#1
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Raising from the dealer position.
Did I figure this out correctly. Assume that three people are dealt two cards from three completely different decks. They can be dealt exactly the same hand. Now I will rank each two card hand all 1326 combo's from high to low. It is unimportant to me what the ranking system is. Suppose I have two cards that are exactly at the 40th percentile.
Is the probability that I have the best hand. 1-((2*.40)-(.4*.4)) = .36 or 36 percent of the time I have the best hand. Thanks, Cobra |
#2
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Re: Raising from the dealer position.
When I say 40th percentile I mean 60 percent of the hands are worse than it and 40 percent are better.
Cobra |
#3
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Re: Raising from the dealer position.
If you strip away all the card references, does this question basically boil down to "There are 100 numbers in a hat, three people pick a number and replace it...what are the odds that if I pick 60, it's the highest number of the three?"
If so, I'd say 34.81% of the time you'll have the best "hand" or whatever. |
#4
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Re: Raising from the dealer position.
Yes LetYouDown, that is what I am asking. If you don't mind how did you come up with that answer. I am more interested in how the answer was derived than the answer itself.
Cobra |
#5
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Re: Raising from the dealer position.
Well, we know you'll pick 60. So you need the other two people to pick 59 or lower. They both have a 59% chance of doing this. .59 * .59 = .3481 or 34.81%.
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#6
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Re: Raising from the dealer position.
Boy that was easy wasn't it. Thanks, Cobra
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