[banrock]

Playing the odds what is the best way to play the following, assuming no bidding from either op and no problems getting between hands. Is it possible to play for 2 tricks or is that only when AK is doubleton with an op?

Q98x opposite Jxxx

[Mbodell]

You'll do better if you can get the opponents to break the suit for you. But you can get 2 tricks if you can get a 3-2 split (or stiff T) as long as you don't lose the first 3 tricks to A, K, T. If you play low to the J and then (after that loses) play low to the 8 you have the best chance to win 2 tricks.

You will win 2 tricks for the following combinations of North's cards (I'm making our x's be 2,3,4,5 and theirs be 6,7):
-
7
6
76
T
T76
K
K7
K6
K76
KT
A
A7
A6
A76
AT
AK
AK7
AK6
AK76
AKT

You will win just 1 trick for the following:
T7
T6
KT7
KT6
KT76
AT7
AT6
AT76
AKT7
AKT6
AKT76

So you win: 1 5-0 split, 6 4-1 splits, and 14 3-2 splits; and lose: 1 5-0 split, 4 4-1 splits, and 6 3-2 splits. So that is about 2/3 of the time you get 2 tricks compared to 1/3 of the time you get 1 trick.

[banrock]

Thanks for that. At least if I play it that way all the time I'll get 2 tricks more times than I'll get 1