[eagles123]

---

thanks,

Eagles

[ArtK78]

Good luck.

Hands such as these - where one member of the partnership has a huge hand and the other member of the partnership has virtually nothing - are extremely difficult to bid. A relay system has a chance to find out the distribution, but how would it distinguish between Jxxxx of clubs and, say, JT9xx of clubs, which would be a crucial distinction for the club slam.

I suspect that most partnerships would arrive in 4♠. 6♣ is not bad, but it probably needs 3-2 clubs (definitely needs 3-2 clubs on diamond lead and continuation) and some additional handling to make. 6♣ would be much better if the long club suit hand had JTxxx of clubs. You could then handle 4-1 clubs if spades broke 3-3 or there was some other extremely lucky lie of the spade suit (in some situations in combination with the lie of the club suit).

[wyman]

Basically this is going to depend on e-w system and whether w rebids 3S setting trumps. A likely auction is:

2C 2H* (bust, if you play that)
3S 4S (3S sets trump, 4S is the worst possible hand even in the context of having a bust -- east had the option of qbidding or bidding 3n to just say "can't cue but I have a max bust")
Pass

After
2C 2D (negative or waiting)
3S 4S
It's possible west makes another try. After all, east can't know how good xxx xxx xxxx xxx is here. And we might go down in 5S.

[Cyberyeti]

2♣-2♦(old fashioned neg)
2♠(we reserve 3♠ for solid suits)-3♣(Herbert 2nd neg not to wrongside NT)
3♠-3N
4♠

I'm not going to find 6♣.

[billw55]

Agree 4♠ is the likely final destination.

In a strong club system, can east show both a bust and 5 clubs?

This post has been edited by billw55: 2013-September-30, 12:02

[Zelandakh]

billw55, on 2013-September-30, 10:34, said:

In a strong club system, can west show both a bust and 5 clubs?

Yes, my auction probably starts 1♣ - 1♦; 1♠ - 2♣; 2♠ - 3♣, where West has shown an unbalanced game force with spades and East has shown a bust with clubs. Even then, it seems likely the bidding will stop in 4♠ or 5♣, for example, clubs can be set by: 3♠ - 3NT; 4♣ - 4♦ (bad in context); but now West would be taking a big leap to commit past game and East has no reason to make a further try. Perhaps IMPrecision can do something clever...

[aguahombre]

Cyberyeti, on 2013-September-30, 10:24, said:

2♣-2♦(old fashioned neg)
2♠(we reserve 3♠ for solid suits)-3♣(Herbert 2nd neg not to wrongside NT)
3♠-3N
4♠

I'm not going to find 6♣.

This looks good on the surface; we find that responder has nothing, and also fewer than 2 Spades. However, I wouldn't be sure 3S is forcing upon a hand with nothing and fewer than 2 Spades. It isn't for us.

[Zelandakh]

aguahombre, on 2013-September-30, 10:48, said:

This looks good on the surface; we find that responder has nothing, and also fewer than 2 Spades. However, I wouldn't be sure 3S is forcing upon a hand with nothing and fewer than 2 Spades. It isn't for us.

I suspect Cyber uses the traditional approach where 2♠ establishes a game force. That is probably still the default if not otherwise stated.

[Cyberyeti]

Zelandakh, on 2013-September-30, 10:51, said:

I suspect Cyber uses the traditional approach where 2♠ establishes a game force. That is probably still the default if not otherwise stated.

Indeed, anything other than 2♣-2♦-2N is GF (OK, since we play that 2N as 22-23 you can probably pass 2♣-2♦-2♥-2♠-2N 24-25 with a 3334 yarborough), but any suit bid is GF.

[aguahombre]

Zelandakh, on 2013-September-30, 10:51, said:

I suspect Cyber uses the traditional approach where 2♠ establishes a game force. That is probably still the default if not otherwise stated.

O.K., I thought 9 tricks for a major 1-suiter was the default minimum. We use 9M or 10m, and were not aware this was unusual.

[ggwhiz]

With us it starts 2♣ - 2♥ (bust) - 2♠ )(only a 1 round force_ - 3♣

Since 2♠ was only a 1 round force west has to pick between bidding 4♠ or 6♣ and since that has no play opposite 5 small with a couple of diamond leads we're gambling the slam or playing 4♠. Can't win em all.

[lowerline]

2♣-2♦ (1)
2♠-3♣ (2)
4♠-p

(1) negative
(2) 2nd negative (i.e. no A/K, no 3crd support)

Steven

[gwnn]

aguahombre, on 2013-September-30, 12:33, said:

O.K., I thought 9 tricks for a major 1-suiter was the default minimum. We use 9M or 10m, and were not aware this was unusual.

I think neither approach is unusual. Keeping it simple as Cyberyeti (i.e. only 2C-2D; 2NT is non-GF, or only 2C-2H; 2NT is non-GF) gets my vote, though. Jumping around with good hands depresses me.