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[lastpost] => 2796107 【星宝】1月11日投注流水21500元 1768204475 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。: z7 P1 Q# L G* ]/ R7 I% C
/ f) h$ L. d& W3 ~8 n1 K# 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?
; M2 r4 X/ f% m' A& p& `: O, I4 c! `. X9 M$ q: {0 e( _4 J9 L
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:
: I: h f4 u! R) d( L; b/ M( E) e' I
铺数 注码
9 J' u9 N- |0 x3 A0 }* Z/ L
2 t4 F. b: Q7 `; i& K1 ¥1000/ P% F8 E0 {. m5 j
6 U6 M. x( S" T; @& Y/ y
2 ¥2000- n' r+ k: M! P5 v0 v2 k
: l8 u/ v. k9 r9 A- ~! f3 r9 A6 F3 ¥4000
# F4 ?; r+ o0 M6 d! B* ?' a; N% w0 W6 A% a4 B- D! M
4 ¥8000; m5 j P- p! i8 s8 c
" a q0 F+ k- D: |" D5 ¥16000, @5 B9 F, b% ]7 m5 x; j
* ?* [. ?8 y5 v+ C( [7 o- X
6 ¥32000. _6 Y% Z S9 L. c9 r, q
+ j" `& k7 X5 ^, N4 i7 ¥640000 |, U# f4 _, I4 ^
' P. k, Y- c. s( d& X8 ¥128000
& n9 H5 i4 H, N1 L! c0 _. {3 c! L
9 _$ v4 ]& x/ e% z" b6 b9 ¥256000
1 @+ J+ G5 `: |. }7 t7 p1 ` s9 v
! f# N7 `2 N" R* Z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。) D8 \+ O- S2 O R9 k+ D
" r. P/ W) K' J4 _
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。
1 I7 ^8 m: e5 R1 z' V# D9 q0 W, f; b% @. P, k$ w, `+ N
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:4 f+ p r' A# ~8 D: |9 _' E
( f, o/ c1 w' [8 Q
举一个简单的方法:
! u5 e7 K! w3 L! Z# r' U9 p. a; G- R
B B B B B+ e, p, g, u9 @* p0 s
) p7 P6 n% s! y0 |/ j庄 庄 庄 庄 庄( ^5 w" r7 i2 Z/ O! x( y
7 ?4 G* R. _, j( L# z5 R, Q
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。
2 @' M+ v& I# R% V9 d/ w% V5 @0 ]. d+ x& C5 E0 B' n3 \0 \9 b$ E+ I
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。; s5 Z6 p) H. }9 S. q
( W0 q! d( D. ?5 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。. f" T: ?+ O/ j Z# U) v
7 |1 v) `/ R. {" V p
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' [/ k- ^& Z( C' L5 i C+ Y% A% y2 i2 B
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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:
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" Q2 V. f$ Q# y6 B& h! o+ I/ _B P P B P4 U! L7 K( }1 c
8 _3 ?& P: X/ {
庄 闲 闲 庄 闲# ?" a3 V" z0 `2 ^: j) X
$ _" }+ Q" C: G7 w$ d- I, ^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。* P4 }, m: u0 t) C9 G' ]2 [
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