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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2 `! s1 d4 L8 H4 t
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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
" ~7 y2 I! ]  ~' O* ?$ Y特定的非对子牌有4*4=16种组合
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特定的同花非对子牌有4*1=4种组合
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2 X3 [7 `9 G' K6 J# B; GAXs有13-1=12种组合& v% R( ?0 E' Q5 p4 A1 J' X
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KXs有13-2=11种组合4 U  O/ G* l0 X9 j2 ~) C7 F
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思考:去牌效应的应用。
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. f# d. `# h+ e8 s2 O: F" v1 ]HERO手上有一张A,那么对手AA的组合还有多少呢?
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答:有C23=(3*2)/(2*1)=3种,与你没有A时的情况相比少了一半。
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" h- I2 a/ q3 K1 l5 I! f/ o3 F' `HERO手上有一张A,那么对手AK的组合还有多少呢?3 n2 A" d/ I  u' k2 [: A
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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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& l$ }* G8 u, q& o0 N手牌组合4 U, i7 Z4 x( Z0 o) V6 f) h
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成牌组合4 Y  s, |4 u- V9 h3 J3 q! N- p

2 u$ D- p- y$ ]% k$ ?" tFLOP没有公对时:! k0 ]  A5 L7 t- \+ S2 J
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三条的组合有C23*C23=9种
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5 U9 `0 b6 _( x7 p指定三条的组合有C23=3种
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( J5 D- o: x7 ~) M+ L, I( _; @% ]* l两对的组合有C23*C23*C23=27种+ m! O5 a3 x# k! @1 E1 w
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指定两对组合有C23*C23=9种/ M: N% c$ [5 I3 U" V0 l. q
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TPTK的组合有3*4=12种$ j! B4 H9 q/ t' Y1 a6 O( w

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1 l& K# i5 k& ^1 H! zFlop有公对时:
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0 }! H+ A( |% A$ F! T- ]明三条顶踢脚的组合有2*4=8种
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8 v8 j; [0 \* J2 x# j2 B葫芦的组合有C23+C13*C12=9种
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听牌组合
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9 D, l6 |4 C: u( d3 p8 hFLOP有两张同花时:
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对手理论上的同花组合有C211=55种
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+ Q6 t7 v4 ]# k  F  N. Q8 E9 u4 Y+ t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7 C( g- f( m; l2 M. k
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, ]# [' D& K; F在类似K34这样的FLOP时:
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! w2 |. {7 G7 t/ e7 n5 `6 W. t& u两头顺组合(25、56)有2*4*4=32种
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卡顺组合有(A2、26、57)3*4*4=48种
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9 d: a* d4 @7 T& f5 L9 ~/ B7 o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 m0 x- O6 I+ N3 Z; q& K
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卡顺组合有(24、37、48、79)4*4*4=64种1 l, x: w, {' E+ s0 F6 J# K# ?

' Y: _8 F) c9 x; A1 d& N% r% l0 A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 M5 E4 x6 U2 z# I; s
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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
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5#
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
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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。
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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。
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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