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起手牌组合
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( `* P' `8 K! e: j7 x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
4 A1 E% `2 v- t' ?1 a, T) t- e特定的非对子牌有4*4=16种组合6 Y* J1 P; M2 W% Z6 n. t- x! |0 L
3 ^- |- R: i4 `2 n3 }9 J$ c
特定的同花非对子牌有4*1=4种组合
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AXs有13-1=12种组合' b& v! `/ C' p" y, O. U
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KXs有13-2=11种组合# W( c, G& C) ^8 h8 e
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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?
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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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HERO手上有一张A,那么对手AK的组合还有多少呢?
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, r$ k0 P9 o( I答:有3*4=12种,与你没有A时的情况相比少了1/4。$ ~3 H: s5 T+ J& c- | N( M; U
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手牌组合+ R/ S( w3 Z! H# E/ U% ^# h# C( J( ^/ Q
D) N! l% G n6 R成牌组合6 I; ~3 A- h. B# t: q9 q7 R/ `: S
; b% B. U. p& |7 sFLOP没有公对时:
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& o. W, ?/ H# P三条的组合有C23*C23=9种1 D2 \6 `+ u6 T: T3 s' Y% c
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指定三条的组合有C23=3种* ^3 f8 [- W! y8 H, m) v. o; i
4 i' b9 T2 d/ V- S7 Q两对的组合有C23*C23*C23=27种
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+ N; k3 a& e) h2 b指定两对组合有C23*C23=9种- s8 g7 h* b3 X& @8 a* ?
' Q" y; X3 D% Y" s8 n$ fTPTK的组合有3*4=12种9 K+ a0 I# ~+ u) ^- u: o
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+ K; f1 v2 m% sFlop有公对时:
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: N+ X9 x- f3 x/ }5 I$ _$ u4 g/ }. G明三条顶踢脚的组合有2*4=8种
7 z" ^. G4 P( C4 q7 N: ^( c' u: _- s1 {
葫芦的组合有C23+C13*C12=9种5 H. X# a5 Z4 K4 r! c2 }
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听牌组合
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& ?: p& s, e9 H. D% X- lFLOP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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( _6 h8 D" E" j# s3 _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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在类似K34这样的FLOP时:
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( n B2 p" {7 G8 X8 ~) V b) d两头顺组合(25、56)有2*4*4=32种1 l( E p6 Z3 w
/ [/ C. }# S5 m: ?- J卡顺组合有(A2、26、57)3*4*4=48种
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9 O- M: R6 ~' Y/ Y: ]4 \ l在类似K56这样的FLOP时:7 S! j! r/ O; H. W
+ n, [; K) z5 C a7 M两头顺组合有(34、47、78)3*4*4=48种! @+ _& }6 G4 @7 p
U8 ]0 [0 r' L4 J2 @* t) Q卡顺组合有(24、37、48、79)4*4*4=64种
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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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我现在还是研究一下理论打法,谢谢你的分享,我也来收藏 |
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