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Stayman All the Time

A bidding question for you. Thanks in advance for your time.

In a 2/1 system, playing a 15-17 No-Trump, with 4-Suit transfers. (2 transfers to , 2NT transfers to ).Since we have no natural 2NT invite, we go through Stayman.

1NT - 2 - 2 - 3 (a)

1NT - 2 - 2 - 2NT (b)

1NT - 2 - 2 - 3 (c)

1NT - 2 - 2 - 2 (d)

1NT - 2 - 2 - 2NT (e)

1NT - 2 - 2 - 2NT (f)

(a) 4-4  fit with 8-9 HCP

(b) 8-9 HCP and no 4 card major. (Alert!)

(c) 3. 4-4  fit with 8-9 HCP

(d) Opener may have two 4 card majors so we bid 2 to cater to this possibility.

(e) 8-9 HCP. May not have a 4 card major. (Alert!)

(f) 8-9 HCP. May not have a 4 card major. (Alert!)

All understood. I am in agreement with all of the above.The sequence below I disagreed with a teacher about. 

1NT - 2 - 2 - 3NT

He states the above auction denies a 4 card major in responder’s hand. I disagreed, stating that the above auction shows a 4 card  suit and game-force values. If opener has 4  as well, he will “correct” to 4.

His method makes no sense to me. Why bother with Stayman? Why give the opponents the opportunity to double what amounts to a “fake Stayman” as lead-directing. In my opinion, this auction should retain the same meaning as a standard auction when not employing 4 suit transfers.

The teacher went on to say “I find it convenient to treat 2 over 2 as forcing 1 round so opener will raise to 3 with 4-4 in the majors or bid 2N without 4 .” Convenient is an underbid (using his system) since, with game-force values, and no 4 card major, all auctions go through Stayman. 2, in his system, MUST be forcing. Additionally, why force to the 3 level when 2 may be the last safe spot? Granted a 4-4 major fit with 23+ HCP should be safe, but why push it unnecessarily? 

Would you be so kind as to render a verdict in our little “disagreement.”

What is standard?

Thanks!

 

 

 

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