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[lastpost] => 2775417 【星宝】10月21日投注流水166330元 1761119021 rainwang
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所谓“缆”者,即一系列的投注,一连几铺的落注。最常见的就是当开了三铺庄之后,就开始买闲。第一铺输了,第二铺又再加倍;第二铺输了,第三铺亦继续加倍,到买中为止。% I( ] I9 b" `" G/ _! S
# l/ \3 c$ A% r8 z* E$ D有一次在赌场内遇到一位大豪客,当时他已输了很多钱,叹气说:“怎样才能赢钱呢?”旁边即有人说:“买缆就可赢钱。”那大豪客马上回应,如有不断的缆,他愿意出任何价钱购买。他是对的。世上没有不会断的缆,总有一次会断,一旦断了就要赔大本。既然如此,买缆岂能必胜呢?! g/ T8 r6 R8 S3 Y. `
* s, W- d% ^2 p- ]要促成买缆必胜。首先我们要了解游戏规则。一般赌场都会设一个注码上下限的比例,最普通的就是1:150。例如最低是100元,最高,就是15000元,视个别赌场来定。在亚洲的赌场,很多是以全赌注总和计算;在美国等地方则以每一位赌客独立注码计算。目的都是一样,因为赌场要保障本身利益,如果不设上限,无止境地复式加倍又加倍投注,始终会给闲家买中,赌场就会不利。那么什么上下限都会介乎最少的1:100到最高的1:200呢?原因是开赌的人也计算过或然率,令闲家即使肯博,长期买缆,也要冒输大钱的险。下表就可以看出买缆的情况:
5 C d2 e5 y& Q
4 K. }$ s: S4 b/ E8 p" P4 \铺数 注码$ J0 B3 X9 N# R% L0 `+ ^9 ?
, F5 x1 [7 U" b. d% ~
1 ¥1000% O, \; _/ i& ]% D; d/ X3 m" z2 |
) y5 o: ~# ^) S5 `9 W( ^& u/ S2 ¥2000& ?! X( y" z y, f
1 n, g1 h% ^5 L: B3 i d
3 ¥4000" m C# }; v3 c- @. {; h
! }: F |$ a9 n/ p, E6 _+ K8 A
4 ¥8000
5 |( v8 C, _4 K8 `$ \! P2 Q2 P% R2 K/ a
5 ¥16000" Y9 R% F/ W/ O3 I5 p s
" `2 S! e" i( C! Q6 ¥320006 |; r8 H$ @$ g+ E3 Q5 M
6 }3 ^, I" B9 _3 F( K9 S
7 ¥640005 B2 @3 w! e9 o2 \* c6 h
, F/ s8 x; |. D' M" c, v4 F9 W
8 ¥128000! b1 {' t! O" T
0 e5 w1 D- f- t9 ¥256000. B; g! G; l* i
8 ?8 t- @, s- _7 `- L. O. d可以看得出最尽只能买到第八铺,第九铺已超额。如是者,必须于八铺之内买中,否则就要断缆,一次断缆,所输的钱,要连续赌四盒牌不断缆才可赢回损失。
" D: `* e$ ~* P4 }4 q5 {( a
G0 k' h9 z0 z, g$ B1 I经统计及电脑分析,要达至长期不会断的缆,至少要十铺为标准才可保障长胜,但十铺又刚好超过了注码,所以是有矛盾,不可行。任何一条公式缆,无论庄闲的次序是如何排列,9铺之内还是有机会断,十铺以外则机会较微,愈长则愈难断,这纯然是溷合或然率道理所在。, k0 T8 A: S" y
' C( O9 H3 N8 h换言之,没有永不断的缆,即是代表有必断的缆。非常正确,想缆不断可能很困难,但想其经常断就不成问题。特别是四铺或五铺的缆,差不多每一盒牌都会断的。因此我们就可运用这个逻辑,或然的东西难赌,必然的东西自然是易赌,下注的方法就是:
& |+ e& j. ]0 W8 f' v8 C* i: s1 \6 y6 Y6 g1 U, ~& [0 I
举一个简单的方法:; d- h" ~4 S7 v0 [$ i
. q# I% e. W$ x2 g
B B B B B
0 \! p* |, K# R1 s) o( O6 v3 ~6 }5 {' O; ~0 f
庄 庄 庄 庄 庄, }0 \0 N2 ^5 @9 l
- r. R) n8 w' J% w0 N) A6 Q5 e- H7 m6 h
正常的买缆会是:输了第一口买庄¥1000,第二口买¥2000,输了第三口买¥4000,输了第四口买¥8000,直输四铺后,第五口买¥16000。这么一旦连开五铺闲,就马上断缆,一输就是¥31000元,亦即要赢回31铺才可打和。这个方法基本上可以断定,长赌必输。因为出现连续五铺闲的情形多的是,一盒牌内断三次都不出奇。
! Y7 [# z# n( |+ X0 |1 L( i6 G+ ~8 {& g G3 S
要赢就要做刚刚相反的动作。每铺都是买闲,如果开庄,输了又继续买闲,反正每次输都只会是输掉一个起步注码。如果开闲的话就不同了,第一铺开闲买中¥1000元,第二铺就买¥2000元,再中第三铺就买¥4000元,中第四铺就买¥8000元,到第五铺就买¥16000元,全部连中五铺闲的话就赢¥31000元,一盒牌内能中一条缆就不会输钱,中超过一条肯定赢大钱。
) C/ z( y5 I- u$ {! K$ x) R& _
! f3 h4 m3 J+ J% F: K6 a经本人统计,平均断得最多的当然是四式缆(四铺连续),大概每盒牌有二到三次。五式缆则平均每盒一到二次,六式缆则每两盒就有一次。七式及以上就不会是每盒会出现。所以我认为五式缆会是比较可行,既是常见,也可保持不俗的利润。
, W. p+ }/ j, }- w+ | K. x) e5 h' q2 Z" ]9 [% ]
赌这种反缆的方式,是需要相当耐性,必须要坚持信念,相信这种现象一定会出现。输了前段的不可气馁,一次遇五闲就可连本带利赢回来。另外,有的时候在一盒牌的前段己经出现连续五次遇了闲的现象,那利润己到手,已经可以再等第二盒。
# M! D& W! w1 p) W. h
& H6 b$ a# {7 F4 r问题是如果长期只赌一方,庄或闲,那会令其他人觉得很奇怪,所以我提议一条公式如下:. B/ c* l: z# x: s, d
% E5 H: N+ h/ `B P P B P6 V- b d" u+ M
/ g6 u) h- M4 G0 s& @
庄 闲 闲 庄 闲
. S' a1 Z, ~) l* n- @2 q% u3 N, ?! N& w+ J+ L$ {
依以上的次序,按序投注,赢了第一铺庄就买闲;赢了第二铺闲就买第三铺闲;之后再买庄,到第五铺又买闲,全部中齐就重新由头开始过。不要担心,这个排列次序,差不多每盒牌都有出现。8 l% P N G, j. b6 {4 b" A( b
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