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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。* y8 }: W- w9 e" [3 W
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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。) F' [% @" b" U8 _" P5 Q
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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。: [/ h, H( C, y5 P7 e5 w3 d- [2 `4 R' G
4 A" M% R6 Z. D! m8 F
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。- Y0 k- ]) N: P- B
4 ]# O; G1 p( j9 s4 T, C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。7 ]4 N5 f- j" Q3 y) K: C
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。* b. V- U/ Q" L* v8 `
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。
$ h" K8 ]3 [; o# b; c$ {( S9 A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。
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PLO如果用类似的排名,做法一样,只不过计算起来稍微麻烦点。
* g2 l6 |3 k% @2 c F, q! c: R3 C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。
8 t) D3 m2 }6 P3 m6 n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。
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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。4 D8 D. W: Q5 v$ \8 y
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。$ o; h2 G- h8 A- ?! l. |" z
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。
7 g c) I$ V, ?& y a若以百分数表示,就是排在15.95%之后,25.66%之前。
' ^1 H9 t& V7 r& d
( |7 f9 I ^; c" B7 u; m如果硬要在Holdem里面找个对应,那么在:/ t0 F* a8 Q5 e: k' y; }
1) 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)
4 N, C$ q4 {5 R7 F, f5 L4 K2) 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)
8 e M$ Q9 P( D4 u5 G$ S2 E) f+ ^# r6 ^" R$ T1 L8 b1 z/ e5 N1 P
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。9 |7 K% \3 X5 ~) V8 I
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