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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/ O1 E3 F0 z' }" ]$ r9 s5 a
( ?8 H9 b  v6 N
发现了一个有趣的现象,
# s9 D+ q( e  @4 _3 S; W每一条大路,
, D" h$ F* ]) G, O3 p按2珠路排列,有2种不同的路数1 _% l2 j  Y8 @" u) {# f
& U9 l/ b/ E; O5 d: {按3主路排列,有3种不同的路数& n( |2 F1 f; ?, q' L3 A: Z4 z7 [1 s* }$ r( m& v. c% n
按4珠路排列,有4种不同的路数
2 O# {' y; J* {# p& H" u6 v按N珠路排列,有N种不同的路数。: n1 f3 R9 @  p% ]* C% q1 l8 ~) k1 s8 m+ d- _$ L
什么意思呢,可能大家一时不清楚,4 \) B$ [' j" L; D
我举一个例子,假设 B= 1 , P=2,T暂且忽略吧。7 f# f3 c1 ~2 l! e- a) u- g4 F- R( T6 S5 Q6 j% o
假设大路开这样的路:! z  ~0 G3 v. K' o5 A4 ]" }5 w# u
: L  d" S+ P- n% u0 ^1 P12121212121212121212121212121212, j0 j; e6 p* q3 Q5 h" ?1 V/ C* b2 a+ U
够极端了吧: i/ u3 Y2 _# v( l7 y8 e
按2珠路,就是BPBPBPBPBP....., m8 f) O! x+ S9 L2 x) F, d* S& y& u- ]) F! e
如果我们去掉第一口,就会出现完全相反的结果:3 }7 Y3 S& a- c2 F- ^
212121212121212121212121212121216 y; t% q5 H- N5 ?
变成了PBPBPBPBPB.......1 u5 g6 c6 F/ L4 g! f7 Z# E+ C
- E* O- ~( d, [3 v如果我们再去掉一口,又返回第一种情况了。! [( g, L- u! O3 @# G4 D) O( D- l3 c/ k& ?8 c( E+ w' w$ H( ~
所以每一条大路,! R6 V# q$ h. c; z% O- m+ [7 P4 Y: A" E) e( z
按2珠路排列,有2种不同的路数。0 s& L  N9 }; d* }5 Z# ?, H0 v4 g
再举一个列子:
6 S8 C0 W: X2 [" ], Z大路:122122122122122122。。。
# o' A+ H: n7 \9 G' j" |: `6 Q按三珠路排列:
' o, p9 f8 Q% h. Y122,122,122,122,122,122。。。8 n# o$ U& y' d1 O+ G# X
去掉第一口,变成:6 G2 V! \" L' c" k: d
/ V* j9 ^7 K0 W* ~$ A+ C# t( F221,221,221,221,221,+ R$ P2 d/ f7 Z, P5 M
3 l- Q! Z5 L4 g6 f7 d去掉前2口,变成:3 v1 q) z( ?2 f! }) U/ W
0 v2 [6 E8 V- I6 i4 s; p  z212,212,212,212,212,: y& l, B; I6 e- m. J# c) `) x* R0 h9 i4 O" L" M
去掉3口,又返回122,122,122,了
* a4 x& B! s) H3 U所以每一条大路,
) _2 n1 [! [6 @/ ^+ G按3珠路排列,有3种不同的路数。
- ?) Z) [4 |7 o% b同理:按N珠路排列,有N种不同的路数。* g9 N1 n, _1 u! x* T; j4 }* z6 q2 V( N
我提出这个的意义在于:" v) X/ h3 }* D1 {5 J! a1 s. O
3 {9 G: n# E+ ~字串83 L: R) G1 V- s2 Y+ Y* i! s4 m" }# V2 L
- p, d3 [6 U. s5 J' [0 {1 A" m' k( @3 p- S7 Z1 N% \0 ~+ h' [
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* |
, \% n7 d8 }0 r+ \2、为三多理论提供了下注的多面性奠定基础。
( k/ f" j" l/ `. ~8 X. X3、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.
% Z% t) L, }0 Y0 F; l  ~

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