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[lastpost] => 1820252 #####给赌友们介绍一种百家樂策略##### 1760925785 一击即中
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通过我几个月来对珠路理论的研究,发现了一个有趣的现象:每一条大路,按2珠路排列,有2种不同的路数;按3主路排列,有3种不同的路数;按4珠路排列,有4种不同的路数,按N珠路排列,有N种不同的路数何解?我举一个例子,假设 B= 1 , P=2,T暂且忽略吧。
7 {' a3 ?- I9 W* }
1 c& S# ^$ `& j3 O& D 假设大路开这样的路:
+ [: Q& g& z, I
0 B) b h( G( S- Q4 X! U8 u 12121212121212121212121212121212。/ n8 _: s r$ P8 T/ U- }* [ a
5 M% |" D z0 y; P& E
按2珠路,就是BPBPBPBPBP。% U$ t2 m+ m6 u4 q; k0 U
: t0 ~) Y( n, I 如果我们去掉第一口,就会出现完全相反的结果:' S4 }% s6 L8 m! ~2 w2 C" H c
( o3 j; l/ S4 L2 \, H1 S) e. U0 j2 N 21212121212121212121212121212121。: b, l- ]8 q; e# P
- a( u+ z- b& j4 ~6 g# O+ _$ {& X2 k
变成了PBPBPBPBPB。
% D/ `# `) c9 ?& L2 K: P
+ h# h; v' |' L8 e K* ~ a 如果我们再去掉一口,又返回第一种情况了。0 t7 h6 S/ v( ~- h
8 j* _7 E+ l, ^; i 所以每一条大路,按2珠路排列,有2种不同的路数。
( `' B: w( x5 G4 `+ a- f) l- z0 U$ w: K9 a8 u7 o, d6 o
再举一个列子:0 q; R% W# F/ g @, T
2 b; s& A h, m7 s' e 大路:122122122122122122。
5 \5 g/ h5 b4 Q5 a/ y8 `
# Q. T4 I+ V# o+ P% [ 按三珠路排列:3 l& ` N. Q v w7 X% m
" s: }. x9 B) K5 Y# m 122,122,122,122,122,122。" S7 L- w5 J2 \1 g: {% k
3 P. D9 }3 h/ a2 y# ^& } 去掉第一口,变成:
, Y; f5 r* E) H8 L* s. P7 p
- Y$ ^7 q; v- Q2 v3 x3 k) ^4 M 221,221,221,221,221,去掉前2口,变成:
6 X4 h$ j- F) k( ^, G$ n- E( y6 K
2 Q: a9 |, w2 E, S 212,212,212,212,212,去掉3口,又返回122,122,122,了所以每一条大路,按3珠路排列,有3种不同的路数。- L# P8 ^& o& I3 {1 ]
: B6 {2 W* f+ y' B! n/ U6 \; r
同理:按N珠路排列,有N种不同的路数。. q3 _+ a. s6 h! U* u
1 W6 {! u2 M- e0 N) @$ K7 E' F 我提出这个的意义在于:
1 z$ h! {5 B3 t& t/ f7 Y+ @" B u- K7 F/ `6 W& M* |
1、字串81、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。
% ~/ V4 r' I6 m: A6 Y1 k) q
0 U* N I! c% @) g7 m- M( r 2、为三多理论提供了下注的多面性奠定基础。
9 }- B( p; ^: N. w- i- z+ x( Z7 M8 r% l" w2 ~$ d2 L
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。
. [; r0 I1 V& O/ A, D) P% @. _3 e) V
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我现在还是研究一下理论打法,感谢你的分享,我也来学习 |
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