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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- y9 [: S  x% f' x+ `! }/ s, I

$ C4 P! R8 |, z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
& Q' H8 n; G5 D6 {7 z/ [* h特定的非对子牌有4*4=16种组合0 z/ G# k5 `5 S  k
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特定的同花非对子牌有4*1=4种组合- M6 k# I4 J$ ^# n  j

' y7 i9 C3 h4 {4 k6 NAXs有13-1=12种组合
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7 f- W/ i8 D4 S$ _KXs有13-2=11种组合' h( F1 L% W4 M( k

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  l0 w2 C" h" F0 F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?  ~3 w( m: S( V  V$ b* z

2 T* V( b+ ^% V  K; F/ Y5 R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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HERO手上有一张A,那么对手AK的组合还有多少呢?- C; u% ~0 D. B- |; L
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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: N# I. n7 j3 J& |4 Q+ W+ B. h
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成牌组合1 u: p3 c- K5 r  u6 o; ~7 X% K, [) p
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FLOP没有公对时:" a  K( i$ \$ s# _( `
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三条的组合有C23*C23=9种7 P9 |7 x6 ]( I) m; W3 j6 B
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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
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指定两对组合有C23*C23=9种* U  ?) p% X, Y# d: l
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TPTK的组合有3*4=12种
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( c7 K. K2 ?: i$ s: m- LFlop有公对时:
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" g! J1 T! G$ C+ O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 P! y: D+ f9 P  Y3 n

7 \; C: T4 v0 G8 N- ]$ H+ R* L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
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% ^5 l- l: T, ^8 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
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0 b$ j3 h1 Q5 J$ L" f4 w' C在类似K34这样的FLOP时:) q4 L6 S: I0 u1 ^/ ?# g

2 {! \& @3 ]* @! o+ D) G$ l, I两头顺组合(25、56)有2*4*4=32种
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% p* w# ]- ?# ?& g# K卡顺组合有(A2、26、57)3*4*4=48种+ N" |7 i; T' Z2 _# s" W1 k: {( H

, Y) m7 t, @/ O8 X8 u+ t在类似K56这样的FLOP时:; E) R2 u* a  ^* A6 \' n( A+ E

7 N7 _1 D3 Z, W( k两头顺组合有(34、47、78)3*4*4=48种# k* [, A( [- ?

2 {7 j5 `, N, u8 ?8 _- T2 K4 T卡顺组合有(24、37、48、79)4*4*4=64种
/ C: m; E. \2 ]! e9 ^  X! ]
; T7 n( m: W- i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
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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
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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。
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。
avatar
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。
avatar
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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