1D-1S; 2C auctions

[metobillc]

Playing Precision with 2♣ showing 6+♣, 1♦ showing 2+♦, what's the best agreement on the 1♦-1♠; 2♣ auction? In the past I've played it as guaranteeing at least 9 cards in the minors, but there are obvious problem hands with this approach, even if one freely raises 1♠ on 3 cards, or freely rebids 2♦ with 5 cards regardless of suit quality. 1=4=3=5, 1=4=5=3, 2=4=5=2, 2=4=2=5 are all still problematic. If you only guarantee 3♦ and 4+♣, how do you handle the followups?

 

Bill Campbell

[SteveMoe]

I like 2♣ to show 9+ cards either way.

With the hands you raise we tend to rebid 1N unless shape and texture define 2♦ as clearly better (5 cards, weak non-♠ doubleton). Always open 2=4=5=2, 2=4=2=5 1N (14-16)

With 5♣ cards I might get to clarify later (Play all 2N in competitive auctions to ask opener to clarify minor length).

[yunling]

Rebid some off-shape 1NT(even with 1-4-4-4/1-4-3-5) to clean up your 2C rebid.

[awm]

Sam and my agreement is that 2C shows 5+D/4+C unless too strong to rebid 1nt (we bid 1nt with 1345/1444/1453/1435 if in range). This means responder will usually preference diamonds and we can bid on with the off-shape too strong for 1nt hands.

[whereagles]

It depends on the rest of your structure, of course. The more you "dump" into the classical 2m openers, the less overloaded 1♦ is.

 

From what I see the present tendency is to 6-card clubs 2♣, so the 1♦ will sometimes be made on 3D-5C, 44, etc.

[Free]

I play it as 8+ cards in the minors (like Meckwell). 2452 and 2425 are considered balanced (= irrelevant), so we only have 8 cards in case of 1444/1435/1453. 2♥ is "4th suit forcing" after which 2♠ shows exactly 8 cards in the minors (so 4♥) or ♠ support. Then 2NT asks. Space enough ;)

[rhm]

Playing Precision with 2♣ showing 6+♣, 1♦ showing 2+♦, what's the best agreement on the 1♦-1♠; 2♣ auction? In the past I've played it as guaranteeing at least 9 cards in the minors, but there are obvious problem hands with this approach, even if one freely raises 1♠ on 3 cards, or freely rebids 2♦ with 5 cards regardless of suit quality. 1=4=3=5, 1=4=5=3, 2=4=5=2, 2=4=2=5 are all still problematic. If you only guarantee 3♦ and 4+♣, how do you handle the followups?

 

Bill Campbell

I have looked into this sequence and have developed my own ideas based on a notrump rebid, which shows 11-13 HCP.

 

Principles:

 

1♦-1♠-2m denies 4 cards in hearts

1♦-1♠-2♣ shows 5+ cards in clubs and 4+ cards in diamonds and this is what partner will assume (see one rare exception below, where I have to fudge)

1♦-1♠-2♦ shows 6+ cards in diamonds and less than 5 cards in clubs and this is what partner will assume (see one rare exception below, where I have to fudge)

 

1♦-1♠-1NT is frequently bid with a singleton spade (but in no other suit), e.g. 1♠=4♥=4♦=4♣, 1♠=3♥=5♦=4♣ 1♠=4♥=3♦=5♣ when in range (assumed to be 11-13).

With 14 HCP and the above distribution decide whether to treat the hand as 13 HCP (1NT rebid) or 15 HCP (see below).

2♠=2♥=5♦=4♣ will either open or rebid 1NT, depending on HCP range. With 2♠=2♥=4♦=5♣ we can choose between 1NT and a 2♣ rebid based on the heart holding.

 

1♦-1♠-2♥ (a reverse heavily underused in most precision variants) shows a minimum of 4 hearts, extra distribution (unsuitable for a notrump rebid or a spade raise, so at most 2 cards in spades), but does not require extra strength

 

The following hand-types rebid 2♥:

 

1) three suiter: 0♠=4♥=4♦=5♣, 0♠=4♥=5♦=4♣ (will bid 3♣ over 2NT (relay)

2) 6+ diamonds 4+ hearts (will bid 3♦ over 2NT or 3♥ with five hearts and six diamonds)

3) 1♠=4♥=4♦=4♣ and 1♠=4♥=3♦=5♣ 15 HCP or worth an upgrade from 14 HCP. (will bid 3NT over 2NT. These hands will not have an honor in spades or would have oepned 1NT(14-16)

 

I will bent my rules above with 15 HCP and precisely 1♠=3♥=5♦=4♣

 

There are 4 options to choose from based on actual hand holding (in order of likelihood):

 

1) if clubs are good treat clubs as a 5 card suit and rebid 2♣

2) if diamonds are good, treat diamonds as a six card suit and rebid 2♦

3) Open 1NT(14-16) with an honor in spades

4) upgrade to 1♣

 

For me this schema accomplishes two objectives

 

1) It avoids minor suit length ambiguities and if 1NT is bypassed it does show an unbalanced hand.

2) It does not miss 4-4 fits in hearts and responder need not try for it after a 2m rebid.

 

Rainer Herrmann

[Shugart23]

Thanks RHM

 

This thread is very timely for me and my wife as we are looking at Fourth Suit Forcing where bidding has gone 1D - (1H or 1S) -2C. For us, Opener has 9+ cards in the Minors and is almost certainly NOT 2-2 in the Majors . With 2-2 in the Majors, we are apt to open 1NT or rebid 1NT depending on point count.

 

SO given opener's distribution is something like 1-3-4-5 or 3-1-4-5, I am thinking there might be subtle differences using 4th suit forcing vs. when the opening is not 1D. At least this is the theoretical track I am thinking about currently.

 

While on the topic of 3145 distributions, has anyone played with mini-Roman being expanded to include 3145, 4441, and 5440 hands (shortness anywhere) with success or is this just way to inclusive ?

[metobillc]

Thanks everyone for your input. I've decided to go with the following:

 

1D-1S; 2C shows 9+ cards in the minors (4=5 or 5=4 or better) with a minimum range hand (11-13), 8+ cards with a max (so 1=4=4=4, 1=4=3=5, and 1=4=5=3 hands rebid 1N with a min, 2C with a max).

 

As for 3-card spade raises, maximum 3=1=4=5 and 3=1=5=4 hands, and all 3=4=5=1 hands, make the 3-card raise; otherwise rebid 2C.

 

Judgment is used in the case of singleton honors, with unbalanced hands sometimes opening or rebidding 1N (along with 2=2=4=5 and 2=2=5=4 hands).