eagles123 Posted September 30, 2013 Report Share Posted September 30, 2013 [hv=pc=n&w=sakt9873hakd3cakq&e=shjt32d9875cj8754&d=s&v=b&b=7&a=p]266|200[/hv] thanks, Eagles Quote Link to comment Share on other sites More sharing options...
ArtK78 Posted September 30, 2013 Report Share Posted September 30, 2013 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). Quote Link to comment Share on other sites More sharing options...
wyman Posted September 30, 2013 Report Share Posted September 30, 2013 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 4SIt'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. Quote Link to comment Share on other sites More sharing options...
Cyberyeti Posted September 30, 2013 Report Share Posted September 30, 2013 2♣-2♦(old fashioned neg)2♠(we reserve 3♠ for solid suits)-3♣(Herbert 2nd neg not to wrongside NT)3♠-3N4♠ I'm not going to find 6♣. Quote Link to comment Share on other sites More sharing options...
billw55 Posted September 30, 2013 Report Share Posted September 30, 2013 (edited) Agree 4♠ is the likely final destination. In a strong club system, can east show both a bust and 5 clubs? Edited September 30, 2013 by billw55 Quote Link to comment Share on other sites More sharing options...
Zelandakh Posted September 30, 2013 Report Share Posted September 30, 2013 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... Quote Link to comment Share on other sites More sharing options...
aguahombre Posted September 30, 2013 Report Share Posted September 30, 2013 2♣-2♦(old fashioned neg)2♠(we reserve 3♠ for solid suits)-3♣(Herbert 2nd neg not to wrongside NT)3♠-3N4♠ 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. Quote Link to comment Share on other sites More sharing options...
Zelandakh Posted September 30, 2013 Report Share Posted September 30, 2013 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. Quote Link to comment Share on other sites More sharing options...
Cyberyeti Posted September 30, 2013 Report Share Posted September 30, 2013 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. Quote Link to comment Share on other sites More sharing options...
aguahombre Posted September 30, 2013 Report Share Posted September 30, 2013 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. Quote Link to comment Share on other sites More sharing options...
ggwhiz Posted September 30, 2013 Report Share Posted September 30, 2013 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. Quote Link to comment Share on other sites More sharing options...
lowerline Posted October 1, 2013 Report Share Posted October 1, 2013 2♣-2♦ (1)2♠-3♣ (2)4♠-p (1) negative(2) 2nd negative (i.e. no A/K, no 3crd support) Steven Quote Link to comment Share on other sites More sharing options...
gwnn Posted October 1, 2013 Report Share Posted October 1, 2013 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. Quote Link to comment Share on other sites More sharing options...
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