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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
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' m9 X9 ~; c: n. P4 A% i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
- l7 f; P7 P6 K0 a& U. A特定的非对子牌有4*4=16种组合6 a, R- r& S0 @, J
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特定的同花非对子牌有4*1=4种组合
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AXs有13-1=12种组合! M, S9 B) _: [5 e* k- g6 q  ~% Y

' |" j0 z6 [) W) @9 [" LKXs有13-2=11种组合4 t5 v4 e/ G& F3 n% }- |* m2 Y
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思考:去牌效应的应用。& j  D- v6 N8 c8 v1 X& g: U

9 W8 T! c4 r8 c# QHERO手上有一张A,那么对手AA的组合还有多少呢?
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. V& l0 D* i4 i# |! J7 J答:有C23=(3*2)/(2*1)=3种,与你没有A时的情况相比少了一半。+ n. t9 d7 O/ }
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HERO手上有一张A,那么对手AK的组合还有多少呢?2 q+ Z9 j; }# f. F+ C
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答:有3*4=12种,与你没有A时的情况相比少了1/4。
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手牌组合0 q2 y! V3 W  Q6 I4 s2 l/ d# A+ h

" \* M1 l7 q* @* M( z# v8 e成牌组合
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; L! z: R- X5 ]! ?FLOP没有公对时:% |+ L2 R5 T* j; e+ s

5 w$ f5 ~" ~6 S& L; \三条的组合有C23*C23=9种
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. K3 j, Z8 |* N7 G指定三条的组合有C23=3种
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两对的组合有C23*C23*C23=27种! o. b  B$ U( V1 \, u8 y
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指定两对组合有C23*C23=9种
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1 V9 q$ D, I& `, w8 bTPTK的组合有3*4=12种
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Flop有公对时:
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明三条顶踢脚的组合有2*4=8种
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葫芦的组合有C23+C13*C12=9种1 A3 @" y2 A: M* o/ A& M+ S/ M8 L: @
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听牌组合5 ]& ^) K: T/ J3 r" t+ h* g0 Y, P# ^

& g* U! N9 ^! c) H/ I9 u8 KFLOP有两张同花时:
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# Z/ [4 {" z1 F: C0 i- o! a对手理论上的同花组合有C211=55种7 I0 o- P3 a1 s& o  H) E

, ~' v8 t( `* H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2 g- [$ a) L+ G, [, e- |+ a. E
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在类似K34这样的FLOP时:
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) P+ O, M% t, k& p" b0 c+ x0 t4 }两头顺组合(25、56)有2*4*4=32种, E8 _- r9 i; ~% a
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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:: C9 C4 g  }  d; g# i2 G

/ |  m7 B1 y8 Y! D5 h两头顺组合有(34、47、78)3*4*4=48种% L3 W+ h5 L7 l

+ z9 \3 p  O- v* C0 J卡顺组合有(24、37、48、79)4*4*4=64种
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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~)
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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。
avatar
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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