Trinidad Posted December 24, 2010 Report Share Posted December 24, 2010 This is where I think you are wrong. After 1♠ 2NT, both hands are known to be "opening values", one hand has 5+ spades and one hand has 4+ spades. Unless you want to quibble about one vacant space, both hands are undefined. That is why neither hand should have captaincy, and both hands should be able to show a shortage (or long good side suit if that is your preference). I pretty much agree with this reasoning why opener should not be captain, but I would state it stronger. Responder should (generally speaking*) be captain, since opener has the first shot at describing his hand. After responder's first bid both side know about the same; after opener's rebid responder knows a lot more than opener. * The Bergen structure is geared to this: All opener's rebids are describing with the exception of 3♦, which is asking responder for shortness. I myself find that I don't need this exception in practice. Rik Quote Link to comment Share on other sites More sharing options...
hansen50 Posted December 25, 2010 Report Share Posted December 25, 2010 This is an easy one 1♠-2NT3NT-4♦4♠-Pas 3NT = 11-15hp and no shortness4♦ = Cuebid4♠= no stop in ♣ Quote Link to comment Share on other sites More sharing options...
tolvyrj Posted December 26, 2010 Report Share Posted December 26, 2010 We play a bit different kind of gadget called Stenberg and in our sys it would have go like this;1S - 2Nt ( actually limit or better but it dosent matter)3C - 3D ( 3C showed minimum and 3D asks shortness and is FG)3S - 4D ( 3S denies short suits and 4D showed lowest cue and is invitation to slam)4S - pass (4S denies control in clubs and pass is very angry one:)) Quote Link to comment Share on other sites More sharing options...
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