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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。' O' v+ O& P2 O7 }0 \
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(1) 首先写「 1 2 3 」的数列。: A# k! r) J/ x
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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) 。( }1 r+ g/ \) Z+ ?
9 B) r" u( ~8 P4 Y* _: T& `0 a(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) 。; h X& w) X% N, n. D! R/ F5 q3 J
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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。- |. m$ u/ P/ Y9 R; E9 u
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(5) 如果数列所有的数字都消了就代表结束。 V3 Q" W$ j# \- ^: ?. G$ C
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(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 ) 。. T0 N* [3 J0 c$ G, B2 |8 c
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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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6 G; S# L1 d$ _. B$ U: j( O: `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？( Z2 x) i, ]' K+ e0 I
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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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! `: t0 d9 b) Z( c这方法我试了不咋地，你们可能可以?5 ]9 k0 \9 t# @6 ^( t2 {! W) P
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