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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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& q, L* ?1 {3 L" \& H* `4 x# L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
* [/ J) ]+ \+ G- D特定的非对子牌有4*4=16种组合
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: i6 r$ J. w6 o. X+ e特定的同花非对子牌有4*1=4种组合
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% a+ {4 P+ `2 e$ f) `# SAXs有13-1=12种组合
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9 {* \' E# w, rKXs有13-2=11种组合
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思考:去牌效应的应用。
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* D) ?( R( D; s4 @5 j1 ]4 cHERO手上有一张A,那么对手AA的组合还有多少呢?1 ]2 ~" }# B! C" J, B7 K

+ x7 \0 ?$ S$ _: ]( k) b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 D, [1 S# U) E9 ?4 l1 F
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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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! S+ d  L- h/ W1 f5 i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! H9 }0 w! Z0 U

3 u4 ~6 l+ F# Y2 q成牌组合, e% J( n0 j, ?3 g; O: b( `
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FLOP没有公对时:
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! K6 e3 z6 ~6 d三条的组合有C23*C23=9种
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  ~7 F  s# t8 S6 c) R9 n% m指定三条的组合有C23=3种6 j  J, F, W1 F8 E) g

& R8 [  a8 y* O两对的组合有C23*C23*C23=27种  G" H. L" M; L9 U1 a9 F  y

/ m% |! c" y0 V, q指定两对组合有C23*C23=9种& b7 Z$ j' j4 |' ]+ u- j+ h# @! _
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TPTK的组合有3*4=12种
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, u4 D, X+ S( j! J" m/ G! M  W, sFlop有公对时:
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4 Q8 H- b1 J! \. T# a% B明三条顶踢脚的组合有2*4=8种1 ~, }( L: m5 c7 Y6 ~0 P: Q+ h9 W

! L8 s. _$ l3 M2 z* T葫芦的组合有C23+C13*C12=9种# B2 Q# g5 a/ V& o( Y
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听牌组合
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FLOP有两张同花时:
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对手理论上的同花组合有C211=55种7 C7 ~' _: F" W4 M) Z% T$ b. F
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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
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/ p4 z( f$ w) ?! F, o/ G在类似K34这样的FLOP时:) B! k( @3 a/ M3 i, I, [$ q
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两头顺组合(25、56)有2*4*4=32种
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' x+ |7 i6 m0 r7 |) w) I卡顺组合有(A2、26、57)3*4*4=48种5 i( m, J; ?4 m( l6 R3 @( j

9 j6 |1 A1 T" q# `在类似K56这样的FLOP时:
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两头顺组合有(34、47、78)3*4*4=48种5 p+ x+ j4 K+ U* _$ X: t  q  T+ u

: l+ t& K- J, f! C1 X卡顺组合有(24、37、48、79)4*4*4=64种3 G& l3 `& q7 b1 v$ O  p
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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
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4#
知道有多少种,肯定是有好处的
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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?
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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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