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[真人] 百家樂終極破解-方法編 第五章 <五手算點法>
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本帖最后由 wlcomp 于 2011-10-16 09:09 编辑
2 K& n7 A! R8 j$ d" Z
, C/ k% w! e) \: t6 @8 z引言:
7 z. `' w5 k( s* _本章是第三章的加強版, 因此讓大家先看看第三章內容.

8 _" }2 I0 S0 F. H) [. R
" w" j" D+ z: {1 q0 U第三章 五手統計法4 v8 r) \5 s6 O9 {1 d
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引言:
0 V. h' [, Q1 K  x- ~) ^' u7 S首先你明了理論編一至四章,並且看了方法編第一章,O其實代表單數列贏,而X其實代表雙數列贏,在第二章用20手來統計,有人說太花時間,所以我就用五手來算兩手. 在理論編第四章提過, <統計每段數據的變化, 預測近幾手變化>, 五手統計變化十分大, 所以不用人手算, 直接用電腦算十萬局(共七百多萬手)來找出結論.
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統計方法:7 r" d$ \6 n4 y" o& ^- M
1. 先要知道那兩張是單數列,那兩張是雙數列9 o- N( k: c% ?" ?5 a
2. 如果V在閒行上,閒頭兩張牌就單數列, 莊的兩張就是雙數列
6 g7 L% p- h; G7 }" K# T! f3. 如果V在莊行上,閒頭兩張牌就雙數列, 莊的兩張就是單數列1 R! g. F* Y" p) r3 i% r- z
4. 把各列頭兩的點數記下來(注:五,六張不用理): T( O3 ]: O1 I- J2 e9 |: e0 g6 D
5. 統計五手, 多少個7~9點(注:開和也要統計點數)
5 {, S: w) J. E; Q6 {; W) x( n6. 用統計的結果來確定買之後二手% o$ f6 y' c( O6 }5 Y& Y* ~, a% N

8 A4 J9 ]* t0 F5 s! j3 k) x& V例子;V在閒上, 閒-3,4=7點 莊-9,7=6 因此單數列是7點,雙數列是6點

0 e' v& J; X  t
* f( a" c  b3 c8 s0 f! x4 D第五章 <五手算點法>; u- |6 s7 r7 G8 T' s

& Z" E1 V" ^3 r分組法:
! o3 P3 m7 N' L4 T8 ^1. 把牌分為四組, 0點,小,中,大
  I$ o# @# W5 ~) P2. 0點就兩張點結果是零
5 S1 J4 X1 S7 \* E3. 小是123,中是456,大是789

% u: J5 O# o0 t4 N
) e" {2 H. \8 u, e以下六十萬局, 四百多萬手的結果:
; y9 w) g1 u4 H2 s  x5 U
7 g& y' \) R. J% ~對本列有優勢的組合- : f* k8 a! I2 L) O, T4 j
(本列= 單數列點數對單數列結果, 雙數列點數對雙數列結果)

- Z  n1 G' {" B<優勢由大至小>
9 R0 w% j- P, {& D$ O% F8 r4 U; f) s零小中大# I/ @7 a4 e2 F. T$ s
10042 K6 O3 P( w% u9 A! j
00050 \( G3 H& ?' o7 H3 C( G
0032) i8 }- q, V. d8 L" e- A0 I* M/ I+ d
0500
5 x- ?% n  U3 k% ~5 T32007 M. ^* A5 z6 A# }
5000
& X$ {) }6 e+ r" J( X. `4001
: m, d+ {8 V: B0 y/ x! O0 o8 R1400. K1 f4 `3 A! T. ?
1022
* q2 Q. f5 t1 |8 f6 z: ]3011& H) W' j5 C# Y2 G& p
1112
( x9 o# {* \7 `7 p, E" W" `0023
& m) R' p1 \* A% e3 Y" A; r& N2102
/ m& |2 m- \: i- j3 o20036 U5 }. Q" y- A) g7 S
20120 B  N) V5 w; }% {- y
1103
5 `4 \& o, J2 C0 Z6 Y( }0113
. l" [  d- `! W9 k' p& n0302( j/ B8 F" g1 a( Y8 a( _* s
3020+ H( L; ?% I, x# D7 e+ F
2030
; Y! m0 M" `! T& F' [( X6 u0410  E3 P) R: s  g$ ~' r
0140
; Q+ H' d8 ^9 G4 \$ h1130
! H1 R% X+ p, ]& V/ c- e/ I8 Y0 E0041
6 u! ]( a& A) T" W% A01228 h3 V4 k) J0 d! R
1031, \" W) {2 r: j  X8 F% \
0401
& C: d9 f$ h% q% b% `& E2210
& p. u( c5 c$ h1 `! \: d5 H* g- Y* L21203 U0 |: m9 X/ t3 f$ X5 ]: b* ^# l  k

