2024年3月11日发(作者:奈一凡)
GAZZILLI打标准体系,开叫一阶高花,应叫人应叫 1NT。如果开叫人持有单套 FG
的牌,他没有方便的再叫,他可用 3 张低花跳叫 3 阶建立逼叫形势,带来的问题
是应叫人不知道低花是 3 张,4 张还是 5 张。如果开叫人 5-4-3-1 进局强牌,那
么如果应叫人弱牌,在开叫人 3 张的花色持有长套,那么很难找到这门花色配合。
跳叫占用了太多的空间,而这些空间对探索正确的定约是非常重要的。类似:S:
AKQTxx H:xx D:AQx C:AQ同样的问题在开叫人持有 15-17HCP 均型的情况,
如果开叫一阶高花,那么再叫存在问题。许多牌手持有 5 张高花开叫 1NT 来解决
这个问题。然而这种处理方式带来了另外一个问题,你可能错过一些边缘的更好的
定约,如 5-4,5-3 高花配合。为了解决这些难题并且允许非逼叫的 3 阶跳叫
(14-16HCP),意大利人用 Gazzilli 方式解决上述问题。AGazzilli 定义Gazzilli 是
1H/1S 后 1NT 或者 1H-1S 后这三种情况下开叫人作一个人工性的 2C 逼叫。2C
用于以下情况:1. 5-3-3-2 牌型,12 -14 15-17 和 18-20 HCP2. 5Mamp4C11-16HCP3.
其他单套或者双套 17HCP 的一手牌4. 采用此约定放弃自然 2C,2D 的定约B开叫
人在 1H 后的再叫Gazzilli 涵盖了绝大多数 17HCP 的一手牌,它可以让我们灵活
的做一个精确的再叫,例如跳叫。这些跳叫不再是逼叫性的,2 阶(除 2C 外)的
叫品,其大牌点是限制在 11-16HCP 范围内。这带来的一些优势,应叫人不必再维
持叫牌进程。以下是开叫人在 1H 开叫,在 1S/1NT 应叫后的再叫进程。1H – 1S 开
叫人再叫如下:1N – 均型,12-14 HCP2C – Gazzilli2D – 自然,4D 11-16 HCP2H – 自
然,6H 11-14 HCP2S – 3 或者 4S 支持 11-14 HCP2N – 6Hamp 3S 14-16 HCP3C –
5Hamp5C 14-16 HCP3D – 5Hamp5D 14-16 HCP3H – 自然,6H 14-16 HCP 无 3S3S –
4S 14-16 HCP 45224C – 4Samp6H单张 C 14-16 HCP4D – 4Samp6H单张 D 14-16
HCP1H – 1NT 开叫人再叫如下:2C – Gazzilli2D – 自然,4D 11-16 HCP2H – 自然,
6H 11-14 HCP2S – 5Hamp4S 17 HCP2N – 6HampD/C 4 张套,17HCP(详见下述进
程)3C – 5Hamp5C 14-16 HCP3D – 5Hamp5D 14-16 HCP3H – 自然,6H 14-16 HCP3S
– 6Hamp5S 14-16 HCP3N – 坚固套 14-16 HCP4C – 6-5 型 14-16 HCP4D – 6-5 型
14-16 HCP1H – 1N2N – 3C –接力,FG?开叫人澄清牌力:3D – 4 个 D3H – 4 个
C3D/H – 打此定约C开叫人在 1S 开叫后的再叫,类似于 1H 的进程。1S – 1NT 开
叫人再叫如下:2C – Gazzilli2D – 自然,3D 11-16 HCP2H – 自然,4H 11-16 HCP2S –
自然,6S 11-14 HCP2N – 6Samp任意 4 张套,17HCP(详见下述进程)3C – 5Samp5C
14-16 HCP3D – 5Samp5D 14-16 HCP3H – 5Samp5H 14-16 HCP3S – 6S 14-16 HCP3N
– 坚固套 14-16 HCP4C – 6-5 型 14-16 HCP4D – 6-5 型 14-16 HCP4H – 6-5 型
14-16 HCP1S – 1N2N – 3C –接力,FG?开叫人澄清牌力:3D – 4 个 D3H – 4 个
H3S – 4 个 C3D/H/S – 打此定约D开叫 1H 再叫的深入进程通常的一个逼叫性的
2D(接力,8HCP),如果开叫人再叫表示 17HCP 后,GF 的局势自动建立,应叫
人必须尽快限制自己手中的牌力。1H – 1S2C – 2D(接力 7-8 HCP)2H – 5 Hamp3 C
11-16 HCP2S – 5 H amp3S 17 HCP2N – 5 H 5-3-3-2 16-17 HCP3C – 5 H amp4C 17
HCP3D – 5 H amp4D 17 HCP3H – 6 H 17 HCP 无 3S3S – 5 Hamp 4 Camp 3S 15-16
HCP3N – 5 H 5-3-3amp 2S 18-19 HCP2H – 2/3 H 要打2S – 5 S H 单缺 要打2N – 5
D,止叫3C – 5 C,止叫3D – 5 Samp 5 D 8-10 HCP3H – 限制性 3 张加叫3S – 6S 邀
请1H – 1NT2C – 2D(接力 7-8 HCP)2H – 5 Hamp3C11-16 HCP2S – 5 H amp4C/D 17
HCP2N – 5 H 5-3-3-2 16-17 HCP3C – 5 H amp5C 17 HCP3D – 5 H amp5D 17 HCP3H
– 6 H 17 HCP3N – 5 H 5-3-3-2 18-19 HCP2H – 2/3 H 要打2S – 5-4 低花,D 长,止叫
2N – 双低花,止叫3C – 5 C,止叫3D – 6 D 止叫3H – 限制性 3 张加叫3S – 5 C,
单张 S,8HCP3NT– 5 C,单张 H,9-10HCP1H – 1S2C – 2D2H1 – 2S(接力)2N –
2-5-2-4 型3C – 5 H amp5C3D – 5 H amp4C amp3D3H – 6 H amp4C3S – 5 H amp4C
amp3S3N – 5 H 5-3-3-2 14-15 HCP15 Hamp3C11-16 HCP1H – 1S2C – 2D2H – ?2N –
inv3C – 要打3D – 4SF(第四花色),GF3H – 同意王牌花色,请求扣叫3S – 6 S3NT–
止叫1H – 1NT2C – 2D2H – ?2S – 4C,10-11HCP2N – 均型,10-11HCP3C – 5C,
8-9HCP3D – 6D,8-10HCP1H – 1N2C – 2D2S 17 4C/D – 2N relay3C – 5 Hamp4C3D –
5 Hamp4D3H – 6 Hamp4C3S – 6 Hamp4D1H – 1N –2C – 2D –2N 16-17 HCP – 3C –
5C GF3D – 5 D GF3H – 双张带一大 H 支持,GF3S – 双低花1H – 1S2C – 2D2S
17HCP amp 3 S – 2NT(relay)3C – 5Hamp3Samp4C 17 HCP3D – 5Hamp3Samp4D 17
HCP3H – 6Hamp 3S 17 HCP3S – 5Hamp 4S 17 HCP3N – 5-3-3-23S 17-19 HCP4C –
