers to this kind of question. Looks like they ought to come back.

[TEKEE]

I apologize BRUCEZ,i stand corrected read the post you mentioned,why does everyone have different figures,for example the odds of making a set on the flop when your hole cards are paired.Everyone says something different.Please give me figures for flop, turn and river.

[Cobra]

The reason people's answers are different is because they are answering different questions. For example suppose you have KK and you ask what the probability of getting a set is. You might see the following answers.

=1-(48c3)/(50c3)=.118 or 1 in 8.5 This is really the probability of getting either a set, four of a kind, or a full house with three kings.

=(2c1)*(48c2)/(50c3)=.115 or 1 in 8.7 This is really the prbability of getting a set or a full house with 3 kings.

=(2c1)*(12c2)*4*4/(50c3)=.108 or 1 in 9.3 This is the probability of getting a set and only a set.

I am not sure what you mean by on the river and turn, explain and I will try to answer it.

Cobra