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[lastpost] => 2808623 【星宝】2月23日投注流水23702.9元 1772342396 舞出精彩
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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, ^( _% u0 T" c
0 R% c: V4 w. ~6 a2 s% j% b+ E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?2 J6 ^8 j; B, e& j3 W$ F. K
) G: f, |0 s- p0 n Q7 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:
$ e, F# Y2 l9 ` F# _: p( Y" U4 N" ?/ B. f8 c, _' W
铺数 注码
" l1 z T1 l* E4 R2 o# q2 B2 F: }
" r! a9 m) ~0 z/ m) e3 B! O0 m' T1 ¥1000
" R& l- b3 b) _+ X6 ^2 O
0 d& Y1 j W7 @! Z2 ¥2000
! m3 A0 J. p; l
7 J" q7 C3 P' N9 C3 ¥4000- p: T/ F& t+ I$ x5 |$ D
5 D6 e N2 `& J Q
4 ¥80005 s% ~* \8 H) M* T3 K$ b( S
2 q5 O; _ w9 Y5 F( n
5 ¥16000+ ^! t7 H) d5 n( g7 C
! q" `; Q6 P$ \3 }5 K1 |6 ¥320004 E2 u3 u* n* K4 e4 S: N
1 \" T B# v( L' ~4 L r
7 ¥64000 M% k. h% |0 S+ Z3 U5 z
# @. i( V: _, J: l
8 ¥128000
4 L7 c+ ?" R5 l, h, Y# H6 D, ~
1 P' w3 |' x8 |+ @7 v9 ¥256000
/ t5 r% j/ Q8 ]- Q4 W
4 {+ j1 T, o' s" D( N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。
& R& X- ~4 p8 @% u# F2 G; Q8 j' G! ^3 X9 P; Y
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。" n! |) t8 z3 D. b! S3 P' {
) `6 _) } b% i) t8 U8 u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:. X6 i D. B p2 C p/ G( [& G
( n2 [5 {1 ~ o) U举一个简单的方法:$ i0 ] i- [' S/ M6 N" D
# _. N, X% U6 J, T, FB B B B B/ V# w7 q$ `+ A/ L
1 g* @9 r1 {' i庄 庄 庄 庄 庄8 B6 ^0 `* \% G' u+ _! `
4 \$ D- g; s4 [, o, r9 V2 }& X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。& U9 P0 o% e9 v& l0 g
% ?7 P1 f* R/ P: J
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% G D9 C3 h4 I" ~5 M [% O3 }
+ S& e1 ]7 C* k# o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 h% Z0 I* o1 `
3 z5 A! J7 T" @4 d- ?; h
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。
2 K; t5 n- `4 \0 Z
( H, x! D; G- b. S, _( w2 f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! x/ b; Z# o# ]! F& A
$ k, A X; Q% D3 X5 t% j w, f# S# vB P P B P0 t% o6 m6 T4 E$ C
1 E& L1 ~/ I: Y/ S$ Q" H3 L% z
庄 闲 闲 庄 闲" i/ {! y- X! O! J, Z/ x/ d
$ t- F) F* |, E% [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。4 Q- M$ Q7 n) s& [
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