5H-4S,C 单缺 17 HCP4D – 5H-4S,D 单缺 17 HCP1H – 1S –2C – 2D –2N 16-17
HCP – 3C – 4 S amp5C3D – 4 Samp5 D3H – 双张带一大 H 支持,GF3S – 6 S GF3N
– 止叫1S – 1N2C – 2D2H – 5S amp 4C 15-16 或者 5S amp 4C/D/H 17 HCP,逼叫一
轮2S – 5S amp3 C 11-14 HCP2N – 5S 5-3-3-2 型 15-17 HCP3S – 6 S 17 HCP 否认 3
H3C – 5 amp 5C, 17 HCP3D – 5 amp 5D, 17 HCP3H – 5 amp 5H, 17 HCP3N – 5S
5-3-3-2 型 18-19 HCP1S – 1N2C – 2D2H – 2S(relay)?2N – 5Samp 4C, 15-16
HCP3C– 5Samp 4C, 17 HCP3D– 5Samp 4D, 17 HCP3H – 5Samp 4H, 17 HCP3S –
6 Samp3H,17 HCPGazzilli 也可以应用于 1C/D 开叫后的进程。E1C/D 开叫后的
进程开叫 1m 后在开叫人再叫 2om 定义为低花的 Gazzilli 进程,它可以表示:1.
单套低花,17HCP2. 4 张高花支持 18-19HCP3. 17 HCP ,5 张低花的非均型牌这意
味着开叫人对直接的 splinter 叫品可以用于 4-4-4-1 型,以下是一些序列:1D –
1H2C – Gazzilli描述包括 4Hamp17HCP 6D 17 HCP 或者 5D amp4C 11 HCP 三种
情况3C – 5 D amp5C 14-16 HCP3D – 6 D 14-16 HCP3H – 4 H 14-16 HCP1C – 1H2D –
Gazzilli 描述包括 4 H amp17 HCP 6C 17 HCP 或者 5C and 4D 17HCP3C – 6 C,
14-16 HCP3D – D Splinter 4-4-4-1 型 18 HCP3H – 4H,14-16 HCP1D – 1H2C – 2D
– 要打2H – 5 H,要打2S – relay,F12N – 4 C,止叫3C – 4 C,inv3D – 4 D,inv3H –
6 H,inv3S – 6 Hamp5SGF1C – 1H –2D – 2H – 5 H,要打2S – relay,F12N – 4 D止叫
3C – 止叫3D – GF 5 Hamp4 D3H – 6 H inv3S – 6 Hamp5 SGF1D – 1H2C – 2S
Relay2N – 5 Damp 4 C 17 HCP3C – 5 Damp4 Clubs 最多 16 HCP3D – 6 D 17HCP3H
– 4 H 18-19 HCP3S – 5 D amp 4H 17 HCP,S 单缺4C – 5 D amp 4H 17 HCP,C 单缺
1C – 1H –2D – 2S Relay2N – 5 Camp 4 D 17 HCP3C – 6 C 17 HCP3D – 6 Camp 5Ds
GF3H – 4 H 18-19 HCP均型3S – 5Camp4H 17 HCP, S 单缺4D – 5Camp4H 17 HCP,
D 单缺1D – 1H2C – 2D2H – 3H S 单缺 14-16 HCP2S – 5 Damp4C 17 HCP2N – 6 D
19 HCP3C – 5 Damp5CGF3D – 6 D 17-18 HCP Inv3H – 4 H 18-19 HCP3S – 5 D amp
4H 17 HCPS 单缺4C – 5 D amp 4H 17 HCPC 单缺1C – 1S2D – 2H – relay2S – 5 S要
打2N – 4 D,止叫3C – 止叫3D – GF 5 Samp4D3H – GF 5 Samp 5H3S – 6 S GF1D –
1S2C – 2D – 要打2S – 要打2H – relay F12N – 4 C止叫3C – 4 C inv3D – 4 D inv3H –
5 Samp5H inv3S – 6 S GF1C – 1S2D – 2H Relay2S – 3 Samp5 C amp 4 D 17 HCP2N –
5 Camp 4 D 17 HCP3C – 6 C 17 HCP3D – 6 Camp 5 D GF3S – 4 S 18-19 HCP均型4D
– 5 C amp4S 17 HCP,D 单缺4H – 5 C amp 4S 17 HCP,H 单缺FGazzilliConvention
与 BartConvention 对比:Example1 :a AJT53 b) AJT53 Q92 A92 J6 T6 AQ4 A54c
QJT54 KQ4 8 AQJ7a)采用 Bart Convention1S - 1NT12C 2 - 2D 32H 4 -1 Forcing2
Bart Convention3 relay4 低限 3 个 H采用 Gazzilli Convention1S - 1NT12C2 -
2D32S 4 -1 Forcing2 Gazzilli3 7 HCP4 5S amp 3C 11-14 HCPb)采用 Bart
Convention1S - 1NT12C 2 - 2D 33H 4 -1 Forcing2 Bart Convention3 relay4 高限 3 个
H采用 Gazzilli Convention1S - 1NT12C2 - 2D32NT 4 -1 Forcing2 Gazzilli3 7 HCP4
5S-3-3-2,15-17 HCPc)采用 Bart Convention1S - 1NT12C 2 - 2NT 33H 4 -1 Forcing2
10-11HCP,inv3 接受邀请,3 个 H采用 Gazzilli Convention1S - 1NT12C2 - 2D32H
4 - 2S52NT61 Forcing2 Gazzilli3 7 HCP4 5S amp 4C 15-16 or 5S amp 4C/D/H 17 HCP5
relay6 5S amp 4C 15-16 HCPGGazzilliConvention 与 MeckstrothAdjunc 对比:
Example1:a AKJT953 b) AKQ96 7 7 AK52 AQ5 A4 A84d AQ9875 c) AQ52 7
AKQ964 A8 7 AKQ7 A8f AKQ84 e) 7 A6 AKJ975 A8 5 AQT6 AKT92a)采用
Meckstroth Adjunc (Jeff Meckstroth)Convention1S - 1NT12NT 2 - 3C 33D 4 -1
Forcing2 Meckstroth Adjunc Convention3 checkback,对 S 无支持4 5S-ampD-4 FG采
用 Gazzilli Convention1S - 1NT12C2 - 2D32H4 - 2S53D61 Forcing2 Gazzilli3 7 HCP4
