## Cardkountr: Re: Question

> Your reasoning may turn out to be correct,
> but I'd just like to point out that having
> the "highest number of outs" can't
> be used in a void. Those outs all have
> probabilities assigned to them, and it is
> the overall probability of ending the
> tourney with more money that determines the
> correct play. For example, pushing all three
> hands (both players) may count as one out,
> but it has a much lower probability than,
> say, all hands winning. So, you can't just
> add up the outs, as you suggest.

> Maybe if I have some time, I'll try to do
> the math.

> Don

Don if you have time I would appreciate you doing the math for each possible outcome. I wasn't referring to "outs" in the traditional sense as you would in poker, and I realize that each scenario has it's own ev, perhaps I didn't express myself well but the point I was trying to make was in response to Parker saying that he would not split the pair of 10's if the other player recieved a low card on his double but would split if the other player recieved a 10 or Ace on the double.

Keep in mind the other player started the hand with \$10 less than you and had to bet/play first. He bet table max of 500 so you matched his bet with 500, he doubled a 9 so now he has 1000 in play. If he received a low card on his double you should certainly split your 10's to match your oponnents 1000 total bet, and then stand on any card you recieve on each of your split hands. That way you both either win or lose the hand together and in doing so you will finish with \$10 more in chips and are guaranteed to win the tourny since no other player was remotely close in chips.

Card.