, {9 U* ^. k3 X5 u
7 b8 }4 U6 V5 o; G% ?# h對相反列有優勢的組合-
: g4 X% s+ L& {8 ^/ M" P<優勢由大至小>& E  o/ ~& z9 g8 i5 i
零小中大
8 m! C4 r8 }" s( m$ p* S; i00142 J. M6 ~; _4 D: `/ E) Z
1301
; \& F6 w* _  V* x5 A- z0050
$ N4 r; c! B7 {5 x) m/ R) [" @0104. u5 i4 R% C* Z& a3 U
4010
7 v5 N9 d2 H5 d1 m* e0221% }* D( K- A; `8 f
1013
+ w+ e1 C1 l7 |( F, B) U0 V20210 ^4 O: l+ M* k1 ~" o8 ]
02120 d1 o$ `3 S! v- C( U/ \
1040
( d* r+ C1 m) o1 J- t0320
' H; E1 c5 Y; A" t" l- J5 P4100
4 }4 K) _( r9 n+ J2 F+ U0311( n; H2 P' W9 y3 S$ ?
2300. F4 v( ~6 q+ r. Z0 D- \
1121
1 [0 t- a! h/ v, b2 n: A# S1211; s; d5 D- U' `  ~6 r
3101
+ ?6 ^2 H5 F& b5 }2 O  p& ?3110
  B% u6 R; A3 \30026 U; x' J) w/ d3 S7 K
0230+ [0 C' h) I8 |5 r: U6 O( t
2111* h6 u! W/ F. E& Q
13108 Y" n, w; M% q# _
1 g- A$ V: w, _: T3 Z2 ?0 O* H! V
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14#
学习中,谢谢分享
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13#
:handshake学无止境,继续顶。
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12#
又來了.1 e5 c  g. O) l; u+ D

0 _2 s/ }, `2 Z- g在頂
) M7 v; B% p  _5 m在在頂
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11#
回复 1# wlcomp
% E) ~! L5 w  R4 C7 _2 x3 T
* N, Q  y0 l0 [. Y) K; V又出错了
6 e, Q& g' f- l) J4 ~4 L应该是 <大中小零>
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10#
回复 1# wlcomp
! {. t4 @( y9 F% u/ a$ c
) {2 M9 U7 `, l* {7 P3 {0 e4 y1 [1 |2 u+ q' k9 X9 r
    如果楼主把你的帖子发表在一个标题里就好了
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又有东西出来了,顶下。
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8#
楼主不如真接出软件吧,我们资质愚纯,越简单越好
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7#
回复 6# c26432196
8 J) D; m" ]  c: t6 i
* M9 c. @( n" r' h* I5 H9 N" ^1 ?$ T
% F( U: r6 X1 M5 Z  }% E. Q8 U   看來有軟件, 你才會明白
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6#
为什五手 只用了4个
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主题回复处广告图案-天策传媒
完全不懂,浪费了菠菜币:'(
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4#
太深奥,,,我的质历看不懂。。。。。。。。。。。。。。。。。。。
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3#
好东西一定要支持
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2#
谢谢老大。。。。。。 :)
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