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[lastpost] => 2804914 【吉祥坊】2月9日投注流水54668元 1770624995 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。5 e# y. U- l1 A& M; n3 k. L
% @% @8 Q z5 g4 Z- @* D/ C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?; F# q4 I2 v x2 q' ?# ~# { J
1 _8 ]1 v+ A' a* K1 O4 {
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:
, |1 r" \2 z8 ~4 Z' j
9 g/ r2 @/ L" E( T1 g. s铺数 注码0 {) s4 m2 ]2 P
2 Y# u) |* n5 O" @6 X% e! Z, E1 ¥10007 V5 n+ v. Y, Y7 O
+ B6 B6 ~# X; K
2 ¥2000
# j: c$ B) O! d) c; p: ^4 v H1 W( d" z. X- v
3 ¥40000 G/ j: l& y( x% J i/ t
6 c! j3 J& h! F! t' W4 ¥8000
+ ~+ ?7 i8 V# ?: g0 T+ N) T3 h" [- {* j* L
5 ¥16000
. |1 ?) Y0 C1 @; E ]6 |9 R. S) ?% P x) F% F3 C p2 a
6 ¥32000
k; L8 y" u3 Z t. s- U
" J; L4 G% B. Z$ F3 W7 ¥64000% k& f5 g: ^# K2 {1 h
# o6 Z q2 X0 b* S: m8 ¥1280003 {( A3 ?& G1 k0 r% |, T. i) [9 G
5 ?: A9 u) k9 [6 H# R% U: K2 q9 ¥256000* R. q( \/ e/ W
, C1 R( z1 b" {* |$ N* E( x' Z, o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。7 `: k6 n* m2 N
3 }/ H3 C0 Q7 ^7 H' L' J4 ~& R
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。
( a! ^) d* h& _% h; T8 O2 g5 @, 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:: p% N6 q5 h( U2 |0 ~& \
, c8 @. ^7 ?0 J) ~1 T- @+ k举一个简单的方法:
. c' W3 p5 P- x! ^& Z+ T8 X4 @8 E0 K9 z
B B B B B
4 q* n p" t) ]% \, d& c5 {2 ~; S1 L8 `. \5 N# ?6 E0 i
庄 庄 庄 庄 庄: w$ a5 H' S& C% H( n
( |! p6 v* K1 F( V( M# G, U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。
+ H6 P2 v W- \4 D, }9 |* W: Y( f8 k. L' Z: o, w! F" B% Q9 O
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。
+ { \! e0 N2 v
( T5 W" }. @( G. 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。7 `2 }. X6 b2 o' F- R- Y5 l
! y7 F0 t! a- |8 J1 F( e2 D6 U8 q
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。
. k% f9 G, {1 T5 D ]
. ^. J: \3 j' c) y. V! v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:- |& e" ~1 _0 }0 b8 A2 o& O/ e+ `
0 |4 @2 X! G% r& F9 K
B P P B P
: I, ~' @! ^5 a: [, |/ N w
- E [8 j/ F# k! u' T庄 闲 闲 庄 闲
/ g# [1 K8 k' r5 T$ b. U- @# b7 C0 n5 S2 }5 Z5 a
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。# }: t8 s4 y7 @& G$ }6 K
5 J, o9 E0 ~) {8 c* o& b9 E8 x
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