madongjun Posted October 7, 2012 Report Share Posted October 7, 2012 [hv=pc=n&s=s85h742dak98caq84&d=s&v=0&b=11&a=1dp2dp]133|200[/hv]体系：哈代2/1.由于考虑可能出现的开叫方的逆叫，选择开叫了1♦.在同伴低花反加叫后，开叫方的叫品是不是只有3♦了？因为，这时叫2NT表示低限（12－13HCP）并承诺双高花有止；叫3NT保证18－19HCP的均型牌；叫3阶新花是斯普林特，至少合格的14HCP以上的牌力；其他再叫都是显示从低到高的止张。当时，我没有理解3♣也是斯普林特，而是误会成从低到高的止张了。结果5♦下二。拿出来与大家分享，以减少再犯同样的错误。 Quote Link to comment Share on other sites More sharing options...
601821297 Posted October 8, 2012 Report Share Posted October 8, 2012 谢谢楼主的分享 :) Quote Link to comment Share on other sites More sharing options...
sleekrain Posted October 11, 2012 Report Share Posted October 11, 2012 这牌属于均型，D和C点力接近，按原则应该开叫1C，至于D套埋没也不可惜，看看同伴实力。 Quote Link to comment Share on other sites More sharing options...
lycier Posted October 11, 2012 Report Share Posted October 11, 2012 我是这样着想的: 1♦----2♦----3♣=非跳叫性splinter加叫,♣单缺。 1♦----2♦----3阶新花=splinter加叫 ，额外牌力。 1♦----2♦----3♦=双高无止,可以考虑pass。 1♦----2♦----2nt=12-14p,双高有止,不否认4张高花。或者18-19p,其额外的强劲牌力将在后续叫牌进程中等待体现。 1♦----2♦----3nt=15-17p,门门有止。 Quote Link to comment Share on other sites More sharing options...
gamebridge Posted October 11, 2012 Report Share Posted October 11, 2012 本人多半考虑开叫1C,13大牌点均型,标准的 1m 1X,1N 的进程。当然在当时的情况下,只能叫出3D来,毕竟 3Y已经定义成是SPL。 Quote Link to comment Share on other sites More sharing options...
hejiang Posted October 18, 2012 Report Share Posted October 18, 2012 学习了 Quote Link to comment Share on other sites More sharing options...
wuhuan Posted October 18, 2012 Report Share Posted October 18, 2012 学以致用——好！ Quote Link to comment Share on other sites More sharing options...
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