For 2 tricks ?

[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