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[lastpost] => 2796704 【星宝】1月13日投注流水30000元 1768440545 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。2 u5 m5 S6 K2 ^9 i' W
9 c' b7 X1 t% z" P% W6 Z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?
) A; M. U" M0 a# @+ m
. s3 j3 _) c% i6 J0 V( m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 K t8 Q, P+ l
6 B+ O. H3 x4 U8 W9 ?! B% A' W4 r铺数 注码1 C! P2 I' I+ ~8 Y( {7 ?
/ I/ @! q" e( d% u I
1 ¥1000
8 D& l) \6 r; e5 g4 J/ `" c6 A g2 k+ q
2 ¥2000" @7 u( m7 B% l- k- n# k' O
: q$ P) V7 l, a, V3 ¥4000: | E; T; r0 l# }# t0 N
. Y+ I( _5 j7 {3 n4 L4 ¥8000
: x" K7 M6 } ]. G! p
3 L3 \/ |4 i; a- J2 X% c5 ¥160009 r% A2 m8 s. D) O
/ C2 u2 t3 P! f
6 ¥32000
1 `# q& i( J' F, k. z( y6 o. N8 w. b
0 q& Q8 o% G' r' Y& F7 ¥640009 @+ L6 w! p8 n! X, E) R5 P- |# J
' _; M1 s4 M" Q4 m( e$ c+ Y, s f
8 ¥128000: z2 [6 `0 t% \
c1 v/ J" g) L
9 ¥256000: h5 ~) }4 F5 A+ Q: R6 k- c; P
8 O5 f+ H1 `- E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。
1 W( q8 }7 L% T( g6 L, [; K
3 d/ W, |4 B$ }; G$ O8 ]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。
' d, A* f0 l3 J, x; X& F5 d; p
% b: j# |% W6 b, 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:
0 }0 ? y/ @1 _+ e4 F; |9 A2 q& ]5 O
举一个简单的方法:
( T0 U O7 b$ k! h; `: D- ^2 J, u$ U. Q& ^* n" R
B B B B B+ B; h" Y- p: Y" i
, z- {2 K0 k# f$ n' b, U4 [; y庄 庄 庄 庄 庄" }: P; P; `$ v
6 ~% i5 b# Z/ o, ^8 `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。9 Q Q# Y- T F: D3 z2 N" @5 Q
* g; x$ C) d! J8 C4 R" V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。# i0 t3 V4 h/ ?
9 J& D0 u8 O3 {' Y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。# J4 i6 ]: ? O/ v H$ T
; c1 Z' W* n! p# f7 l: T4 K- t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。
/ f$ o" j" @. q y/ S& I+ s ~' G: e |" e6 y9 s0 S
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:. ?$ F! N' \9 R9 D/ Q
/ ~0 Z2 p, N7 W# \% i, vB P P B P
9 F4 o4 c( H+ }: {2 X$ N0 J9 C1 X# m
庄 闲 闲 庄 闲# ] x& C6 `( }+ ?# q5 Y
! C* o+ t% c$ K, O6 f
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。1 `; B; }+ _ u, y: M
" \) Q. V- y) n! V# S4 T& B) d
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