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  #1  
Old 01-19-2005, 11:33 PM
Jon Poker Jon Poker is offline
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Default What is the answer to this question?

I am holding Kh, Jd, the board is showing Qs, As, 7s. There are four other players.

How would you calculate the probability that one of the opponents has a pair of Aces?

How would you calculate the probability that one of the opponents has a flush?


Thanks in advance for your help.
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  #2  
Old 01-19-2005, 11:59 PM
deacsoft deacsoft is offline
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Default Re: What is the answer to this question?

You can mathematically, but besides having a general idea of how they play preflop what else would be more reliable?
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  #3  
Old 01-20-2005, 12:04 AM
Jon Poker Jon Poker is offline
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Default Re: What is the answer to this question?

True, but I am just looking for the mathematical answer.

I am aware that starting hand rankings, and how the opponents have played the hand also tell a bigger part of the story.

If anyone is good at statistics please reply.
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  #4  
Old 01-20-2005, 01:08 AM
Kaz The Original Kaz The Original is offline
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Default Re: What is the answer to this question?

You would need a complete range of their preflop standards to have an accurate answer.
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  #5  
Old 01-20-2005, 03:44 AM
bisser bisser is offline
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Default Re: What is the answer to this question?

If you are willing to completely ignore any information you may have obtained from their preflop play then the math is very easy.

I will calculate the pair of Aces and leave you the flush as homework [img]/images/graemlins/smile.gif[/img] ... the flush is slightly more difficult.

There are 47 unseen cards in the deck. Three of them are Aces.The four players hold 8 cards. So at least one of them will have pair or trips of Aces if at least one of the 8 cards is A. There are C = (47*46*45*44*43*42*41*40)/(2*3*4*5*6*7*8) total 8 card combinations. Of these B = (44*43*42*41*40*39*38*37)/(2*3*4*5*6*7*8) do not contain an Ace. The probability that none of them has a pair or trips of Aces is B/C = (39*38*37)/(47*46*45) = 0.5636. The probability that they have it is therefore 1 - 0.5636 = 0.4364. If you want exactly a pair (no trips) you need to add the number of combinations giving trips to B.
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  #6  
Old 01-20-2005, 12:03 PM
Indiana Indiana is offline
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Default Re:The flush part

Good question...If three cards to a flush are out there, there is about a 4% per person chance...so with 4 people , about 16% chance...I would worry about the A being out there, but would have little worries until a fourth flush card came down...This is not to say that someone doesn't have the flush....Just not all that likely.

Indiana
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  #7  
Old 01-20-2005, 03:37 PM
citizenkn citizenkn is offline
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Default Re: What is the answer to this question?

And the answer to the big question is yes, you should fold.
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