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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4 H* s0 C: @7 a8 L6 G0 n
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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
/ H$ v, a$ m1 h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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# @, Z! W, C/ C. p  x4 F. @/ V; [AXs有13-1=12种组合
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KXs有13-2=11种组合( H! T( i" ~! T+ \
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- k8 Y% u* ^! f: {. @思考:去牌效应的应用。, l! `2 l" |& n& E* Q% c/ B* d3 C. m

1 J. f4 ^. R# |HERO手上有一张A,那么对手AA的组合还有多少呢?! y" a3 u6 W& |. l4 C% Q5 K# E6 s

0 ^. N7 p- b) t5 x/ j2 ?4 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。$ w/ |" e1 h0 [

" M, u! W, L' e  o; l% ^$ q! u2 {HERO手上有一张A,那么对手AK的组合还有多少呢?/ T$ M! g; _+ s6 H' X  h

$ Z4 L1 E7 N9 X  g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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' a1 e( L0 d9 Q/ o7 ~手牌组合: O& ~) j( _' c
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成牌组合
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0 G& _& U7 F: P4 q/ m" x& `FLOP没有公对时:
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三条的组合有C23*C23=9种$ t/ W  o; _) `5 w( _* D9 w

0 C6 A6 O* x( G+ k6 ^; d指定三条的组合有C23=3种5 C( Q) L+ v" V1 S, S0 D
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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
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4 m: E4 g: }! F" i- C3 Q2 D1 xTPTK的组合有3*4=12种
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Flop有公对时:6 V7 F% c3 t; o

9 N" s) Q( w4 \1 n# C1 m; X* t) T! k明三条顶踢脚的组合有2*4=8种4 X  @$ k( Z6 t, A9 ?+ R
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葫芦的组合有C23+C13*C12=9种
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听牌组合  N" T) Y3 C  }
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FLOP有两张同花时:
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9 X$ ]0 {& F; h3 w% a; B对手理论上的同花组合有C211=55种
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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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7 G! t) {1 ]- W4 L* w& K) [在类似K34这样的FLOP时:5 L& L! g& C8 e) T" W; M  }

$ ]% v5 M3 g8 J0 b两头顺组合(25、56)有2*4*4=32种# v" i% F' |8 ^* r9 {, k

! E8 U+ W( i* J! z# C卡顺组合有(A2、26、57)3*4*4=48种4 p  s( n3 A0 i* _3 A
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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" v& \$ |( {- d; n9 [
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卡顺组合有(24、37、48、79)4*4*4=64种, X3 M8 ~( G+ A+ x

. ?; t. p8 r& ~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~)# X; w8 z: y) |( h
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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
avatar
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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