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[打法交流] 百家乐珠路理论技巧(转)
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通过我几个月来对珠路理论的研究,5 R+ y* d$ ?  k6 I7 G/ O! V1 z. T' N4 C. ?# c7 L

3 l5 O" ]3 n2 V发现了一个有趣的现象,& g; `9 r/ a# C
每一条大路,
: t% `. X7 v% Y按2珠路排列,有2种不同的路数1 _% l2 j  Y8 @" u) {# f
3 J% b( D  Z% y# j* }/ i按3主路排列,有3种不同的路数& n( |2 F1 f; ?, q' L3 A: Z4 z" G" ^0 g$ z6 n% k
按4珠路排列,有4种不同的路数2 F/ h# j& f% m& J& P% p  [+ f
按N珠路排列,有N种不同的路数。: n1 f3 R9 @  p% ]* C7 \% r7 b4 D3 y2 ~, n
什么意思呢,可能大家一时不清楚,6 j) R- O0 i2 K* ^2 N! @
我举一个例子,假设 B= 1 , P=2,T暂且忽略吧。7 f# f3 c1 ~2 l! Z! O; ?/ D; `
假设大路开这样的路:! z  ~0 G3 v. K' o5 A4 ]" }5 w# u
  X5 _  J0 `4 S12121212121212121212121212121212, j0 j; e6 p* q3 Q5 h
' ]) D5 H& o  Q1 L够极端了吧; j% j( W% \  R8 a7 T
按2珠路,就是BPBPBPBPBP....., m8 f) O! x+ S9 L2 x) F
% R4 w1 h- |! w  Z- D如果我们去掉第一口,就会出现完全相反的结果:
) y* X/ _" Z& e3 o3 m( d. X8 F0 ?21212121212121212121212121212121
6 p/ w! h4 g2 U7 u4 P3 A- r  r8 ^1 x) [变成了PBPBPBPBPB.......1 u5 g6 c6 F/ L4 g! f7 Z# E+ C0 L- {0 b2 a2 ?! M* ?% ^
如果我们再去掉一口,又返回第一种情况了。! [( g, L- u! O3 @# G4 D) O1 B& k8 Q& `: P1 i  I8 u/ [
所以每一条大路,! R6 V# q$ h. c; z% O5 S  S# v8 N8 w0 K2 p4 O) O  O1 E7 z
按2珠路排列,有2种不同的路数。
0 y4 b8 @9 {% U$ a$ l! j再举一个列子:
* n2 h5 O6 D; i1 P- n大路:122122122122122122。。。4 S& Q+ m  h" o7 C7 M# i2 R
按三珠路排列:
. Y% q$ T: z. [122,122,122,122,122,122。。。
& S: _0 A5 u/ b$ }9 k  C* v去掉第一口,变成:6 G2 V! \" L' c" k: d% S0 i1 u: }( i; y$ r
221,221,221,221,221,+ R$ P2 d/ f7 Z, P5 M6 }6 B- V( |2 h9 W4 F3 G/ i
去掉前2口,变成:3 v1 q) z( ?2 f! }) U/ W
8 u7 i+ @6 d1 P212,212,212,212,212,: y& l, B; I6 e- m. J# c) `$ f* G9 B6 K) R: A; t2 I- A  D
去掉3口,又返回122,122,122,了$ w1 M) e( \! v% z( y
所以每一条大路," R( Y+ l' C& Q
按3珠路排列,有3种不同的路数。
: g. E7 z' t1 u& ^3 E同理:按N珠路排列,有N种不同的路数。
1 e6 l, Q- b9 N. J( @我提出这个的意义在于:" v) X/ h3 }* D1 {5 J! a1 s. O9 y5 {$ m, L4 d' s$ l6 [2 d% ~
字串83 L: R) G1 V- s2 Y+ Y, \( b1 \6 f: L$ l
- p, d3 [6 U. s5 J' [0 {1 A" m' k( @: S, W0 s6 j) T& |3 u
1、每靴的第一口为起点来编排二三珠路,与第2口,第3口开始的排列是不同的结果。2 g: X/ W# h7 N" o7 |9 l* |
- ~# D3 X( r) Y5 x3 Y  @5 D$ z& f1 w+ L( Z2、为三多理论提供了下注的多面性奠定基础。
# ]$ ?5 b+ p$ X3 {$ E+ ^3、某一靴牌,按第一口开始的珠路可能是 烂路,按第2口,第3口开始的珠路可能是 上上路.
% `; M3 ~: Y4 x5 ?" R+ Z$ D

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