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1C-1D, 1S-2S


straube

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1C-1D strong, weak or GF

1S-2S 4+ spades, weak with exactly 3 spades

 

If opener has 5 spades, it's easy. If opener has 4S/5m I have a problem.

For continuations looking at

 

2N-5+C, forcing

.....3C-no fit, 0-2

..........3D-4-1-3-5, f

..........3H-4-3-1-5, f

..........3S-?

..........3N-?

.....3D-no fit, 3-4

.....3H-fit, 0-2

.....3S-fit, 3-4

3C-5+D, forcing

.....same steps

3D-4-1-5-3, f

3H-5+S, inv

3S-5+S, f

 

I'd like something better and simpler.

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Maybe drop the fit/no fit responses

 

2N-GI, 5+C

.....3C-club preference, weak

.....3D-5D, f

.....3H-5H, f

.....3S-short C, weak

.....3N-other (2+C then)

 

3C-GI, 5+D

.....3D-diamond preference, weak

.....3H-5H, f

.....3S-short D, weak

.....3N-other

 

3D-5S, game try

 

3H-4-1-5-3

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Why do you say

"If opener has 5 spades, it's easy."?

 

It's not. Opener needs some sort of trial bid here.

We play

 

1 - 1

1 - 2

 

2NT = start of a short suit trial, usually 5 s

3m = natural, not forcing

3 = long suit trial, balanced-ish

3 = invite, usually 6s

 

Note that we play opener's 1 as limited, 16-19.

Stronger hands rebid 1 as 19+ any

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Why do you say

"If opener has 5 spades, it's easy."?

 

It's not. Opener needs some sort of trial bid here.

We play

 

1 - 1

1 - 2

 

2NT = start of a short suit trial, usually 5 s

3m = natural, not forcing

3 = long suit trial, balanced-ish

3 = invite, usually 6s

 

Note that we play opener's 1 as limited, 16-19.

Stronger hands rebid 1 as 19+ any

 

Unlike your method, our 1S is unlimited, so we have a problem holding something like 23+ 4-1-3-5. We have to look consider that we might belong elsewhere than spades and we have to be able to both invite and force game opposite a hand that is 0-4 hcp and not 0-7 hcps.

 

It would have been more precise to say that when opener holds five spades, it's "comparatively" easy because reaching a fit at the point of 2S is much easier than restarting to find one.

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