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[打法交流]
近期在网上研究到了个百家乐著名战术 蒙特卡洛战术
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百家乐蒙特卡洛战术,需要纸和笔。但是不很难。1 m0 T& t2 b# O& e
% t9 u" t: T! V+ y(1) 首先写「 1 2 3 」的数列。
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) L$ ^3 Y; p: p) S* p3 L5 E, p D(2) 赌本就是左端的数字 (1) 加右端的数字 (3) 即是 (4) 。* ?1 y% V9 }1 i
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(3) 输了的话就在数列右端之外又加上顺序的下一个数,现在这个例子要加的数字 (=4) ->( 「 1 2 34 」 ) 。如此类推,并且重複上面的第 (2) 点 ( 但今次会是 (1)+(4)=5) 。1 o/ l/ J# B1 P" W
7 ^: T) U+ q- h' U(4) 胜了的话,倍率 2 倍的游戏就把左端的数字 (1) 和右端的数字 (3) 消去。倍率 3 倍的游戏从数列上去掉左端的二个数字和右端的二个数字。( w% u% [, o2 z, i# z
+ i) [/ b9 d* e2 D: n: I
(5) 如果数列所有的数字都消了就代表结束。2 B) k; v' b. A) q) P. Y
; I3 Z6 ~' q2 J5 w- K, s3 |+ J' f(6) 想继续的话,回到 (1) 。$ `/ O. u* O7 X7 ]7 t
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蒙特卡洛战术的问题,在于如果没有充分而源源不绝的赌金,就会出现途中被迫停止的可能。而且连败的话,亦有机会超越 Max Bet ( 赌注的上限额 ) 。& t0 u6 ?8 J5 q/ M! K8 k
9 c" C/ X* h4 V! [1 `& [- @3 k二合一战术. }$ X( _: G' F8 T: [$ I
! G. X4 D2 ~& i+ k5 _+ V这个可说是蒙特卡洛战术的一种,是以 1 局取回 2 局所输的方法。方法简单。写下输了的金额,赌注就那个左端和右端的数字的共计金额。
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" z( L+ t9 N" L) E0 O: X譬如,先写下「 1 」。所以第一局的赌注是 1 。如这局输了,右侧添写 1 ,成为「 1 1 」。那个下一局赌注将会是左端和右端的共和 =2 。
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反复这样做,如四连败,数列将成为「 1 1 2 3 」。因此下一赌注会是 4 ,而且胜了的话,消去左端和右端的数字,数列将成「 1 2 」,下一赌注就等于 1 和 2 的总数 3 。再胜的话,会变成?
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9 x" P0 q( m) o- N0 |* W* j儘管结果是 2 胜 4 败,共计收支是一加一减等于零。
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这方法我试了不咋地,你们可能可以?! V1 f! v3 ]/ k1 R/ f
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