5S amp 4C 15-16 or 5S amp 4C/D/H 17 HCP5 relay6 5S amp 4D17 HCPb)采用
Meckstroth Adjunc (Jeff Meckstroth)Convention1S - 1NT12NT 2 - 3C 33S 4 -1
Forcing2 Meckstroth Adjunc Convention3 checkback,对 S 无支持4 6S FG采用
Gazzilli Convention1H - 1NT12C2 - 2D33S 4 -1 Forcing2 Gazzilli3 7 HCP4 6S 17
HCP,否认 3 个 Hc)采用 Meckstroth Adjunc (Jeff Meckstroth)Convention1H -
1NT12NT 2 - 3C 33S 4 -1 Forcing2 Meckstroth Adjunc Convention3 checkback,对 H
无支持4 5H-amp4Sslam try采用 Gazzilli Convention1H - 1NT12S2 -1 Forcing2
5Hamp4S17HCPd)采用 Meckstroth Adjunc (Jeff Meckstroth)Convention 方法1S -
1NT12NT 2 - 3C 33NT 4 -1 Forcing2 Meckstroth Adjunc Convention3 checkbac,对 S
无支持4 4 个草花,同时表示如果开叫人有 6 个 S,不是独立可打的好套采用
Gazzilli Convention1S - 1NT12C2 - 2D32H4 - 2S53C61 Forcing2 Gazzilli3 7 HCP4 5S
amp 4C 15-16 or 5S amp 4C/D/H 17 HCP5 relay6 5S amp 4C 17 HCPe)采用
Meckstroth Adjunc (Jeff Meckstroth)Convention1H - 1NT12NT 2 - 3C 34C 4 -1
Forcing2 Meckstroth Adjunc Convention3 checkback,对 H 无支持44 个草花及 6 张
高花独立可打的好套采用 Gazzilli Convention1H - 1NT12C2 - 2D32S4 - 2NT53H61
Forcing2 Gazzilli3 7 HCP4 5H amp 4C/D17 HCP5 relay6 6H amp 4Cf)采用
Meckstroth Adjunc(Jeff Meckstroth)Convention1S - 1NT13C2 -1 Forcing2 5Samp5C
FG采用 Gazzilli Convention1S - 1NT12C2 - 2D33C 4 -1 Forcing2 Gazzilli3 7 HCP4 5S
amp 5C17 HCPH 逼叫 1NT 后的 Gazzilli牌例取自 cozofu《逼叫性 1NT 后的一些
难点》Eric Kokish and friends《重新探讨逼叫性 1NT》翻译:刘京 蒋又新Example1:a
AQT964 b) 32 AK3 76 2 T9876 J84 AKT9一般 2 over 1 进程:1S - 1NT12S -
pass1Forcing采用 Gazzilli Convention:1S - 1NT13S2 - 4S1 Forcing2 6S,14-16HCP
也许队友在这副牌中采用如下进程1S - 1NT12C! - 2S3S - 4S但是同伴偶尔会 pass
你的 2C 比如他拿了 S:K H:Qxxxx D:xxx C:Qxxx显然 2S 要比 2C xx xxx稳
妥得多.其次 在同伴叫回 2S 之后 你叫 3S 也可能有会太高.比如同伴拿了 S: H:
D:KQxxx C:Qxx 3S 大概难免要下 1 或者下 2!Example2:a AQJ964 b) AQJ9642
AK2 A3 K2 T6 84 T9传统方法很难处理这种牌!应叫人在拿 8-9 点时 对开叫人的
加叫很困难。比如:Example3:a 32 b) JT964 KQ76 A73 T6 2 QJT32 AK98一般 2 over
1 进程:1S - 1NT12C - 2S?3C?1Forcing如果你示弱回叫到 2S 那么恭喜你,你错
了!如果你加叫到 3C,同样存在问题,你的牌似乎没有那么强!采用 Gazzilli
Convention:1S - 1NT12C2 - 2D32S4 - 3C1 Forcing2 Gazzilli3 relay7HCP4 5S amp3 C
11-14 HCP5 pd 低限,最后是修正定约Example4:开叫人持: KQT96 KQ3 2 AJT65
应叫人持有:JT 64 A2 J64J9876 A973 J9873 73 K982 AJ92 A982 7652 KQ 872 98
9876一般 2 over 1 进程对于所有的应叫人持牌情况:1S - 1NT12C - 2S1Forcing btw:
开叫人的牌是不是一定可以再叫 3C?采用 Gazzilli Convention:1S - 1NT13C2 -1
Forcing2 S5ampC5,14-16 HCP对于牌 3 和 4,完全可以叫 4S!Example5:开叫人
持: AJT KQT96 43 AJ4应叫人持有: 32 K 4 643 J983 3 76 JT982 AJ92 AT9852
AK8K9872 KQT9 KQ876 QT8532开叫人开 1NT 么?还是 1H?That is a question开
叫 1NT,再此假设采用 1NT。1.1NT - 2NT13C2 - pass1 询问低花情况?2 C 好于 D
采用 Gazzilli Convention:1H - 1NT12C2 - 3C31 Forcing2 Gazzilli2 要打,示弱2.1NT
- 2C12H2 - 4H采用 Gazzilli Convention 并无优势!3.1NT - 2NT13C2 - ?1 询问低
花情况?2 C 好于 D? 怎么处理!采用 Gazzilli Convention:1H - 1NT12C2 -
2D32NT4 - 3S51 Forcing2 Gazzilli3 relay4 16-17HCP,5-3-3-25 双低花4.1NT -
2NT13C2 - ?1 询问低花情况?2 C 好于 D? 3NT?5C?采用 Gazzilli
Convention:1H - 1NT12C2 - 2D32NT4 - 3C53S6 - 3NT71 Forcing2 Gazzilli3 relay4
16-17HCP,5-3-3-25 5C GF6 S 上有实力7 partner,不用担心 D,我很强!Example6:
开叫人持: A54 应叫人持:KT9862 KJ8762 2 AK Q98 97 863 一般 2 over 1 进程
对于所有的应叫人持牌情况:1H - 1S2H - 2S?? pass 或者继续 2S 是一个选择问
题,因为 S 套的质量稍差!采用 Gazzilli Convention:1H - 1S2NT1 -1 Forcing2
