Re: over-flush odds
[ QUOTE ]
If I get it correctly, you want the probability for your opponent to get an over-flush in the moment when all community cards are dealt (7 cards seen). In this situation, the only possible over-flush is a diamond suit with an ace.
So, 45 unseen cards, 8 diamonds left with one ace.
If you want the probability for one opponent (a specific one) to get that over-flush, this is 7/C(45,2) = 7/990 = 0.70%.
If you want the probability for at least one opponent to get this, the formula is: 7*n/990, where n is the number of your opponents (maximum one opponent could hold Ax of diamond)
[/ QUOTE ]
Ok, bear with me here, (still trying to learn the lingo). You answered my the right question but now I have questions about your answere [img]/images/graemlins/tongue.gif[/img].
What is 7/C(45,2)(where do these numbers come from?)
45 is the number of unseen cards i believe. 2 is the number of cards that they are holding maybe? And 7/C I just have no clue. And what do you do with (45,2) to turn it into 7/990 = 0.70%.
thanks for any input. [img]/images/graemlins/blush.gif[/img]
|