shevek Posted February 19, 2015 Report Share Posted February 19, 2015 Playing Standard, 15-17 NT & better minor,I believe the chance of a 1♣ opener having 4+ clubs is about 85%. Right? What is the figure when responder has four clubs?So the auction goes:1♣ - (pass or bid) - ? Responder has four trumps, no major to show and is thinking of raising to 2♣ or 3♣.What is the chance of an 8+ card fit? Quote Link to comment Share on other sites More sharing options...
PhilKing Posted February 19, 2015 Report Share Posted February 19, 2015 Obviously the fact that we have 4 clubs increases the chance that partner dos not hold length, but there is another huge factor at work - RHO's pass over 1♣. Particularly if we are very weak, then RHO's pass over 1♣ becomes the equivalent of the dog that didn't bark. If partner had long clubs, the chances of RHO not being able to overcall are pretty small, so the profile of hands becomes skewed towards partner holding a balanced hand with relatively short clubs. Quote Link to comment Share on other sites More sharing options...
Bbradley62 Posted February 19, 2015 Report Share Posted February 19, 2015 The stats depend on your partnership agreements as to what you open when you are 3-3 or 4-4 in the minors. Quote Link to comment Share on other sites More sharing options...
1eyedjack Posted February 19, 2015 Report Share Posted February 19, 2015 Not to be dismissed also is the possibility of agreeing that a 1N response shows 4 card Club support. That consults opener on whether or not to retreat, and removes broadly the same amount of bidding space. Quote Link to comment Share on other sites More sharing options...
Bbradley62 Posted February 19, 2015 Report Share Posted February 19, 2015 Not to be dismissed also is the possibility of agreeing that a 1N response shows 4 card Club support. That consults opener on whether or not to retreat, and removes broadly the same amount of bidding space.This is, of course, a very common old-fashioned understanding. If responder shows four of any of ♦♥♠, then responding NT must be showing four ♣. If it is your plan to raise ♣ with only four of them, then it should be your partnership understanding that responding NT shows four ♦; otherwise, you'd never bid NT over a 1♣ opening. 1 Quote Link to comment Share on other sites More sharing options...
fromageGB Posted February 21, 2015 Report Share Posted February 21, 2015 Raising clubs on four implies there is a lower bid that is put to some artificial use, as Bbradley points out. It could be that 1♦ shows 5, or 1♦ shows a 4 card major while 1M shows 5, but without anything unusual, I would have thought raising clubs shows 5. Quote Link to comment Share on other sites More sharing options...
suokko Posted February 21, 2015 Report Share Posted February 21, 2015 About 71% for 8+ cards fit in clubs. But I would only raise with 4 card support after overcall or t/o double. But the raise with 4 feels a routine action in a competitive auction. northbal = shape(north, any 4432 + any 4333 + any 5332 - 5xxx - x5xx) c1_unbal = hcp(north) > 10 && hcp(north) < 22 && !northbal && ((clubs(north) > diamonds(north) && clubs(north) > hearts(north) && clubs(north) > spades(north)) || shape(north, 4414)) c1_bal = ((hcp(north) > 11 && hcp(north) < 15) || (hcp(north) > 17 && hcp(north) < 20)) && northbal && diamonds(north) < 4 c1 = c1_unbal || c1_bal condition c1 && hcp(south) >= 5 && shape(south, 2344 + 3244 + 3334) action average "8+ club fit" clubs(north) >= 4, 8+ club fit: 0.708293 Generated 10000000 hands Produced 47870 hands Initial random seed 1424543431 Time needed 5.484 sec Quote Link to comment Share on other sites More sharing options...
Bbradley62 Posted February 21, 2015 Report Share Posted February 21, 2015 About 71% for 8+ cards fit in clubs. But I would only raise with 4 card support after overcall or t/o double. But the raise with 4 feels a routine action in a competitive auction. northbal = shape(north, any 4432 + any 4333 + any 5332 - 5xxx - x5xx) c1_unbal = hcp(north) > 10 && hcp(north) < 22 && !northbal && ((clubs(north) > diamonds(north) && clubs(north) > hearts(north) && clubs(north) > spades(north)) || shape(north, 4414)) c1_bal = ((hcp(north) > 11 && hcp(north) < 15) || (hcp(north) > 17 && hcp(north) < 20)) && northbal && diamonds(north) < 4 c1 = c1_unbal || c1_bal condition c1 && hcp(south) >= 5 && shape(south, 2344 + 3244 + 3334) action average "8+ club fit" clubs(north) >= 4, 8+ club fit: 0.708293 Generated 10000000 hands Produced 47870 hands Initial random seed 1424543431 Time needed 5.484 sec I'm not really sure how to read the code, but are you stipulating that Opener would open 1♣ with (32)44 shape (with appropriate HCP)? Quote Link to comment Share on other sites More sharing options...
TylerE Posted February 21, 2015 Report Share Posted February 21, 2015 I'm not really sure how to read the code, but are you stipulating that Opener would open 1♣ with (32)44 shape (with appropriate HCP)? No. diamonds(north) < 4 Quote Link to comment Share on other sites More sharing options...
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