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

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

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.

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

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%.

Re: Raising from the dealer position.
 

Boy that was easy wasn't it. Thanks, Cobra