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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0 }5 N# a% b( `8 d, }4 R! E3 \

. d: c' Y/ y, k0 u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
! a  a" F0 |, U" V) T2 F; S特定的非对子牌有4*4=16种组合/ T2 m3 Q# b* T! w

4 w5 x3 @0 c, T. n& U' {7 Q特定的同花非对子牌有4*1=4种组合% K8 w$ t$ e' Y1 @

1 Q) m$ p% ?' v% S+ ^3 R. @* d4 sAXs有13-1=12种组合
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6 P  d6 F/ M) uKXs有13-2=11种组合
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思考:去牌效应的应用。
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HERO手上有一张A,那么对手AA的组合还有多少呢?
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8 w4 _: R( P0 t% y" G" X答:有C23=(3*2)/(2*1)=3种,与你没有A时的情况相比少了一半。  O( a' ^; I" y' |+ F* T
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HERO手上有一张A,那么对手AK的组合还有多少呢?& N: t- D& |! M) k0 Y4 \4 d. e
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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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( N# Q  F, T6 P) L0 ?手牌组合  z4 N9 ^+ i$ ^/ U6 Y+ D7 E
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成牌组合
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9 |0 n" X  T/ g3 Z5 @5 ?/ J, I" mFLOP没有公对时:
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三条的组合有C23*C23=9种3 H% {+ j: r1 `
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指定三条的组合有C23=3种
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/ N0 }% s6 W! S% }# F两对的组合有C23*C23*C23=27种
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指定两对组合有C23*C23=9种
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$ Z' b8 N/ {6 ZTPTK的组合有3*4=12种0 i1 x. m$ }5 r4 p
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* _6 h; T2 t: @# I0 RFlop有公对时:
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明三条顶踢脚的组合有2*4=8种+ O  L, ]* D( Y
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葫芦的组合有C23+C13*C12=9种
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听牌组合% ^0 l+ _2 c/ ]# {$ A

0 Y7 i- z0 k8 v+ ]& J+ g0 gFLOP有两张同花时:
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对手理论上的同花组合有C211=55种- {0 R4 }# ?, o+ b

7 u5 K& f) _9 Z  g" o5 O. q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
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) H% x5 j6 m9 U. L在类似K34这样的FLOP时:# X* b5 E  }/ H6 t4 @# u$ ~
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两头顺组合(25、56)有2*4*4=32种
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- u' m4 B7 s7 s) X卡顺组合有(A2、26、57)3*4*4=48种- H; ?, K) R4 D0 M6 y( S, T
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在类似K56这样的FLOP时:
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. w3 _; j; @8 i$ ~7 r4 C0 c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9 m' T4 H0 _/ M/ q
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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~)( ^# I" X+ E5 [% Z, d. B
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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。
avatar
理论实践得好,收米的时机就来了。
avatar
你估计也是一个理论很内行的玩家,感激楼主的分享。
avatar
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
avatar
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
avatar
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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