The control bids by classical bidding and by majeure précisée

The explicit and explicit fits, and the controls

A fit is explicit when a color has been explicitly bidded by a player then supported explicitly by his partner. Examples :

1	1
3 (with 4)	4 (control )

The opener may bid his first control 4 ou 4 if he has a slam hope.

1	1
1 (with 4 4)	3

A fit is implicit when a color has been explicitly bidded by a player, then supported implicitly by his partner. Some examples.

1NT	2 (Stayman)
2NT (2 majors)	3 (Transfer ) with 4
3 (completing)	4 (control ) Slam hope

or :

1NT	2 (Stayman)
2 (with 4)	3 (with 4) Slam hope
4 (control )

or more :

1NT	2 (Transfer )
2 (completing)	3 (natural)
3 (fit )	4 (control ) Slam hope

and

1	1
4 (splinter) *

* With 4, singleton , forcing game

By all theese events of implicit fit, the control bid is used both as fitting bid and as slame hope. And whatever the fit is explicit or implicit, the controls operate perfectly. It would be possible to show many other examples, and generally the control bids overeach the level of 3NT.

Why do theese controls perfectly operate ? For a simple and single reason : The fitted color to play is always designated before the explicit or implicit control.

But if a control is bidded before the color to play has been designated, that brings many problems, even if this control shows clearly the frame of a hand

The 3 critical cases of controls by classical bidding

Opening 1NT when the responder has 5 4

1NT	2 (Stayman)
2 (no major 4°)	3 (5 4)
? (with fit )

If the opener has no fit no problem, he bids 3NT. But if he has fit he must bid 4 because he does not know if the responder has just 5HCP or at the contrary he has 13HCP and a slam hope. The controls are not possible.

Furthermore, if the opened bidded a control 4 or 4, the control process would not or badly operate, because the color is not yet designated.

Opening 2NT when the responder has 5 4

2NT	3 (Stayman)
2 (no maj. 4°)	3 (5 4)
? (with fit )

The problem is exactly the same.

Opening 2NT with two-suiter

2NT	3 (Stayman)
3NT (2 maj. 4°)	?

There is a fit in a least one major. After 3NT from opener The responder bids his major by transfer at level 4. The opener must complete to the major of responder, because this last may have only 5HCP. The controls are not possible.

The 3 same cases, but by playing majeure précisée

For our visitors who have not yet read the "majeure précisée", main subject of our site, the openings 1 and 2 are exactly and respectively equivalent to classical openings 1NT and 2NT, that is to say balanced and with 15-17HCP and with 20-21HCP.

Opening 1NT when the responder has 5 4

1	2 (with 5 4, 5 4, or 4 4) *
2(no major 4°)
2 (with 4) *
2 (with 4) *
2NT (with 4 4) **

* After the answer 2 the responder closes to 4 and after answer 2 to 4, for game. With slam hope he bids his first control at level 4.

** The responder bids his 4cards-major by transfer, 3 for or 3 for . After completing by opener in the responder's major the responder closes to game in the designated major, but he bids his first control if he has a slam hope.

The answer 2 from opener :

1	2 (with 4 4, 5 4, or 4 4)
2(no major 4°)	2 (with 4 4) and 8HCP *
	2 (with 4 4) at least 9HCP *

	3 (with 5 4 ) at least 9HCP **
	3 (with 5 4 ) at least 9HCP **

* The opener corrects to 2NT or 3NT according to his strenght. If corrected to 2NT, the responder closes to 3NT with at least 10HCP.
(If the responder has a five-cards major with 8HCP that he will show via a Stayman bidding, he has at his disposal 2 other disrinct auctions. Read if necessary Stayman bidding with five-cards major. about this matter.)

** If fitted the opener cmpletes in five-cards major of responder. If not he completes to 3NT.
Theese economical biddings 3 et 3 are quite distinct from mixed 4M 5m two-suited which are included in the indirect Stayman 2.

Opening 2NT when the responder has 5 4

Theese same mechanisms cannot be used due to the higher bidding level. But the economical nature of the opening 2 enables biddinngs whith same efficiency.

The responder has 5 4 or 4 4 forcing game

2	3 (bicolore 5 4 ou 4 4)
3 (avec 3 ou 4) *
3 (avec 4) **
3SA (avec 2 et 3)

* The response 3 denies 4, because the response 3 has always priority over response 3. With 4 the opener woukd have bidded 4.
Therefore, after 3 the responder closes to 4 for game, but bids his first control if he has a slam hope.

If the responder has 4 4, he bids 3 what requires completing to 3NT by opener.

** The responder closes to 4 for game. With slam hope he bids his first control.

The responder has 5 4 forcing game

2	3 (bicolore 5 4)
3 (avec 4) *
3 (avec 3 ou 4) **
3SA (avec 2 et 3)

* The responder closes to 4 for game. With slam hope he bids his first control. ** The responder closes to 4 for game. With slam hope he bids his first control.

NB : After the respose 3 ou 3, if the opener notes 2 major fits he bids the major he has rather to play.

When there is a fit in a major, its color is always designated by the opener at level 3. That enables the respondre to bid his first control if he has a slam hope.

If the responder has 4 4 he must not bid 3 because he could receive the response 3 from opener if this last has 3 2 ou 3. This bid 3 would force him to bid and play 3NT. At the contrary if he bids 3the contact of 3NT will be played by opener.

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