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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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( O; Q7 r$ d8 |3 D' l6 {+ O: F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, r( A; M2 l0 i. F/ D
特定的非对子牌有4*4=16种组合
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/ h! F' S8 F+ x特定的同花非对子牌有4*1=4种组合# z' a0 U# s7 ~0 \! Q
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AXs有13-1=12种组合: x9 q" A* u. L2 j  z: m8 g
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KXs有13-2=11种组合  c" V) b! L0 m, M5 |
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思考:去牌效应的应用。! @+ b$ m2 y/ ?
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HERO手上有一张A,那么对手AA的组合还有多少呢?; ?/ d9 L4 I* J* ^& ~; `2 [* L! j

% n  K, D  S* ~  b% p  f  s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 u- s- ~/ v( m  `# J1 H
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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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9 `5 v: W  X8 a0 `& r9 c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+ m% Z; N3 W/ n: G. D7 A, z
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成牌组合& m0 B! y4 o6 n7 ^! i6 Z
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FLOP没有公对时:
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) V& X/ o6 A# x$ Q2 t三条的组合有C23*C23=9种
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指定三条的组合有C23=3种" P" n" j/ ^1 W4 Z. E

+ r( a) O) h( R; n两对的组合有C23*C23*C23=27种
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: \: o+ _2 {! R7 ]  [* f指定两对组合有C23*C23=9种; d( O# `4 H1 w0 h% p

8 _$ \* B% {/ S( e8 m3 |TPTK的组合有3*4=12种
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Flop有公对时:
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/ \- X. v/ ]3 }* x7 j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4 w' Z/ H/ `, T# i, r( x

/ w# q: q# ]2 n  t听牌组合; a4 O& g3 [7 f2 V

! k! `$ @0 h, N3 S) m9 H, WFLOP有两张同花时:+ Y" L+ \/ H' K+ V( V* n( G

2 y: ~9 @* G3 Z9 W. c对手理论上的同花组合有C211=55种% p- [' H) C' @

. g4 Q5 F9 e  v, |+ y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 H# M$ B$ M8 H$ Z; m4 }

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) m6 B+ @8 e% u( p在类似K34这样的FLOP时:
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两头顺组合(25、56)有2*4*4=32种
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7 S3 w$ k9 v7 E0 F卡顺组合有(A2、26、57)3*4*4=48种; G  P" T- X- Y+ _5 X9 z. b* K' r
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在类似K56这样的FLOP时:
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1 `; q/ Y6 O; E; s两头顺组合有(34、47、78)3*4*4=48种
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0 J4 X6 ], W+ p: i9 C1 m& H" D! e卡顺组合有(24、37、48、79)4*4*4=64种
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  @2 P$ W5 [; A1 G1 |+ P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#
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。
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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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