H6ampS3,14-16 HCP现在的问题是应叫人可以很放心的叫 3S,总墩数定律!
GazzilliHow to handle openers rebid after the Kaplan/Granville Inversion and the related
subject ofhow to handle 1SP-1NT forcing in particular how to combine Bart and Gazzilli
- one attemptis in ETM Gold Premium holds great interest for me - however a recent
posting asking about Bart/Gazzilli had zero replies so most do not find it
so absorbing so Illunderstand that most will skip the rest of this or skim unread
avoid repetition below all sequences that begin 1HE-1SP are using the
KaplanvilleInversion where 1SP acts like a proxy quotforcing notrumpquot. In this style
1HE-1NT of the design work in this area involves mapping hand types
to openers rebid anddeciding what works or not. For example I considered
1HE-1SP--2CL to show either4SPs or 6HEs openers 2HE rebid shows CLs - thus a
remapping. Now responder cansignoff in 2HE or can bid 2DI to ask. However some
hands want to invite opposite 6HEsbut pass opposite if 4SPs so the scheme has
an experts Vincent Demuy and Gavin Wolpert play or played a method
where1HE-1SP--1NT shows a balanced hand or DIs. This allows them to play
1HE-1SP-2DIto show 4SPs sort of a delayed Flannery. The problem with this mapping is
that the range of1NT is considerable forcing responder to move on when a pass would be
better. Also thedistributional HEs amp DIs hands might place a notrump contract by the
wrong tly I favour this approach:After 1HE-1SP:1NT: Maximum of 14.
Balanced or with exactly a four card minor not 0-5-4-4. Now 2m is toplay 2SP is an
artificial game force 2HE and invite with 2HEs and or poor invite with3HEs 3X natural
invite.2CL: Four or longer minor and 14/15 or five or longer minor or any 18/19.
Structure below.2DI: 4SPs up to 16 not 5SPs if 14-16.2SP: 4SPs 16/17-18 if 4SPs 14-16
if 5SPs.3CL/DI: 5-5 13/ fairly 1HE-1SP--2CL:2DI: Up to 9
4-5DIs or 3DIs and short HEs. Now: Pass: 4DIs no game interest. 2HE: 4CLs and 15-18
or 5CLs up to 13. 3CL to play. 2SP asks 2NT shows 15-163CL 5-5 above 3CL if 17-18
2SP: 4DIs and game interest forcing just to 2NT. 2NT: 18/19 Puppet to 3CL or CL
signoff with 6CLs. After responder bids 3CL 3DIshows 4DIs 3HE and 3SP show 4SPs
the latter with 3/4DIs 3NT is flat with 6HEs. 3CL: 18/19 with 4CLs. 3DI asks hand type.
3DI: 17/18 with 5DIs. 3HE and higher: Hands with 6HEs and a singleton/void.2HE: Up
to 9 2HEs amp 4-5CLs. Now: Pass: 4DIs no game interest. 2SP: 4DIs and game interest
forcing just to 2NT. 2NT: 18/19 Puppet to 3CL or CL signoff. As 2NT puppet above.
3CL: 17/18 with 4CLs. 3DI asks hand type. 3DI: 18/19 with 5DIs. 3HE and higher:
Hands with 6HEs and a singleton/void.2SP: Most hands with 9/10 see 2NT alternative.
Now: 2NT transfer to 3CL with CLs “ opener can pass 3CL if 5-5 but will bid again
if14/15 to show shape. 3SP over 3CL shows 4SPs and short DIs. Responder does nothave
to accept the transfer - in particular can bid 3NT with 12 and can bid 3DIs with 6DIs.
3CL shows 5-5 in the reds not forcing - responder can pass with long CLs. 3DI is natural
and forcing. 3HE and 3SP are natural 18/19. 3NT to play. 4CL/DI with big 6-5. 3NT .
