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Detail specifications 61BA - Level 2 ★

2♣ Stayman

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The 2♣ Stayman is a conventionnal bid.
In response to a 1NT opening, a 2♣ bid is a Stayman, looking for an eight-card fit in a major suit.It promises:

  • at least 8 HCP
  • and at least 4 cards in one of the majors. It is also used with both majors, 4-4 or 5-4.
Avoid using this convention when you hold a 4333 hand with scattered honors. These hands are too balanced to work well in a trump game. You'd rather play a NT game (9 tricks instead of 10)
2♣ Stayman is forcing for one round, indeed !
SWNE
1NT
Pass
3NT
K
5
4
3
Q
4
3
Q
10
3
Q
J
5
Problem E2062 3NT Even though you've got 4 ♠ cards, don't use the Stayman 2♣ conventional bid. You have a balanced hand, no ruffing added value. Your pair's points are 25-27 HCP. Try a NT game.
 
SWNE
1NT
Pass
2
4
3
Q
10
9
3
A
Q
J
9
4
3
4
Problem E2068 2♣ First, try to find a ♥ fit. With at least 8 HCP, start bidding 2♣ (Stayman) to look for a fit in one of the majors. That's a priority !
 
 
Bidding development : the opener's rebid after his partner's 2♣.
  • 2♦ when he has no major 4th
  • 2♥ with 4 ♥ cards (or 2♠ with 4 ♠ cards)
  • 2♥ with both 4 ♠ and 4 ♥ cards. The opener has the option to bid ♠ later if his partner's 4 carder happens to be ♠. Many players bid 2NT to indicate both majors, but this is giving useful (but unnecessary) information to the opponents.

The responder's 2nd bid after the opener has named a major suit.
  • With 8 HCP the responder proposes game in NT or in the major suit (when there is a fit, of course). The opener will pass with a minimum hand (15 HCP) or go to game if maximum.
  • With 10 HCP or more the responder goes directly to game either in NT or in the fitted major. The pair's combined points are enough for game (minimum 15 + 10 = 25)

A
10
9
K
Q
10
9
7
6
A
Q
10
8
SWNE
1NT
Pass
2
Pass
2
Pass
3
K
5
A
J
8
4
5
4
3
7
6
5
3
Problem E4899 With a minimum Stayman (8 HCP), propose game and let your partner decide.
 
A
10
9
K
Q
10
9
7
6
A
Q
10
8
SWNE
1NT
Pass
2
Pass
2
Pass
4
Pass
K
Q
8
A
J
8
4
4
2
J
6
5
3
Problem E5006 With a support and at least 10 HCP (here, you have 11 HCP), go to game.
 
 
Bidding development : when the opener bids 2♥ with 4 cards in both majors
   If the responder has 8 HCP but no ♥ support, he bids 2NT, proposing a NT game. Then, the opener bids:
  • 3♠ if minimum
  • 4♠ if maximum.

   If the responder has 9 HCP or more, but no ♥ support, he bids 3NT. All the opener has to do is shift to 4♠.
With this system, when there is a ♥ fit, the opponents have no information about the opener's hand. With a 2NT response to the 2♣ Stayman, the opponents know exactly what the opener's distribution is... 4332
Q
10
9
4
A
K
8
6
A
Q
5
7
4
SWNE
1NT
Pass
2
Pass
2
Pass
4
K
J
10
7
4
3
4
3
2
A
K
6
3
Problem E8008 With 10 HCP and a ♥ support, go to game. The opener will be the declarer and his ♣ tenace will be protected. With both majors, don't use the 2NT bid (it's both unnecessary and dangerous!)
 
 
Response to the opener's 2♦ rebid, with both majors 5-4.
   With less than 10 HCP, the responder bids :
  • 2♥ with 5 ♥ cards
  • 2♠ with 5 ♠ cards.
The opener will go to game when he is maximum.
   With 10 HCP or more, the responder uses the Smolen conventional method. He bids his four-card major at level 3 :
  • 3♥ with 4 ♥ cards and 5 ♠ cards
  • 3♠ with 4 ♠ cards and 5 ♥ cards

The opener can then declare 4♠ or 4♥ with support, 3NT without.
Q
5
K
6
4
Q
J
5
4
A
K
9
7
SWNE
1NT
Pass
2
Pass
2
Pass
2
A
J
10
6
4
Q
J
7
5
7
2
6
4
Problem E8036 With 5 ♠ cards and 8 HCP, the responder bids his major 5th. The opener will decide accordingly, and maybe pass.
 
 
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