lycier Posted May 18, 2013 Report Share Posted May 18, 2013 我们常遇到这样的一类普通牌面: [hv=pc=n&s=sk85ha982d9876c98&d=e&v=b&b=10&a=ppp1np]133|200[/hv] 一个需要灵动的问题来了: 你要pass吗?为什么? Quote Link to comment Share on other sites More sharing options...
ambetter Posted May 18, 2013 Report Share Posted May 18, 2013 三家pass,第四家1NT,大概所有人都是均型牌。pd高限有成局可能,但成功概率大概也不会高,低限2NT也不一定成。pass但依然会很忐忑。 Quote Link to comment Share on other sites More sharing options...
lycier Posted May 19, 2013 Author Report Share Posted May 19, 2013 这牌最大的毛病就是双张小,应该视为"不均型",采取垃圾斯台曼等方式,总体上是占便宜的。比如: [hv=pc=n&s=sk85ha982d9876c98&w=sj6432hq654d5cqj2&n=saq7ht7dkqj43ca75&e=st9hkj3dat2ckt643&d=e&v=b&b=10&a=ppp1np2cp2dppp]399|300[/hv] Quote Link to comment Share on other sites More sharing options...
00master Posted May 19, 2013 Report Share Posted May 19, 2013 1NT也没什么危险。 Quote Link to comment Share on other sites More sharing options...
ar5810 Posted May 20, 2013 Report Share Posted May 20, 2013 采取垃圾斯台曼是个好选择,最差的情形是打4-3配的2S定约,由于短将一方有C的将吃价值,因此2S也可打。 Quote Link to comment Share on other sites More sharing options...
601821297 Posted May 20, 2013 Report Share Posted May 20, 2013 请教一个问题,第四家的1NT很多朋友用弱NT的,也就是说其大牌点是12-14P,我们用吗? Quote Link to comment Share on other sites More sharing options...
wuhuan Posted May 20, 2013 Report Share Posted May 20, 2013 垃圾斯台曼——不错的选择。 Quote Link to comment Share on other sites More sharing options...
zyp163 Posted May 20, 2013 Report Share Posted May 20, 2013 可以使用垃圾STM的体系当然2C不会有问题了。有一些NT开叫的框架不使用垃圾STM,因为2C后,NT开叫人会叫出2N及以上的叫品。 Quote Link to comment Share on other sites More sharing options...
haikuo Posted May 23, 2013 Report Share Posted May 23, 2013 垃圾STM 叫2C 也有点问题吧,如是3325 打的是42配的王。 Quote Link to comment Share on other sites More sharing options...
xuefeier Posted May 23, 2013 Report Share Posted May 23, 2013 S三张带一大牌未尝不可,如果是4441会更好一点 Quote Link to comment Share on other sites More sharing options...
000ffj Posted May 25, 2013 Report Share Posted May 25, 2013 你是否也碰到过这样的困惑:例如同伴开叫1S,你持有黑桃三张,另外三门是541牌型,实力足够进局,你正常用5张套应叫2盖一。同伴回叫一个4张新花,你发现双套配合,满贯处于边缘,但是由于配合太好,你不知道如何描述了,你的53配不是最佳组合,该套中的Q至关重要,同时你还有一个5张套和单张,所有这些信息有办法告诉同伴吗?此时相当于联手有3套牌,为了阐述方便,假定为开叫人5张为A套,四张为B套,应叫人5张是C套,四张也是B套,第四花为D套。我们发现不一定要强求AB两套打通的,假如C套打通,就有5墩,A套53配合打通,也是5墩,B套和D套只要2墩就可以完成小满贯,基于此,控制问叫的过程中不能简单地针对开叫人的AB两套问控制。就是说当C套可以打通的时候,A,B两套哪个能打通,那个做将牌是合理的,44配合的B套不一定是将牌。C套能否打通往往与AB2套能否打通同等重要。 Quote Link to comment Share on other sites More sharing options...
zyp163 Posted May 25, 2013 Report Share Posted May 25, 2013 1.如果是用2H做的2/1应叫,开叫人第二个花色是在第三阶叫出来的,则可以先对S做一次加叫的逼叫,在四阶进行扣叫,然后试探满贯。可以停在53配合的5S或者44配合的5m,也可以6阶叫出44的m.2.如果开叫人第二花色是在二阶叫出来的,则可以先在二阶对S进行加叫,然后在三阶或者四阶对44的花色进行加叫,再使用4NT进行6关键张问叫。 总之,采取2/1后再对开叫人进行加叫,总感觉是个问题。 Quote Link to comment Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.