gnasher Posted August 31, 2009 Report Share Posted August 31, 2009 I have an auction that starts with opener showing a balanced hand, and responder bidding 3♦ to show a balanced slam try. After that, I want opener to show all of his four-card and five-card suits. After each of opener's bids, responder relays by bidding the next step. Depending on the sequence, that may happen several times. If the sequence ends with opener bidding 4♦ or 4♥, the next step is a range/control enquiry. If the sequence ends with 4♠ or 4NT, the two sequences show the same shape, with 4♠ being a minimum and 4NT a maximum. I have 20 hands that I want to be able to show: 4333 (4 hands) 5332 (4 hands) 4432 (6 hands) 5422 with a 5-card minor (6 hands) And I have 21 sequences available: 3♥-3♠;3N-4♣;4♦ 3♥-3♠;3N-4♣;4♥ 3♥-3♠;3N-4♣;4♠/NT 3♥-3♠;4♣-4♦;4♥ 3♥-3♠;4♣-4♦;4♠/NT 3♥-3♠;4♦-4♥ 3♥-3♠;4♥-4♠ 3♥-3♠;4♠/NT 3♠-3N;4♣-4♦;4♥ 3♠-3N;4♣-4♦;4♠/NT 3♠-3N;4♦ 3♠-3N;4♥ 3♠-3N;4♠/NT 3N-4♣;4♦ 3N-4♣;4♥ 3N-4♣;4♠/NT 4♣-4♦;4♥ 4♣-4♦;4♠/NT 4♦ 4♥ 4♠/NT Can anyone suggest a good way to map the hands to the sequences? Quote Link to comment Share on other sites More sharing options...
awm Posted August 31, 2009 Report Share Posted August 31, 2009 I'm a bit confused, in that there are 12 4432 patterns and 12 5332 patterns (six if the five card suit must be a minor). Is there some additional constraint here? Quote Link to comment Share on other sites More sharing options...
cherdanno Posted August 31, 2009 Report Share Posted August 31, 2009 Given there are 8 sequences starting with 3♥, it seems natural to have 3♥ show a 4333 or 5332 hand.3♠ could show spades and a second suit.etc... Quote Link to comment Share on other sites More sharing options...
Trinidad Posted August 31, 2009 Report Share Posted August 31, 2009 I'm a bit confused, in that there are 12 4432 patterns and 12 5332 patterns (six if the five card suit must be a minor). Is there some additional constraint here?The "constraint" is that Adam is only interested in the 4+ card suits. He is not interested in the whole distribution. For 44, the combinations are:M'sm'sroundspointsredsblacks I think you get the point that there are 4 suits that can hold the 5 carder in the 5332 pattern. ;) Rik Quote Link to comment Share on other sites More sharing options...
awm Posted August 31, 2009 Report Share Posted August 31, 2009 Here's a fairly symmetric style approach. The general ideas are: 3♥ shows 4-5♣; if 4♣ will not have a five-card diamond suit...... continuations are symmetric with the direct 3NT+ bids3♠ denies a four-card minor...... then 4♣ shows a 4333 hand, 4♦ shows 4432, 4♥ shows 5♥-332 and 4♠/NT 5♠-3323NT shows 4♦ and another 4-card suit4♣ shows diamonds only (3343 or 3352)4♦ shows 5♦ and lowest side suit (club)4♥ shows 5♦ and middle side suit (heart)4♠/4NT shows 5♦ and highest side suit (spade) Stopping early normally shows four-card length whereas zooming on shows five-card length. Via your sequence list: 3♥-3♠;3NT-4♣-4♦ = 2344 (primary clubs, only four, low side suit)3♥-3♠;3NT-4♣-4♥ = 2434 (primary clubs, only four, middle side suit)3♥-3♠;3NT-4♣-4♠/4NT = 4234 (primary clubs, only four, high side suit)3♥-3♠;4♣-4♦; 4♥ = 3334 (primary clubs, one suited, only four)3♥-3♠;4♣-4♦; 4♠/4NT = 3325 (primary clubs, one suited, five clubs)3♥-3♠;4♦ = 2245 (primary clubs, five carder, low side suit)3♥-3♠;4♥ = 2425 (primary clubs, five carder, middle side suit)3♥-3♠;4♠/4NT = 4225 (primary clubs, five carder, high side suit)3♠-3NT;4♣-4♦;4♥ = 3433 (no 4m, 4333, low suit)3♠-3NT;4♣-4♦;4♠/4NT = 4333 (no 4m, 4333, high suit)3♠-3NT;4♦ = 4432 (no 4m, 4432)3♠-3NT; 4♥ = 3532 (no 4m, low five card suit)3♠-3NT; 4♠/4NT = 5332 (no 4m, high five-card suit)3NT-4♣;4♦ = 2344 (primary diams, only four, low side suit)3NT-4♣;4♥ = 2443 (primary diams, only four, middle side suit)3NT-4♣;4♠/4NT = 4243 (primary diams, only four, high side suit)4♣-4♦;4♥ = 3343 (primary diams one suited, only four)4♣-4♦;4♠/4NT = 3352 (primary diams one suited, five diams)4♦ = 2254 (primary diams five carder, low side suit)4♥ = 2452 (primary diams five carder, middle side suit)4♠/4NT = 4252 (primary diams five carder high side suit) Note that there are two ways to show 2344 (depending on whether you want to show primary clubs or primary diamonds). This could be used to show "better minor" perhaps. Quote Link to comment Share on other sites More sharing options...
