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When Our Stayman Is Doubled

There are several possible methods after an opponent doubles Stayman. I invented this new solution as an alternative method. It allows us to show how strong we are in clubs, allows us to play 2 redoubled, and to right-side suit contracts.

It can also be used after "Garbage Stayman”, allowing us still get out at the 2 level, where 1NT-P-2 can hold a 3-suiter weak hand without clubs, or a weak hand with both majors. 

The main idea is the following:

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N
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1NT
P
2
X
?

  • We redouble when we have light penalty redouble, for example AQ9x, and some other values.
  • We bid our normal answer when we have stopper, but we don’t want to redouble.
  • Otherwise we pass which is a 2-way bid, showing either:
    • strong redouble (5+ strong clubs), 
    • or a lack of club stopper

In case of redouble or bid, answerer has the same system as before, keeping in mind that opener has a control in clubs, and that the redouble can be passed having a game invitational hand or better with 3+, or 3+ tricks and 2+ (good tolerance).

Pass shows either the weakest or the strongest holding in clubs. Opener's pass needs further elaboration:

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1NT
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2
X
P
P
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Responder reopens with a redouble only with tolerance to play 2 redoubled: we can do it with all at least game invitational hands suitable to play 2 redoubled with at least 2.5 tricks and 2 clubs. 

After this, Opener passes with what would normally be a strong redouble, and otherwise Opener's bids are transfer responses to Stayman, all of which show no stopper. 

The reason to play this is that the doubler will be on lead. Therefore, playing any suit contract from Responder's side will better protect his club honors.

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1NT
P
2
X
P
P
XX
P
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  • P: penalty, 5+ good clubs
  • 2: 4+, no club stopper
  • 2: 4+, no club stopper
  • 2: no 4-card major, no club stopper
  • 2NT: Both majors, minimum hand, no club stopper
  • 3: Both majors, maximum hand, no club stopper
  • 3: 5, maximum hand, no club stopper
  • 3: 5, maximum hand, no club stopper
  • 3: 5+, maximum hand, no club stopper 

In this case, we can answer higher than 2 as Opener, since partner’s redouble showed a strong hand with light tolerance (at least invitational values).

 

How shall we reopen 2 doubled as responder?

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1NT
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  • XX: at least invitational, tolerance to play 2 redoubled in case of strong redouble of Opener
  • 2: 3 suiter weak hand, followed by pass or correction from Opener (it shows 4+♦, limit hand, (no penalty tolerance, when we use limit+ Stayman)
  • 2: both majors (4+ , 4+) weak hand, pass or correction from Opener (it shows 4+, exactly game invitational hand hand, no penalty tolerance, when we use limit+ Stayman)
  • 2: 4+, exactly limit hand (8-bad 10), no penalty tolerance, not forcing, both cases
  • 2NT: denies 4+, limit hand (8-bad 10), not forcing! We can have 4. If Opener has 4 and a maximum hand, he will bid 3 to ask. Then Responder will bid 3 with 4 hearts, thereby allowing to be played from Responder's side. 
  • 3: game forcing relay without tolerance for playing 2 redoubled.  The continuations are:

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1NT
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2
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P
P
3
P
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  • 3: 4+, no club stopper (either 4 or weak SI hand, 5, than we can bid 4 again)
  • 3: 4+, no club stopper (either 4 or weak 5, than we can bid 4 again)
  • 3: no major, no club stopper
  • 3NT: To play: we had a penalty redouble for , and are willing to play 3NT even if partner is void in
  • 4: Both majors, no club stopper
  • 4: 5, no club stopper, maximum hand
  • 4: 5, no club stopper, maximum hand
  • 4: 5+, no club stopper, maximum hand 

In this case, we can respond above 3NT as Opener, because Responder’s 3 showed a strong hand with game-forcing values and no tolerance to sit for a redouble.

(VS stands for Vikor Solution, which is also used in the name of several of my other new methods available at bridgemagic.net)

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