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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。
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+ n, @# ?. |2 r(1) 首先写「 1 2 3 」的数列。! H; C2 I1 [0 A/ Y
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(2) 赌本就是左端的数字 (1) 加右端的数字 (3) 即是 (4) 。
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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 k8 v; I- x/ F+ b+ S' g/ f. R
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(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。" {' W9 g q6 I
. G$ H# v3 W8 j. n. P, y* K( {0 S(5) 如果数列所有的数字都消了就代表结束。$ \/ p% N7 b) d1 e$ A1 o% d- Y
2 I: ^# n. I+ y( a: E(6) 想继续的话,回到 (1) 。
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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 ) 。
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$ n ?4 `- g& C, Y+ I二合一战术
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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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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 。7 l' w* k3 g2 w( U' ]
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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?( _( M4 s4 \% n" F( o2 g
+ J6 ]4 G6 W6 p3 Y6 i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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2 r, D* b! z; ^1 Y这方法我试了不咋地,你们可能可以?5 I( N* c6 d: h" A! u
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