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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。
% s& H S) L; L6 ]! @, S, b( W1 ]
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。
. {& O6 i& d( H9 `0 x
! V% u' M8 F; x/ m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。
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1 X5 t0 s0 e) H9 m7 O- E$ z3 B, R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。/ F0 a) N; ~' d; \' A
4 T ~; i: T) [5 s! W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。% J' h/ O& y" U/ O+ p7 }1 l/ B" @
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。
3 x) n3 Y% Q& `* n$ l' t% {2 @% 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。. A" E+ F& y% J6 h) N" @( l' t
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。! D: O7 M- b; _4 n' R3 f a
# f9 @* N4 T# j. o) y+ o- W
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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。( a* x Z9 Y y
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。0 @+ t# ?% F- _5 ~
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。$ N0 t: D) l2 V9 r
4 T5 \5 K' F$ B8 c. `; _3 `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。
! o! n t" `' A: B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。
6 }4 G( Q$ B: d% y) ?5 s* s8 Y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。
, }3 s" d* A1 q; e若以百分数表示,就是排在15.95%之后,25.66%之前。
# Y/ B/ w) W8 r9 C1 o
4 p$ }3 Y N3 [8 }/ A$ W如果硬要在Holdem里面找个对应,那么在:6 Z5 ^8 R. r; m0 e8 z4 b0 P8 p; U
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)
" |* G8 V" }2 c% p4 s3 M$ s2) 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)$ h3 e9 w$ j1 ~5 E9 {
9 u% t/ G7 ~1 g& Q1 B$ D: [2 R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。! k9 k! X6 C: s: v+ W
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