Trinidad Posted August 31, 2009 Report Share Posted August 31, 2009 How about 3♥-3♠;3N-4♣;4♦ 4(+) clubs; 4432; with diamonds: (23)44 3♥-3♠;3N-4♣;4♥ 4(+) clubs; 4432; with hearts: 2434 or 3424 3♥-3♠;3N-4♣;4♠/NT 4(+) clubs; 4432; with spades: 4(23)4 3♥-3♠;4♣-4♦;4♥ 4(+) clubs; no side suit; (233)5 3♥-3♠;4♣-4♦;4♠/NT 4(+) clubs; no side suit; 3334 3♥-3♠;4♦ 4(+) clubs; (224)5; with diamonds: 2245 3♥-3♠;4♥ 4(+) clubs; (224)5; with hearts: 2425 3♥-3♠;4♠/NT 4(+) clubs; (224)5; with spades: 4225 3♠-3N;4♣-4♦;4♥ no 4(+) m; single suited hearts; 5332 3♠-3N;4♣-4♦;4♠/NT no 4(+) m; single suited hearts; 3433 3♠-3N;4♦ no 4(+) m; both M's: 44(32) 3♠-3N;4♥ no 4(+) m; single suited spades; 5332 3♠-3N;4♠/NT no 4(+) m; single suited spades; 4333 3N-4♣;4♦ 4(+) diamonds; 4432; with clubs: (23)44 3N-4♣;4♥ 4(+) diamonds; 4432; with hearts: 2443 or 3442 3N-4♣;4♠/NT 4(+) diamonds; 4432; with spades: 4243 or 4342 4♣-4♦;4♥ 4(+) diamonds; no side suit; (233)5 4♣-4♦;4♠/NT 4(+) diamonds; no side suit; 3343 4♦-4♥ 4(+) diamonds; (224)5; with clubs: 2254 4♥-4♠ 4(+) diamonds; (224)5; with hearts: 2452 4♠/NT 4(+) diamonds; (224)5; with spades: 4252 The basis is that 3♥ shows clubs, 3♠ denies a minor and the rest shows diamonds. The clubs and diamond showing auctions are symmetric and so are the 4333's and 5332's. There is one shape that is mapped twice (because of the symmetry in the minors): (23)44: 3♥-3♠;3N-4♣;4♦ 4(+) clubs; 4432; with diamonds: (23)44 3N-4♣;4♦ 4(+) diamonds; 4432; with clubs: (23)44 Perhaps you can use that to differentiate between 2344 and 3244 (and find the magic 7♠ in the Moysian ;) )? Rik Quote Link to comment Share on other sites More sharing options...
Trinidad Posted August 31, 2009 Report Share Posted August 31, 2009 Adam beat me. I had an excuse. My internet connection failed. ;) Rik Quote Link to comment Share on other sites More sharing options...
gnasher Posted August 31, 2009 Author Report Share Posted August 31, 2009 Thanks guys ;) I wish I'd asked you earlier, instead of torturing myself for hours trying to think of a logical structure. Quote Link to comment Share on other sites More sharing options...
Echognome Posted August 31, 2009 Report Share Posted August 31, 2009 Here's a more "natural" way to do it. You can probably change it around to be transfer oriented. The only awkward one was 5S332. As we do not know the shape of the off suits, I just put in 3 as the length for all of them. So 4=3=4=2 or 4=2=4=3 will be represented as 4=3=4=3. 3H 4+ hearts->3N Both Majors or Hearts--->4D 4=4=3=3--->4H 4=5=2=2--->4S 3=5=3=3 min--->4N 3=5=3=3 max->4C clubs--->4H 3=4=3=4--->4S 2=4=2=5 min--->4N 2=4=2=5 max->4D 3=4=4=3->4H 2=4=5=2->4S 4=3=3=3 min->4N 4=3=3=3 max3S 4+ Spades, <4 Hearts->4C Clubs--->4H 4=3=3=4--->4S 4=2=2=5 min--->4N 4=2=2=5 max->4D 4=3=4=3->4H 4=2=5=2->4S 4=3=3=3 min->4N 4=3=3=3 max3N Both minors->4D 3=3=4=4->4H 2=2=4=5->4S ->4N 4C Clubs->4H 3=3=3=5->4S 3=3=3=4 min->4N 3=3=3=4 max4D 3=3=5=34H 3=3=4=3 4S 5=3=3=3 min4N 5=3=3=3 max Quote Link to comment Share on other sites More sharing options...
dake50 Posted August 31, 2009 Report Share Posted August 31, 2009 WHY? Hope to avoid duplication of HH opposite HH or HHx w HHx? Ask for those specifically. Magical 4-4 or 4-3 (even once saw 4-2) grand? Quote Link to comment Share on other sites More sharing options...
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