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  #1  
Old 09-20-2002, 03:44 PM
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Default Probability ? from Sitting Bull

Hello,Analysts!
If the probability of winning on the river is P1 and the probability of your opponent beating U is P2,is the formula correct:

P*=P1(1-P2) where P* is your "correct" winning probability??
just wondering,
Sitting Bull
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  #2  
Old 09-21-2002, 09:37 AM
irchans irchans is offline
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Default Re: Probability ? from Sitting Bull

I don't really understand your question, but I can ask a different question that has the same answer.

If
1) there were two possible hands on the river,
2) the probability of you getting the top hand is P1,
3) the probability of your opponent getting the top hand is P2, and
4) those probabilities are independent,
then the probability of you winning is

P = P1 (1-P2).
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  #3  
Old 09-21-2002, 05:11 PM
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Default Thanks,Erchins! You\'ve answered...

my question.
But a question on "independence".
Let's assume the board is:

x x Q 4

U have: QQ
Your opponent has: 44

If the board pairs with another 4,U have a full-house,but the 4 helped your opponent to obtain quads.

Are these two events "independent"? and does the formula
p=p1(1-p2) still apply? If U don't have "independence",then how would U modify the formula to allow for "dependence"??
Just wondering,
Happy pokering
Sitting Bull
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  #4  
Old 09-23-2002, 12:51 AM
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Default Re: Probability ? from Sitting Bull

"If
1) there were two possible hands on the river,
2) the probability of you getting the top hand is P1,
3) the probability of your opponent getting the top hand is P2, and
4) those probabilities are independent,"


JRChan, wouldn't the possibility of you winning be P1? Since to win you have to get the top hand? Then P1 + P2 + P3 = 1, with P3 being the possiblity of a tie.

- Tony
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  #5  
Old 09-23-2002, 04:39 PM
heihojin heihojin is offline
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Default Independence

The word "independent" means, when describing events in the context of probability, that the information that one event occurs does not affect the probability that another event occurs. In mathematical notation, event A is independent of event B if and only if P(A|B) = P(A), i.e. the probability of event A occurring given that event B occurs is equal to the probability of event A occurring.

I assume that the two events to which are you referring are:

A: You make a full house on the river.
B: Your opponent makes quads on the river.

These two events are certainly not independent:

P(A) = 7/44
P(A|B) = 1

If you are asking about the independence of these two events:

C) You hold the highest-ranked hand on the river.
D) Your opponent holds the highest-ranked hand on the river.

then these events are still not independent.

P(C) = 43/44
P(C|D) = 0


heihojin
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  #6  
Old 09-23-2002, 04:50 PM
heihojin heihojin is offline
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Default Re: Independence

Keep in mind that the exact calculations for those events I defined are only correct if neither player has a flush draw in the given example, including the possibility of a four-flush on board. The events, however, would still not be independent.


heihojin
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  #7  
Old 09-25-2002, 03:08 AM
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Default Hello! Thanks Tony,Hei... from Sitting Bull!

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