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[lastpost] => 2775417 【星宝】10月21日投注流水166330元 1761119021 rainwang
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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 d" n2 q! w1 N; f9 Y2 u( q
; {" E3 \; a0 o* v G5 O3 u" c! g. W 假设大路开这样的路:5 w! Q: ^ E( G7 q+ Q
7 t& K+ K4 I0 e4 `1 z 12121212121212121212121212121212。/ G$ ?: I5 ^* m3 S* a- U' _% i9 {
' N6 x$ F$ ^6 k! @* k
按2珠路,就是BPBPBPBPBP。
: O# H R( ^* g1 R- `, J% G( M- v% ~/ v, I: r. k2 y$ w
如果我们去掉第一口,就会出现完全相反的结果:
( d4 T: ^: i) a$ C% j7 E
1 R( x( A2 |! @$ _$ J 21212121212121212121212121212121。
0 y" \& j: t# m- ?; g0 U
5 T0 C/ E" q+ ]8 ~6 F: k* f- k 变成了PBPBPBPBPB。1 C7 m7 k o' e% T* S- Z
, L# Y$ P: P9 @: Z" n+ \ 如果我们再去掉一口,又返回第一种情况了。
% H0 b9 o4 x, a7 a
: s) c( ^% f% W6 g2 ]' d! q! H* J 所以每一条大路,按2珠路排列,有2种不同的路数。
1 G+ M- Z0 u& i' |6 |* \8 l7 x4 }
* U% p$ @( q$ v 再举一个列子:! j# N7 J7 n' O$ n3 a/ u
( ?# \ b* B9 D. f U 大路:122122122122122122。* w8 b" e" K8 @
( U" d! \9 ^5 n$ ]1 d; L' g 按三珠路排列:4 y p h7 T0 k" E( k# r8 e, H
7 c: U, {/ z S: r: j# w% W/ L 122,122,122,122,122,122。( D$ ~9 p& q; y" V' J
7 \1 C3 w( v6 y" C" I+ K0 M 去掉第一口,变成:% Y5 t7 K$ h! p# D w6 f" | K
' l8 t9 z; X3 ~/ t4 Z! l; f 221,221,221,221,221,去掉前2口,变成:
' o) k' E/ |. N \+ F% p0 A& E) M# ^7 P8 z( }1 A+ O4 M* g1 ]6 l
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。) ~5 L6 k A7 {1 L8 x
7 d W% i& \! X; e 同理:按N珠路排列,有N种不同的路数。
0 b# {8 Y/ P& g+ _' i# G' b! x" e) m7 n
我提出这个的意义在于:( k9 S" z: [3 {4 l; k
: E( J/ o* N6 W0 X2 I# k% S
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。
1 m' o9 s! B& X9 ]9 ~" v& J5 _
3 I9 G5 {+ ]% T: y, W# S. F/ { 2、为三多理论提供了下注的多面性奠定基础。7 s0 [2 ?- ?/ r3 T \2 p( X
6 b7 Z! g; }/ a6 z% r/ q+ e7 T 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。7 [- c1 N2 f9 i2 @, r4 |) i+ g- z
' P0 }& i, x3 q7 _
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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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