Timer
01-01-2005, 06:57 PM
If the board reads k-9-5-3 aren't there 10 ways you can have a set with a k-j already out?
I think I can understand how he comes up with the 27 ways you can make two pair. In order to do this he is eliminating the possibility that the opponent has 95 or 93.
But the set calculation I don't understand. I'm obviously overlooking something simple.
Thanks
I think I can understand how he comes up with the 27 ways you can make two pair. In order to do this he is eliminating the possibility that the opponent has 95 or 93.
But the set calculation I don't understand. I'm obviously overlooking something simple.
Thanks