Re: pocket aces probability
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ok so i got them again on hand 881, so does that mean the chances of that happening are (1-((220/221)^(881-795)=67.7%)=32.3%?
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I don't understand that expression.
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[edit] I got them again in hand 911,....
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So 3 times in 911 hands? If you want to understand this you may want to read up on the binomial probability distribution. The calculation is easy in Excel:
=binomdist(3,1/221,911,true) gives .409492, i.e. a 40.9% probability of getting AA 3x or fewer in 911 hands
=binomdist(3,1/221,911,false) gives .189411, i.e. an 18.9% probability of getting AA exactly 3x in 911 hands
The most likely number of times you would get AA in 911 hands is 4, and that will happen 19.5% of the time.
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....though they were called and sucked out on by some moron with 93s. Dont play at bugsys club[/edit]
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You've got this wrong. Even if you never learn to calculate probabilities, you should understand this.
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