sungh Posted August 12, 2014 Report Share Posted August 12, 2014 [hv= sn=机器人&s=SJ86543HQ6DA876C4&wn=机器人&w=S7HA8743DK942CJ32&nn=lycier&n=SKT2HK52DT3CAKQ65&en=sungh&e=SAQ9HJT9DQJ5CT987&d=e&v=n&b=50&a=PPP1N(notrump%20opener.%20Could%20have%205M.%20--%202-5%20%21C)P2H!(Jacoby%20transfer%20--%205+%20%21S%3B%2011-%20HCP%3B%2012-%20total%20points)P2S(Transfer%20completed%20to%20S%20--%202-5%20%21C%3B%202-5%20%21)P3S(Invitational%20to%204S%20--%206+%20%21S%3B%209%20total%20points)P4S(2-5%20%21C%3B%202-5%20%21D%3B%202-5%20%21H%3B%202-5%20%21S%3B%2015-17%20HC)PPP&p=CTC4C2CQCAC7H6C3CKC8HQCJDTD5D6DKS7S2SQS3DJDAD2D3D7D4STDQH2H9S4H3D8D9SKSAHJS5H4H5SJH7C5S9S8H8C6C9S6HAHKHT]499|350[/hv]假如2H后,西家的机器人加倍,我的首功肯定是打宕定约的H。于是,选择了其他的花色。机器人不打首攻指示性加倍的话,我在红花色中如何选择出H的首功来。 Quote Link to comment Share on other sites More sharing options...
lycier Posted August 12, 2014 Report Share Posted August 12, 2014 哦?您敢说首攻♥J打宕4♠定约?Ok,就让我们演练一下吧。你首攻♥J,明手垫♥6!此刻您拿不拿?1-♥A拿掉这一顿,♠再交你两墩,我真的看不到您的第四枚赢墩长在哪里?2-不拿忍让一墩,那就没你的♥赢墩的份啦,♠交两墩,♦交一墩,您的第四枚赢墩长在哪里? 别忘了,你和搭档的全部边花都是至少分布三张,--额外将吃无缘没份儿!只有3枚赢墩。 Quote Link to comment Share on other sites More sharing options...
sungh Posted August 12, 2014 Author Report Share Posted August 12, 2014 哦?您敢说首攻♥J打宕4♠定约?Ok,就让我们演练一下吧。你首攻♥J,明手垫♥6!此刻您拿不拿?1-♥A拿掉这一顿,♠再交你两墩,我真的看不到您的第四枚赢墩长在哪里?2-不拿忍让一墩,那就没你的♥赢墩的份啦,♠交两墩,♦交一墩,您的第四枚赢墩长在哪里? 别忘了,你和搭档的全部边花都是至少分布三张,--额外将吃无缘没份儿!只有3枚赢墩。 嗯,我看错了,4♠是铁的。♥A拿后,打给♦A,兑现♥Q,♣X到手上的A,用♥K和♣KQ可以垫光明手的♦。 只是庄家要注意次序,♥K一定要垫走一张♦,不然♣KQ大牌带不完明手的♦。 谢谢老师指点。 Quote Link to comment Share on other sites More sharing options...
dvd Posted August 20, 2014 Report Share Posted August 20, 2014 4s那里铁了?攻h,hA后换c就简单宕了。当然,打出这样的防守需要想象力,通常是防不出来,但至少不是铁牌。 Quote Link to comment Share on other sites More sharing options...
bgm Posted August 20, 2014 Report Share Posted August 20, 2014 The first lead does not matter here. In the actual play line actually quite accurate to let W get in at trick 4 and return ♠ at trick 5. All E need to deduce is that N hand has no more developed tricks to discard S hand loser (from the play that ♥ is discarded as losers and ♣ is well guarded). Therefore all E need to do is to draw all declarer's remaining trump and left dummy with 2 inevitable ♦ losers. Just as simple as that. Quote Link to comment Share on other sites More sharing options...
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