2024年3月11日发(作者:奈一凡)
GAZZILLI打标准体系,开叫一阶高花,应叫人应叫 1NT。如果开叫人持有单套 FG
的牌,他没有方便的再叫,他可用 3 张低花跳叫 3 阶建立逼叫形势,带来的问题
是应叫人不知道低花是 3 张,4 张还是 5 张。如果开叫人 5-4-3-1 进局强牌,那
么如果应叫人弱牌,在开叫人 3 张的花色持有长套,那么很难找到这门花色配合。
跳叫占用了太多的空间,而这些空间对探索正确的定约是非常重要的。类似:S:
AKQTxx H:xx D:AQx C:AQ同样的问题在开叫人持有 15-17HCP 均型的情况,
如果开叫一阶高花,那么再叫存在问题。许多牌手持有 5 张高花开叫 1NT 来解决
这个问题。然而这种处理方式带来了另外一个问题,你可能错过一些边缘的更好的
定约,如 5-4,5-3 高花配合。为了解决这些难题并且允许非逼叫的 3 阶跳叫
(14-16HCP),意大利人用 Gazzilli 方式解决上述问题。AGazzilli 定义Gazzilli 是
1H/1S 后 1NT 或者 1H-1S 后这三种情况下开叫人作一个人工性的 2C 逼叫。2C
用于以下情况:1. 5-3-3-2 牌型,12 -14 15-17 和 18-20 HCP2. 5Mamp4C11-16HCP3.
其他单套或者双套 17HCP 的一手牌4. 采用此约定放弃自然 2C,2D 的定约B开叫
人在 1H 后的再叫Gazzilli 涵盖了绝大多数 17HCP 的一手牌,它可以让我们灵活
的做一个精确的再叫,例如跳叫。这些跳叫不再是逼叫性的,2 阶(除 2C 外)的
叫品,其大牌点是限制在 11-16HCP 范围内。这带来的一些优势,应叫人不必再维
持叫牌进程。以下是开叫人在 1H 开叫,在 1S/1NT 应叫后的再叫进程。1H – 1S 开
叫人再叫如下:1N – 均型,12-14 HCP2C – Gazzilli2D – 自然,4D 11-16 HCP2H – 自
然,6H 11-14 HCP2S – 3 或者 4S 支持 11-14 HCP2N – 6Hamp 3S 14-16 HCP3C –
5Hamp5C 14-16 HCP3D – 5Hamp5D 14-16 HCP3H – 自然,6H 14-16 HCP 无 3S3S –
4S 14-16 HCP 45224C – 4Samp6H单张 C 14-16 HCP4D – 4Samp6H单张 D 14-16
HCP1H – 1NT 开叫人再叫如下:2C – Gazzilli2D – 自然,4D 11-16 HCP2H – 自然,
6H 11-14 HCP2S – 5Hamp4S 17 HCP2N – 6HampD/C 4 张套,17HCP(详见下述进
程)3C – 5Hamp5C 14-16 HCP3D – 5Hamp5D 14-16 HCP3H – 自然,6H 14-16 HCP3S
– 6Hamp5S 14-16 HCP3N – 坚固套 14-16 HCP4C – 6-5 型 14-16 HCP4D – 6-5 型
14-16 HCP1H – 1N2N – 3C –接力,FG?开叫人澄清牌力:3D – 4 个 D3H – 4 个
C3D/H – 打此定约C开叫人在 1S 开叫后的再叫,类似于 1H 的进程。1S – 1NT 开
叫人再叫如下:2C – Gazzilli2D – 自然,3D 11-16 HCP2H – 自然,4H 11-16 HCP2S –
自然,6S 11-14 HCP2N – 6Samp任意 4 张套,17HCP(详见下述进程)3C – 5Samp5C
14-16 HCP3D – 5Samp5D 14-16 HCP3H – 5Samp5H 14-16 HCP3S – 6S 14-16 HCP3N
– 坚固套 14-16 HCP4C – 6-5 型 14-16 HCP4D – 6-5 型 14-16 HCP4H – 6-5 型
14-16 HCP1S – 1N2N – 3C –接力,FG?开叫人澄清牌力:3D – 4 个 D3H – 4 个
H3S – 4 个 C3D/H/S – 打此定约D开叫 1H 再叫的深入进程通常的一个逼叫性的
2D(接力,8HCP),如果开叫人再叫表示 17HCP 后,GF 的局势自动建立,应叫
人必须尽快限制自己手中的牌力。1H – 1S2C – 2D(接力 7-8 HCP)2H – 5 Hamp3 C
11-16 HCP2S – 5 H amp3S 17 HCP2N – 5 H 5-3-3-2 16-17 HCP3C – 5 H amp4C 17
HCP3D – 5 H amp4D 17 HCP3H – 6 H 17 HCP 无 3S3S – 5 Hamp 4 Camp 3S 15-16
HCP3N – 5 H 5-3-3amp 2S 18-19 HCP2H – 2/3 H 要打2S – 5 S H 单缺 要打2N – 5
D,止叫3C – 5 C,止叫3D – 5 Samp 5 D 8-10 HCP3H – 限制性 3 张加叫3S – 6S 邀
请1H – 1NT2C – 2D(接力 7-8 HCP)2H – 5 Hamp3C11-16 HCP2S – 5 H amp4C/D 17
HCP2N – 5 H 5-3-3-2 16-17 HCP3C – 5 H amp5C 17 HCP3D – 5 H amp5D 17 HCP3H
– 6 H 17 HCP3N – 5 H 5-3-3-2 18-19 HCP2H – 2/3 H 要打2S – 5-4 低花,D 长,止叫
2N – 双低花,止叫3C – 5 C,止叫3D – 6 D 止叫3H – 限制性 3 张加叫3S – 5 C,
单张 S,8HCP3NT– 5 C,单张 H,9-10HCP1H – 1S2C – 2D2H1 – 2S(接力)2N –
2-5-2-4 型3C – 5 H amp5C3D – 5 H amp4C amp3D3H – 6 H amp4C3S – 5 H amp4C
amp3S3N – 5 H 5-3-3-2 14-15 HCP15 Hamp3C11-16 HCP1H – 1S2C – 2D2H – ?2N –
inv3C – 要打3D – 4SF(第四花色),GF3H – 同意王牌花色,请求扣叫3S – 6 S3NT–
止叫1H – 1NT2C – 2D2H – ?2S – 4C,10-11HCP2N – 均型,10-11HCP3C – 5C,
8-9HCP3D – 6D,8-10HCP1H – 1N2C – 2D2S 17 4C/D – 2N relay3C – 5 Hamp4C3D –
5 Hamp4D3H – 6 Hamp4C3S – 6 Hamp4D1H – 1N –2C – 2D –2N 16-17 HCP – 3C –
5C GF3D – 5 D GF3H – 双张带一大 H 支持,GF3S – 双低花1H – 1S2C – 2D2S
17HCP amp 3 S – 2NT(relay)3C – 5Hamp3Samp4C 17 HCP3D – 5Hamp3Samp4D 17
HCP3H – 6Hamp 3S 17 HCP3S – 5Hamp 4S 17 HCP3N – 5-3-3-23S 17-19 HCP4C –
