Tuesday, October 19, 2010

Possible Stayman Tweak?

Just an idea...

I think that it might be possible to re-structure Stayman to have a 2S rebid by Opener (1NT-P-2C-P-2S) show four spades and a minimum.  2D, then, would show either no 4-card major OR if spades a maximum.

For examples of how this would unwind...

1NT-P-2C-P-
2S-P-P-P  (stopping at 2S is good)

1NT-P-2C-P-
2D-P-2NT-P-
3S = four spades, accepts game try

1NT-P-2C-P-
2D-P-2NT-P-
3H = five spades, maximum

1NT-P-2C-P-
2D-P-3S(Smolen)-P-
4S(four spades maximum)

1NT-P-2C-P-
2D-P-3H(4-4 or 4-5)-P-
3S = 3 spades
4C+ = 4 spades

1NT-P-2C-P-
2D-P-3NT (not four spades; spade situation kept secret)

This type of tweak might allow some stops at 2S and might further allow some degree of knowing whether pener has minimum or maximum, in some sequences.

I am not sure whether this is worth it, but ideas are ideas.

2 comments:

Anonymous said...

Lauria-Versace do the reverse.
A Stayman response of two diamonds denies four spades and amaximum. I've never asked them why...

Barry

Kenneth Rexford, Esq. said...

Good company to keep. Glad to hear that others have thought of this problem.

I also wonder why the reverse is used. It seems like the 2NT problem is solved if Opener buries spades with a maximum, as he can show them on an acceptance but not on a declining, and as then 2S is passable. The other benefits seem present, though.