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4 card majors 2/1 GF

#1 User is offline   Kungsgeten 

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Posted 2014-June-13, 01:46

If playing 4-card majors and strong 1NT, how would you continue after a GF 2/1? We currently play natural (so opener's 2NT rebid might have 4 or 5 cards in the opening suit, rebidding the major shows a 6+ suit and bidding a new suit is 5-4, not showing extras). This leads to some problems, since it is hard to know when to bypass 3NT and sometimes its a bit cumbersome to find a 5-3 major fit. It is also a problem to know when opener has extras. An alternative approach might be to rebid the major with 5+ suit, bid 2NT with a 4-card suit and make bids at the 3-level show extras.
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#2 User is offline   helene_t 

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Posted 2014-June-13, 02:10

I don't think making the 2NT rebid deny a 5-card major is any good. It is quite an accurate bid even if including 5332. I don't see why finding the 5-3 fit is a problem - just bid
1-2
2NT-3.

Using
1M-2
2
as a waiting bid solves some of the issues with resolving opener's strength. Obviously far from all of them.
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#3 User is offline   Kungsgeten 

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Posted 2014-June-13, 06:57

Yes, that is possible. Then what should the rebid be? Promising a 6+ suit or just 5+ and unbalanced? I had a 5-3-1-4 the other day and the bidding went like this: 1S-2D; 2NT-3D; 3NT. Perhaps I should have rebid the spades after partner's 3D, but I had a heart stopper and partner might bypass 3NT in case of fearing that hearts is wide open. I could have bid 3C with this hand instead of 2NT, but it is hard for partner to know if I'm min or if I have extras.

We do not open 1M with 4441 btw. We open 4432 in the order hearts, clubs, spades, diamonds (so yeah, 1D never is 4432) and with 4441 we open our lowest suit.
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#4 User is offline   whereagles 

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Posted 2014-June-13, 08:25

kung, I don't recommend the style you're playing because even if you open 1NT on 5-card majors (which I think you don't), there is no way to accurately split 4- form 5-card major balanced hands with the available bids.

"Demonstration":

1. bal 15-17 5CM opens 1NT

1M-2x
2M = 12-14 5CM bal or unbal
2NT = 12-14 4CM bal
3NT = 18-19 4CM bal
(& no bid for 18-19 5CM bal)

2. bal 15-17 5CM opens 1M

1M-2x
2M = 12-14 5CM bal or unbal
2NT = 12-14 4CM bal
3NT = 18-19 4CM bal
(& no bid for 15-17 5CM bal and 18-19 5CM bal)


If you insist on the structure, a possible way out might be

1M-2x
2M = 5+CM. If bal then range can be 12-14 or 15-17. If unbal range is only 12-14.
2NT = 12-14 or 18-19 4CM bal
3NT = 18-19 5CM bal

Not very mnemonic though
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#5 User is offline   Zelandakh 

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Posted 2014-June-13, 09:09

Whereagles, would it not be easiest just to add the 5cM 18-19 to the 2M rebid in your first structure? This hand might then raise 3NT to 4 if no fit is found along the way. Indeed you could combine that with your split-range 2NT rebid and free up 3NT for anything you wanted.
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#6 User is offline   ArtK78 

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Posted 2014-June-14, 11:32

Playing 5 card majors, it is common for the rebid of opener's major to be just a mark-time bid, not promising extra length. Playing 4 card majors that would seem to be even more important. While you might require 5 cards for the rebid, any requirement of 6 for a rebid is just too much. It creates a number of rebid problems on all hands with less than 6 cards in the major suit.
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#7 User is offline   whereagles 

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Posted 2014-June-14, 11:46

View PostZelandakh, on 2014-June-13, 09:09, said:

Whereagles, would it not be easiest just to add the 5cM 18-19 to the 2M rebid in your first structure? This hand might then raise 3NT to 4 if no fit is found along the way. Indeed you could combine that with your split-range 2NT rebid and free up 3NT for anything you wanted.


That's indeed a possibility, but you'd lose the mnemonic "2M rebid is catch-all with min hands".
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#8 User is offline   Kungsgeten 

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Posted 2014-July-09, 02:15

View Postwhereagles, on 2014-June-13, 08:25, said:

kung, I don't recommend the style you're playing because even if you open 1NT on 5-card majors (which I think you don't), there is no way to accurately split 4- form 5-card major balanced hands with the available bids.


We open 1NT with 15--17 even if we hold a 5 card major. 1M may be a four card suit if 4432/4333 and 12--14 or 18--19 hcp. 1H may be any 4432, 1S may only have 4D.

I do not really get why it would not be possible to distinguish the 4 card suits from the 5 card suits, except when opener supports responder's suit. If you do something like this:

1M--2x;
2M = 5+ suit catch-all, Lawrence style (but may have be 5332)
2NT = 4 card major, 12--14
3NT = 4 card major, 18--19

You basically have a standard 5 card majors strong club system, pretty similar to one if you play strong club (but there 2NT and 3NT would probably be artificial). Now the question would ofcourse be if it is possible to untangle the 2M rebid, which may be a bit hard.

Let's say you change it a bit:

1M--2x;
2M = 5+ suit, unbalanced catch-all or 18--19 balanced
2NT = Balanced 12--14, may have 5 card suit (responder can ask with 3M)
3NT = 18--19, 4 card suit

Here the 2M rebid is even more similar to Lawrence. The main problem is probably 1M--2x; 2NT if responder want to search for a fit for his suit and opener's suit (wanting to know if opener has 3 card support or 5+ major).

As stated above I would guess that the biggest problem is 1M--2x; 3x. Playing 5 card majors, and requiring 4 cards for opener to raise, responder now knows that opener is unbalanced. Here opener may in some sequences be 4-4. This problem is solved if you open 4 card suits up the line, since then 1M can not hold a lower ranking suit if 4-4. This is not the style we currently play, however.

It may also be possible to play something like 1M--2C as artificial, and use a method like the one helene_t suggested.
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#9 User is offline   whereagles 

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Posted 2014-July-09, 07:12

We open 1NT with 15--17 even if we hold a 5 card major. 1M may be a four card suit if 4432/4333 and 12--14 or 18--19 hcp. 1H may be any 4432, 1S may only have 4D.

I do not really get why it would not be possible to distinguish the 4 card suits from the 5 card suits, except when opener supports responder's suit. If you do something like this:

1M--2x;
2M = 5+ suit catch-all, Lawrence style (but may have be 5332)
2NT = 4 card major, 12--14
3NT = 4 card major, 18--19


Sure, you can use 2M as two-way. Personally I prefer to keep the 2M rebid as the min catch-all, but if two-way works for you, then it's a good way out yes.


Let's say you change it a bit:

1M--2x;
2M = 5+ suit, unbalanced catch-all or 18--19 balanced
2NT = Balanced 12--14, may have 5 card suit (responder can ask with 3M)
3NT = 18--19, 4 card suit


I think the first scheme might be simpler. I feel the auction after a 2NT rebid might get messy. Just a hunch though.


As stated above I would guess that the biggest problem is 1M--2x; 3x.

There are two ways to play 3x: (a) min or (b) med+ hand. I strongly believe (b) is superior. And together with scheme #1 above, it solves your problem:

- 44s can always bid 2/3NT and have 3/4x as reserve bids if pard shows slam interest.
- min 54s bid 2M and follow-up with 3x.
- med+ 54s bid 3x directly, implying a 5-card suit in the major.

How do you like that? I think it's very precise. (Might have a bug though.. I just pulled it off my head :) )

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