5H-4S,C 单缺 17 HCP4D – 5H-4S,D 单缺 17 HCP1H – 1S –2C – 2D –2N 16-17
HCP – 3C – 4 S amp5C3D – 4 Samp5 D3H – 双张带一大 H 支持,GF3S – 6 S GF3N
– 止叫1S – 1N2C – 2D2H – 5S amp 4C 15-16 或者 5S amp 4C/D/H 17 HCP,逼叫一
轮2S – 5S amp3 C 11-14 HCP2N – 5S 5-3-3-2 型 15-17 HCP3S – 6 S 17 HCP 否认 3
H3C – 5 amp 5C, 17 HCP3D – 5 amp 5D, 17 HCP3H – 5 amp 5H, 17 HCP3N – 5S
5-3-3-2 型 18-19 HCP1S – 1N2C – 2D2H – 2S(relay)?2N – 5Samp 4C, 15-16
HCP3C– 5Samp 4C, 17 HCP3D– 5Samp 4D, 17 HCP3H – 5Samp 4H, 17 HCP3S –
6 Samp3H,17 HCPGazzilli 也可以应用于 1C/D 开叫后的进程。E1C/D 开叫后的
进程开叫 1m 后在开叫人再叫 2om 定义为低花的 Gazzilli 进程,它可以表示:1.
单套低花,17HCP2. 4 张高花支持 18-19HCP3. 17 HCP ,5 张低花的非均型牌这意
味着开叫人对直接的 splinter 叫品可以用于 4-4-4-1 型,以下是一些序列:1D –
1H2C – Gazzilli描述包括 4Hamp17HCP 6D 17 HCP 或者 5D amp4C 11 HCP 三种
情况3C – 5 D amp5C 14-16 HCP3D – 6 D 14-16 HCP3H – 4 H 14-16 HCP1C – 1H2D –
Gazzilli 描述包括 4 H amp17 HCP 6C 17 HCP 或者 5C and 4D 17HCP3C – 6 C,
14-16 HCP3D – D Splinter 4-4-4-1 型 18 HCP3H – 4H,14-16 HCP1D – 1H2C – 2D
– 要打2H – 5 H,要打2S – relay,F12N – 4 C,止叫3C – 4 C,inv3D – 4 D,inv3H –
6 H,inv3S – 6 Hamp5SGF1C – 1H –2D – 2H – 5 H,要打2S – relay,F12N – 4 D止叫
3C – 止叫3D – GF 5 Hamp4 D3H – 6 H inv3S – 6 Hamp5 SGF1D – 1H2C – 2S
Relay2N – 5 Damp 4 C 17 HCP3C – 5 Damp4 Clubs 最多 16 HCP3D – 6 D 17HCP3H
– 4 H 18-19 HCP3S – 5 D amp 4H 17 HCP,S 单缺4C – 5 D amp 4H 17 HCP,C 单缺
1C – 1H –2D – 2S Relay2N – 5 Camp 4 D 17 HCP3C – 6 C 17 HCP3D – 6 Camp 5Ds
GF3H – 4 H 18-19 HCP均型3S – 5Camp4H 17 HCP, S 单缺4D – 5Camp4H 17 HCP,
D 单缺1D – 1H2C – 2D2H – 3H S 单缺 14-16 HCP2S – 5 Damp4C 17 HCP2N – 6 D
19 HCP3C – 5 Damp5CGF3D – 6 D 17-18 HCP Inv3H – 4 H 18-19 HCP3S – 5 D amp
4H 17 HCPS 单缺4C – 5 D amp 4H 17 HCPC 单缺1C – 1S2D – 2H – relay2S – 5 S要
打2N – 4 D,止叫3C – 止叫3D – GF 5 Samp4D3H – GF 5 Samp 5H3S – 6 S GF1D –
1S2C – 2D – 要打2S – 要打2H – relay F12N – 4 C止叫3C – 4 C inv3D – 4 D inv3H –
5 Samp5H inv3S – 6 S GF1C – 1S2D – 2H Relay2S – 3 Samp5 C amp 4 D 17 HCP2N –
5 Camp 4 D 17 HCP3C – 6 C 17 HCP3D – 6 Camp 5 D GF3S – 4 S 18-19 HCP均型4D
– 5 C amp4S 17 HCP,D 单缺4H – 5 C amp 4S 17 HCP,H 单缺FGazzilliConvention
与 BartConvention 对比:Example1 :a AJT53 b) AJT53 Q92 A92 J6 T6 AQ4 A54c
QJT54 KQ4 8 AQJ7a)采用 Bart Convention1S - 1NT12C 2 - 2D 32H 4 -1 Forcing2
Bart Convention3 relay4 低限 3 个 H采用 Gazzilli Convention1S - 1NT12C2 -
2D32S 4 -1 Forcing2 Gazzilli3 7 HCP4 5S amp 3C 11-14 HCPb)采用 Bart
Convention1S - 1NT12C 2 - 2D 33H 4 -1 Forcing2 Bart Convention3 relay4 高限 3 个
H采用 Gazzilli Convention1S - 1NT12C2 - 2D32NT 4 -1 Forcing2 Gazzilli3 7 HCP4
5S-3-3-2,15-17 HCPc)采用 Bart Convention1S - 1NT12C 2 - 2NT 33H 4 -1 Forcing2
10-11HCP,inv3 接受邀请,3 个 H采用 Gazzilli Convention1S - 1NT12C2 - 2D32H
4 - 2S52NT61 Forcing2 Gazzilli3 7 HCP4 5S amp 4C 15-16 or 5S amp 4C/D/H 17 HCP5
relay6 5S amp 4C 15-16 HCPGGazzilliConvention 与 MeckstrothAdjunc 对比:
Example1:a AKJT953 b) AKQ96 7 7 AK52 AQ5 A4 A84d AQ9875 c) AQ52 7
AKQ964 A8 7 AKQ7 A8f AKQ84 e) 7 A6 AKJ975 A8 5 AQT6 AKT92a)采用
Meckstroth Adjunc (Jeff Meckstroth)Convention1S - 1NT12NT 2 - 3C 33D 4 -1
Forcing2 Meckstroth Adjunc Convention3 checkback,对 S 无支持4 5S-ampD-4 FG采
用 Gazzilli Convention1S - 1NT12C2 - 2D32H4 - 2S53D61 Forcing2 Gazzilli3 7 HCP4
