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[打法交流] 德州扑克手牌组合有多少种?(转载)
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1#
起手牌组合/ T+ R' F5 Y/ I5 Z6 N/ Y9 ~
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德州扑克中可能的手牌组合数有C254=(54*53)/(2*1)=1326种特定的对子牌有C24=(4*3)/(2*1)=6种组合
2 L" b$ t; c$ e' P' C4 ^0 P特定的非对子牌有4*4=16种组合
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8 n: t" S- ?4 u" ?9 ]特定的同花非对子牌有4*1=4种组合3 S2 p( S  i4 t* q% w

& R/ }! b4 l6 C& d$ `AXs有13-1=12种组合
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4 n7 {7 E) D* r$ u' }/ E  G: z' yKXs有13-2=11种组合& ?) ~0 r$ G5 [4 g

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思考:去牌效应的应用。
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5 B" O& N" r6 o+ b' FHERO手上有一张A,那么对手AA的组合还有多少呢?4 @. }! q, a6 y6 n* v

/ E& W$ V7 f8 g2 d0 T; Q/ \答:有C23=(3*2)/(2*1)=3种,与你没有A时的情况相比少了一半。6 y" m. p$ m/ d# ?7 d. P
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HERO手上有一张A,那么对手AK的组合还有多少呢?
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1 u3 Z) A+ x8 w( I; Y, ~2 @; e1 m答:有3*4=12种,与你没有A时的情况相比少了1/4。8 H- I( Q5 M. W5 r% z4 Z

3 R9 k  H/ B8 L* Q手牌组合( |: V% y, M: N- p2 ~+ S, f0 P
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成牌组合/ r( R. G/ ^0 m) D; l% ^

% K/ V0 [4 D2 `, q+ N! WFLOP没有公对时:# L8 \6 r, k9 j  \  S6 B) c
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三条的组合有C23*C23=9种
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指定三条的组合有C23=3种- z+ |% z0 l0 }+ z$ r8 `
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两对的组合有C23*C23*C23=27种% {% r- ?+ {+ y) m+ U

- x; @* v$ Q( R* L3 ]指定两对组合有C23*C23=9种# u7 Q' ], d: B* a- t% M
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TPTK的组合有3*4=12种6 V; `2 }/ q: e0 b+ V4 ~$ M
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" |2 ~) {) q& k$ b; w5 `Flop有公对时:
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; X: @8 S( o  t( i) J明三条顶踢脚的组合有2*4=8种- ?8 `& n/ P. C- V. D
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葫芦的组合有C23+C13*C12=9种
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听牌组合9 C! X, g' I  i  p! d) |

9 h. S. A  K- }; v! F( RFLOP有两张同花时:
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8 |8 {  y' S/ r对手理论上的同花组合有C211=55种
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而当你手上有一张同花时,对手理论上的同花组合有C210=45种' ?1 h& I4 {: K

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在类似K34这样的FLOP时:
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6 e- I% C$ ^7 ^! ?& y两头顺组合(25、56)有2*4*4=32种) K8 V! M* V  }6 B  _9 y2 W

( {: q! [% Y, n卡顺组合有(A2、26、57)3*4*4=48种
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在类似K56这样的FLOP时:2 I. D9 x$ k$ I$ W, M/ }# s% o
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两头顺组合有(34、47、78)3*4*4=48种
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卡顺组合有(24、37、48、79)4*4*4=64种
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, U$ }6 b; O7 w2 n3 ^; y思考:当你知道对手只玩同花连牌时,组合数会发生怎样的变化呢?(这个请大家自己做个计算吧~)5 v' R, e5 F5 d% @4 H, l; n  Z8 n

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参与人数 1贡献值 +3 收起 理由
avatar 天策-瑶娥 + 3 理论区转载发帖奖励

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2#
这个组合应该是很多的了啊。
avatar
这个组合也是要看一下的了
avatar
4#
知道有多少种,肯定是有好处的
avatar
5#
主题回复处广告图案-天策传媒
你的理论打法不错的呀,赢钱了应该是打法对你有用哦~来收藏下了解下了
avatar
6#
感谢楼主的理论分享,也是可以了解到好多的。
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理论实践得好,收米的时机就来了。
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你估计也是一个理论很内行的玩家,感激楼主的分享。
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9#
老哥的理论打法也是有在记录中的呀?
avatar
10#
这个理论打法是需要向这个朋友多学习一下的。
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11#
基本理论也是要学学的了哦
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我的成果的功劳都是楼主像这样的理论分享。
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理论也许看着很牛逼,但是我还是不行
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这样的理论还是需要多多看看一下。
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这样的理论打法是十分科学好有道理哦,我也来收藏
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16#
老哥的理论打法还是很好啊,来收藏下了解下了
avatar
理论看看一下,肯定都是不会吃亏的。
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