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比如Holdem里面 牌面 Q9742四张黑桃。你有黑桃A那肯定是坚果,有黑桃K是传统意义上的“第二坚果”,但是这个“第二坚果”意义不大。因为“第二坚果”之间是不一样的。# X9 T) Y" I5 I
) U$ m S# e# z3 o2 {在比如AT755牌面你有AA,也是“第二坚果”。但这个“第二坚果”却比上面的K-hi同花要厉害得多。事实上,就算第四坚果77,第五坚果A5,第六坚果T5,都比上例中的第二坚果要结实。
7 [9 N& J2 l C% p# o
: y' X7 [$ O, f8 J6 y0 U d4 U: }你看到了自己2张牌面5张一共7张牌,还有45张没看到。对手总共的可能就是(45 choose 2)=990种。你的牌在这990种之中排名即可。3 N; m, `/ ]' g5 b
' j; ~+ p2 S* { F( Z/ V2 X1 ?" b
第一例中K-hi花,因为对手AsX有44种,我们的牌最高只能排到敌人的第45名之后,第46名之前。
0 H2 y# N) y% S/ v$ z( C/ Y0 _
. h; t0 t" M# i' N( x$ j第二例中第二坚果AA,却能排到对手第一名和第二名之间,因为他只有一种55。第三坚果TT排名第4之后,第四坚果77排名第7之后。
8 j V2 w6 j; U# E! F: ~ x第五坚果A5排名也是第7之后,而不是第10,因为对手已经没有55,只剩下一种AA。所以我们A5和77的厉害度是完全一样的。4 |) W4 I4 t3 L. Y
第六坚果T5排名第10之后,对手0种55、 3种AA、 1种TT、 3种77、 3种A5。; M$ t8 |4 u6 k3 `5 N/ O
所以即使是第六坚果,其结实度也会比某些情况下的第二坚果结实的多。% R5 i. ]7 U) T
! S. z- k$ j; p& _
) [0 A$ A2 ]: B1 W: [3 gPLO如果用类似的排名,做法一样,只不过计算起来稍微麻烦点。
$ B1 @6 n/ S8 \* v% R你自己4张牌,牌面5张,还有43张未见。对手总共有(43 choose 4)= 123410种。- p6 J( ?: o: j- ]; j/ z7 Y
比如我们是9622,牌面是T8743无花。我们是所谓“第二坚果”。坚果是J9XY。5 C0 J0 P+ y3 f2 g
2 _& @, E% \: t9 y我们的第二坚果,其结实度相当于Holdem中的什么情况?一共有4个J,3个9,41个X和40个Y,所以对手的第一坚果有4×3×41×40=19680种。
( Z3 Q& b6 J$ l* Z0 \对手的第二坚果是96XY 其中X和Y都不是J。有3×3×37×36=11988种。! \3 N8 y5 G9 e
所以我们的牌在123410副牌的大排名中,位列19680名之后,跟另外的11988种并列。2 R5 b& G/ f, c9 R' T
若以百分数表示,就是排在15.95%之后,25.66%之前。
g% k5 }0 c, t( s8 b5 x# o
* ^+ T( e' S9 {2 H如果硬要在Holdem里面找个对应,那么在:
5 L' g; i# [' a* w" w$ V& M1) Q9742四张黑桃 牌面,相当于8sX或者6sX,其中X不是黑桃。(在990名中排名170之后,或者17.17%之后)
}2 x8 u; ]4 V; y" w2) T8743无花 牌面,只能相当于上至AA,下至T2左右的牌力。(所有顺子J9、96、65共16×3=48种,所有暗三共3×5=15种,所有两对共10×9=90种,AA,KK,QQ,JJ共6×4=24种,AT~T2不含两对各6种。其中略有些deadcard removal效应未考虑,如果考虑可能要下至99或者A8)
6 c6 o3 \1 k- I- _) P/ T, i* l3 f* O9 h: a1 d
当然本排名只考虑河牌,没有计入action,形象,deadcards,发展过程等因素。* f4 Z2 @4 H% [" ~) T0 a
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