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困难的一副牌:4441牌型21hcp


lycier

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[hv=sn=|lin=pn|S,W,N,E|st%7C%7Cmd%7C1S79KH4569TD37C568%2CS5H37JKD49KACJQKA%2CS236QH28D568TJC7T%2C%7Crh%7C%7Cah%7CBoard%2019%7Csv%7Ce%7Cmb%7Cp%7Cmb%7C1C%7Cmb%7Cp%7Cmb%7C1S%7Cmb%7Cp%7Cmb%7C2D%7Cmb%7Cp%7Cmb%7C3N%7Cmb%7Cp%7Cmb%7Cp%7Cmb%7Cp%7Cpc%7CH5%7Cpc%7CH7%7Cpc%7CH8%7Cpc%7CHQ%7Cpc%7CC2%7Cpc%7CC5%7Cpc%7CCJ%7Cpc%7CC7%7Cpc%7CCQ%7Cpc%7CCT%7Cpc%7CC3%7Cpc%7CC6%7Cpc%7CCK%7Cpc%7CH2%7Cpc%7CC4%7Cpc%7CC8%7Cpc%7CCA%7Cpc%7CS2%7Cpc%7CC9%7Cpc%7CH4%7Cpc%7CH3%7Cpc%7CD5%7Cpc%7CHA%7Cpc%7CH6%7Cpc%7CDQ%7Cpc%7CD3%7Cpc%7CD4%7Cpc%7CD6%7Cpc%7CD2%7Cpc%7CD7%7Cpc%7CDK%7Cpc%7CD8%7Cpc%7CHJ%7Cpc%7CS3%7Cpc%7CS4%7Cpc%7CH9%7Cpc%7CHK%7Cpc%7CS6%7Cpc%7CS8%7Cpc%7CHT%7Cpc%7CDA%7Cpc%7CDJ%7Cpc%7CST%7Cpc%7CS7%7Cpc%7CS5%7Cpc%7CSQ%7Cpc%7CSA%7Cpc%7CS9%7Cpc%7CSJ%7Cpc%7CSK%7Cpc%7CD9%7Cpc%7CDT%7C]499|350[/hv]

 

4441 with 21hcp,maybe it is very hard to describe rightly,I have no idea.now how to bid up to slam reasonablly?

thank you very much.

4441牌型21hcp,也许很难正确描述,现在我是没辙呀,请问您如何合理地叫上满贯呢?实战中,还是有一些人凭借这良好的牌感,定约6nt的。

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2d逆叫后封到3n没什么道理,c套已经配合了,总是应该试探满贯的;

4441强牌,出现的机率还是很小的,也是hardy2/1体系有缺陷的地方。这幅我会选择开1d,有机会先逆叫2h,再叫2次c,尽可能的描述自己1444的强牌。

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flying_cat:

您好,欢迎您来到华人论坛讨论。

其实在2逆叫后,应叫人直接加叫3=9P+,4张以上支持了。如此,满贯在望:开叫人会再叫4C做低花木问叫。

而你如果再叫2,开叫人可以直接加叫3和4,很麻烦了。

 

[hv=d=s&v=e&b=3&a=p1cp1sp2dp3cp3d(cuebid)p3h(cuebid)p3s(cuebid)p3np4c(低花木)p4s(143022)p5n(关键张齐备,还有额外的牌力吗?)p7c(有啊,好几枚Q啊)ppp]499|350[/hv]

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个人观点,首先,这牌没叫到满贯,其实跟4441牌型21点,关系不大,东持1个A,4张,且上有Q的情况下,对同伴的逆叫竟然没有满贯想法,有点不可思议。方块上不仅有大牌,且有将吃价值,差不多已经可以数到11个赢墩了。

此外,不知道用不用英格伯曼,如果用的话,开叫人可以知道应叫人有点牌力,4NT邀请就必须了。当然,这是下策,主要问题出在东家。

4441是比较难叫,我个人不喜欢有单张的情况下开叫NT,同伴一个转移叫,把你扔在51配的高花上,并且还没有大牌,真是欲哭无泪。

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Moscito:

 

 

1......... 15+ any

......1... FG any

1......... 18+ any

......2... 6+ 1-suiter / (54)+ +

2......... Relay for shape

......3... 5224

3......... Relay for QP. A = 3, K = 2, Q = 1

......4... 8QP (+1 = Base = 6QP, +3 = Base+2 = 8QP)

 

W is missing AKQ AQ Q a total of 11 QP. E shows 8 QP.

That means at this point W know that they are missing 3 QP only, which could be 1A, 1K1Q or 3Qs.

 

4......... Relay, initiate DCB with K parity

......4NT... Even K, 1/2 Hs, 0 H.

 

Now know that E is missing K and 1Q. If that is Q there is grand.

 

5......... Relay, DCB

......5... 1 H

 

E confirm 1 Honor which has to be A.

So W know that E holds Axxxx AQ Qx xxxx

But there is no room to investigate the presence of J so can only signoff in 7

 

7......... Signoff

 

 

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Moscito 看了一下,感觉非常喜欢,很优秀的一个体系,很适合固定pd间使用;

应该说是现代改良版jq里面的一个非常好的选择吧,可惜国内使用者太少,想找个pd一起练习是非常困难的事;

不管如何,喜欢jq又喜欢复杂体系的朋友值得一试。

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Moscito 看了一下,感觉非常喜欢,很优秀的一个体系,很适合固定pd间使用;

应该说是现代改良版jq里面的一个非常好的选择吧,可惜国内使用者太少,想找个pd一起练习是非常困难的事;

不管如何,喜欢jq又喜欢复杂体系的朋友值得一试。

确实,非常复杂,但叫的非常准确。感觉我初学牌时的篮梅花叫牌法也有这个特征,只不过没有这么细致。

 

另外,我记得好象zyp163(风扬)在学习这个体系,你有兴趣可以与他联系,试试。

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好一个香港桥牌专家bgm !

长期以来,你以积极的心态支持我们华人论坛。

 

好一个香港桥牌专家bgm !

你的牌技感染鼓舞着大陆人。

 

好一个香港桥牌专家bgm !

你不居门户,思想清明,慧光闪烁。

 

好一个香港桥牌专家bgm !

如有可能,你将是我们China BBO门户中中国桥牌新闻栏目第一推介的中国桥牌专家!

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