5S amp 4C 15-16 or 5S amp 4C/D/H 17 HCP5 relay6 5S amp 4D17 HCPb)采用
Meckstroth Adjunc (Jeff Meckstroth)Convention1S - 1NT12NT 2 - 3C 33S 4 -1
Forcing2 Meckstroth Adjunc Convention3 checkback,对 S 无支持4 6S FG采用
Gazzilli Convention1H - 1NT12C2 - 2D33S 4 -1 Forcing2 Gazzilli3 7 HCP4 6S 17
HCP,否认 3 个 Hc)采用 Meckstroth Adjunc (Jeff Meckstroth)Convention1H -
1NT12NT 2 - 3C 33S 4 -1 Forcing2 Meckstroth Adjunc Convention3 checkback,对 H
无支持4 5H-amp4Sslam try采用 Gazzilli Convention1H - 1NT12S2 -1 Forcing2
5Hamp4S17HCPd)采用 Meckstroth Adjunc (Jeff Meckstroth)Convention 方法1S -
1NT12NT 2 - 3C 33NT 4 -1 Forcing2 Meckstroth Adjunc Convention3 checkbac,对 S
无支持4 4 个草花,同时表示如果开叫人有 6 个 S,不是独立可打的好套采用
Gazzilli Convention1S - 1NT12C2 - 2D32H4 - 2S53C61 Forcing2 Gazzilli3 7 HCP4 5S
amp 4C 15-16 or 5S amp 4C/D/H 17 HCP5 relay6 5S amp 4C 17 HCPe)采用
Meckstroth Adjunc (Jeff Meckstroth)Convention1H - 1NT12NT 2 - 3C 34C 4 -1
Forcing2 Meckstroth Adjunc Convention3 checkback,对 H 无支持44 个草花及 6 张
高花独立可打的好套采用 Gazzilli Convention1H - 1NT12C2 - 2D32S4 - 2NT53H61
Forcing2 Gazzilli3 7 HCP4 5H amp 4C/D17 HCP5 relay6 6H amp 4Cf)采用
Meckstroth Adjunc(Jeff Meckstroth)Convention1S - 1NT13C2 -1 Forcing2 5Samp5C
FG采用 Gazzilli Convention1S - 1NT12C2 - 2D33C 4 -1 Forcing2 Gazzilli3 7 HCP4 5S
amp 5C17 HCPH 逼叫 1NT 后的 Gazzilli牌例取自 cozofu《逼叫性 1NT 后的一些
难点》Eric Kokish and friends《重新探讨逼叫性 1NT》翻译:刘京 蒋又新Example1:a
AQT964 b) 32 AK3 76 2 T9876 J84 AKT9一般 2 over 1 进程:1S - 1NT12S -
pass1Forcing采用 Gazzilli Convention:1S - 1NT13S2 - 4S1 Forcing2 6S,14-16HCP
也许队友在这副牌中采用如下进程1S - 1NT12C! - 2S3S - 4S但是同伴偶尔会 pass
你的 2C 比如他拿了 S:K H:Qxxxx D:xxx C:Qxxx显然 2S 要比 2C xx xxx稳
妥得多.其次 在同伴叫回 2S 之后 你叫 3S 也可能有会太高.比如同伴拿了 S: H:
D:KQxxx C:Qxx 3S 大概难免要下 1 或者下 2!Example2:a AQJ964 b) AQJ9642
AK2 A3 K2 T6 84 T9传统方法很难处理这种牌!应叫人在拿 8-9 点时 对开叫人的
加叫很困难。比如:Example3:a 32 b) JT964 KQ76 A73 T6 2 QJT32 AK98一般 2 over
1 进程:1S - 1NT12C - 2S?3C?1Forcing如果你示弱回叫到 2S 那么恭喜你,你错
了!如果你加叫到 3C,同样存在问题,你的牌似乎没有那么强!采用 Gazzilli
Convention:1S - 1NT12C2 - 2D32S4 - 3C1 Forcing2 Gazzilli3 relay7HCP4 5S amp3 C
11-14 HCP5 pd 低限,最后是修正定约Example4:开叫人持: KQT96 KQ3 2 AJT65
应叫人持有:JT 64 A2 J64J9876 A973 J9873 73 K982 AJ92 A982 7652 KQ 872 98
9876一般 2 over 1 进程对于所有的应叫人持牌情况:1S - 1NT12C - 2S1Forcing btw:
开叫人的牌是不是一定可以再叫 3C?采用 Gazzilli Convention:1S - 1NT13C2 -1
Forcing2 S5ampC5,14-16 HCP对于牌 3 和 4,完全可以叫 4S!Example5:开叫人
持: AJT KQT96 43 AJ4应叫人持有: 32 K 4 643 J983 3 76 JT982 AJ92 AT9852
AK8K9872 KQT9 KQ876 QT8532开叫人开 1NT 么?还是 1H?That is a question开
叫 1NT,再此假设采用 1NT。1.1NT - 2NT13C2 - pass1 询问低花情况?2 C 好于 D
采用 Gazzilli Convention:1H - 1NT12C2 - 3C31 Forcing2 Gazzilli2 要打,示弱2.1NT
- 2C12H2 - 4H采用 Gazzilli Convention 并无优势!3.1NT - 2NT13C2 - ?1 询问低
花情况?2 C 好于 D? 怎么处理!采用 Gazzilli Convention:1H - 1NT12C2 -
2D32NT4 - 3S51 Forcing2 Gazzilli3 relay4 16-17HCP,5-3-3-25 双低花4.1NT -
2NT13C2 - ?1 询问低花情况?2 C 好于 D? 3NT?5C?采用 Gazzilli
Convention:1H - 1NT12C2 - 2D32NT4 - 3C53S6 - 3NT71 Forcing2 Gazzilli3 relay4
16-17HCP,5-3-3-25 5C GF6 S 上有实力7 partner,不用担心 D,我很强!Example6:
开叫人持: A54 应叫人持:KT9862 KJ8762 2 AK Q98 97 863 一般 2 over 1 进程
对于所有的应叫人持牌情况:1H - 1S2H - 2S?? pass 或者继续 2S 是一个选择问
题,因为 S 套的质量稍差!采用 Gazzilli Convention:1H - 1S2NT1 -1 Forcing2
