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09-20-2002, 03:44 PM
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

irchans
09-21-2002, 09:37 AM
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).

09-21-2002, 05:11 PM
my question.
But a question on "independence".
Let's assume the board is:

x x Q 4

U have: QQ

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

09-23-2002, 12:51 AM
"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

heihojin
09-23-2002, 04:39 PM
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.

These two events are certainly not independent:

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

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

heihojin
09-23-2002, 04:50 PM
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

09-25-2002, 03:08 AM