Zach6668
Thursday, January 25th, 2007, 1:15 PM
QUOTE (nima @ Wednesday, January 24th, 2007, 2:50 PM)

The only outs to beat two pair are the 3 Kings left and the two Aces, more than likely neither one will be good enough, given everyone has called and the fact that the guy to our right check-raised from a bad position.
I would consider the Tens as outs (of course, since we get the nut straight) but you can't count them all as full outs as you probably will end up splitting with KQ or something.
Nima
Quick Lesson in Hidden Outs:
5JQ A board.
We hold AK.
Opponent holds two pair, say QJ.
We have 3 Ks, 2 As, 3 5s to beat QJ.
Change his hand to Q5.
3s K, 2 As, 3 Js beat his two pair.
Again, this time J5.
3sK, 2 As, 3 Qs beat his two pair.
(and of course the Ts for each example)
********
Moral - Don't forget the counterfeiting outs.
If we take the first example, where he has QJ, we have AK, and the board comes 5QJA5, his hand is QQJJA, we have AA55K. We win.
Cheers.
- Zach
EDIT -
I didn't even think about A5 hands where we can hit any Q, J, K or T to beat him.... yeah, 13 outs.
If he has AJ, 3 Qs, 4 Ts, 3 Ks = 10 outs.
Aside from the straight, or a set, AQ is the only two pair hand we are in a bit of trouble against, and even then we have 7 outs.