H6ampS3,14-16 HCP现在的问题是应叫人可以很放心的叫 3S,总墩数定律!
GazzilliHow to handle openers rebid after the Kaplan/Granville Inversion and the related
subject ofhow to handle 1SP-1NT forcing in particular how to combine Bart and Gazzilli
- one attemptis in ETM Gold Premium holds great interest for me - however a recent
posting asking about Bart/Gazzilli had zero replies so most do not find it
so absorbing so Illunderstand that most will skip the rest of this or skim unread
avoid repetition below all sequences that begin 1HE-1SP are using the
KaplanvilleInversion where 1SP acts like a proxy quotforcing notrumpquot. In this style
1HE-1NT of the design work in this area involves mapping hand types
to openers rebid anddeciding what works or not. For example I considered
1HE-1SP--2CL to show either4SPs or 6HEs openers 2HE rebid shows CLs - thus a
remapping. Now responder cansignoff in 2HE or can bid 2DI to ask. However some
hands want to invite opposite 6HEsbut pass opposite if 4SPs so the scheme has
an experts Vincent Demuy and Gavin Wolpert play or played a method
where1HE-1SP--1NT shows a balanced hand or DIs. This allows them to play
1HE-1SP-2DIto show 4SPs sort of a delayed Flannery. The problem with this mapping is
that the range of1NT is considerable forcing responder to move on when a pass would be
better. Also thedistributional HEs amp DIs hands might place a notrump contract by the
wrong tly I favour this approach:After 1HE-1SP:1NT: Maximum of 14.
Balanced or with exactly a four card minor not 0-5-4-4. Now 2m is toplay 2SP is an
artificial game force 2HE and invite with 2HEs and or poor invite with3HEs 3X natural
invite.2CL: Four or longer minor and 14/15 or five or longer minor or any 18/19.
Structure below.2DI: 4SPs up to 16 not 5SPs if 14-16.2SP: 4SPs 16/17-18 if 4SPs 14-16
if 5SPs.3CL/DI: 5-5 13/ fairly 1HE-1SP--2CL:2DI: Up to 9
4-5DIs or 3DIs and short HEs. Now: Pass: 4DIs no game interest. 2HE: 4CLs and 15-18
or 5CLs up to 13. 3CL to play. 2SP asks 2NT shows 15-163CL 5-5 above 3CL if 17-18
2SP: 4DIs and game interest forcing just to 2NT. 2NT: 18/19 Puppet to 3CL or CL
signoff with 6CLs. After responder bids 3CL 3DIshows 4DIs 3HE and 3SP show 4SPs
the latter with 3/4DIs 3NT is flat with 6HEs. 3CL: 18/19 with 4CLs. 3DI asks hand type.
3DI: 17/18 with 5DIs. 3HE and higher: Hands with 6HEs and a singleton/void.2HE: Up
to 9 2HEs amp 4-5CLs. Now: Pass: 4DIs no game interest. 2SP: 4DIs and game interest
forcing just to 2NT. 2NT: 18/19 Puppet to 3CL or CL signoff. As 2NT puppet above.
3CL: 17/18 with 4CLs. 3DI asks hand type. 3DI: 18/19 with 5DIs. 3HE and higher:
Hands with 6HEs and a singleton/void.2SP: Most hands with 9/10 see 2NT alternative.
Now: 2NT transfer to 3CL with CLs “ opener can pass 3CL if 5-5 but will bid again
if14/15 to show shape. 3SP over 3CL shows 4SPs and short DIs. Responder does nothave
to accept the transfer - in particular can bid 3NT with 12 and can bid 3DIs with 6DIs.
3CL shows 5-5 in the reds not forcing - responder can pass with long CLs. 3DI is natural
and forcing. 3HE and 3SP are natural 18/19. 3NT to play. 4CL/DI with big 6-5. 3NT .